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06Capacitance, dielectrics, and the energy stored in an electric field
A camera flash puts out about three joules in two milliseconds. That is one and a half kilowatts, from a battery that can barely manage two watts on a good day. The trick is to fill something slowly and empty it fast, and the something is a pair of metal foils held a hundredth of a millimetre apart by a film of plastic. Nothing in the first five sections tells you how much charge that pair of foils will accept, or how much energy is sitting in the gap when it is full.
By the end of this section you can take any pair of conductors, say how much charge they hold per volt from their geometry alone, combine several of them, slide an insulating slab into the gap, and state exactly what happens to the charge, the field, the voltage and the stored energy in each case.
In 60 seconds
Two conductors carrying equal and opposite charge take a charge proportional to the voltage between them; the constant of proportionality is fixed by the shape of the gap and by what is in it, and half the product of the two is the energy stored.
pick the form built from whatever is being held fixed while something else changes
of an electric field
$$u = \tfrac12\varepsilon_0 E^{2}\quad(\text{vacuum}),\qquad u = \tfrac12 K\varepsilon_0 E^{2}\quad(\text{in a dielectric})$$
you want the energy per cubic metre wherever the field is, not just inside a capacitor
in the gap
$$C = K C_0,\qquad E = \frac{E_0}{K},\qquad Q_{\rm ind} = Q\left(1-\frac{1}{K}\right)$$
an insulating slab fills the gap; the first relation is geometry, the second and third need the charge to be held fixed
Three most common mistakes
Adding capacitors in series the way resistances add. End to end the reciprocals add, and the answer is always smaller than the smallest one in the chain.
Forgetting to ask whether the charge or the voltage is being held fixed before changing a capacitor. Every quantity except the capacitance itself depends on that one decision.
Dropping the factor of one half in the stored energy. The charge does not all cross the full voltage: the first drop crosses nothing at all.
The published weights for this course are Midterm 1 twenty per cent, Midterm 2 twenty per cent, quizzes ten per cent in total, homework five per cent, the final twenty five per cent and the laboratory twenty per cent. Nothing finer than that is published, so no claim is made here about how many marks this particular material carries.
How much time do you have?
10 minutes
You leave able to do the two things every question on this material starts with: turn a geometry into a capacitance, and turn a capacitance and a voltage into a charge and an energy.
The 60 second card · Formula card · Capacitance: the charge a pair of conductors takes per volt · The parallel plate capacitor, from Gauss's law in two lines · The energy stored, and the factor of one half · Mistake ledger
45 minutes
You add the parts that separate a set-up mark from a full mark: reducing a network, and deciding what is held fixed when a slab goes into the gap.
The 60 second card · Capacitance: the charge a pair of conductors takes per volt · The parallel plate capacitor, from Gauss's law in two lines · Capacitors side by side and end to end · The energy stored, and the factor of one half · A slab in the gap: what the dielectric constant multiplies · Method box: reducing a network of capacitors · Method box: something changed, what was held fixed · Exam level example · Practice set C · Mistake ledger
full read
You can build the capacitance of any geometry you can apply Gauss's law to, account for every joule when a slab is inserted either way round, and say where the induced charge on the slab came from and how big it is.
The opening pages · Prerequisites and pretest · Notation · Conventions · All seven concept blocks · Method boxes · Contrast pair · Scaffolding comes off · Exam level example · Practice sets A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Compute the capacitance of a pair of conductors from its definition, and use it in both directions to go between the charge on the conductors and the potential difference across them.
Derive the capacitance of a parallel plate arrangement from Gauss's law and the potential difference, and say when the small gap condition that the derivation rests on has been broken.
Apply the same three step recipe to a coaxial cable and to a pair of concentric spheres, and recover the capacitance of a single isolated sphere as a limiting case.
Reduce a network of capacitors to one and then work back through it to the charge and the voltage on every individual capacitor.
Calculate the energy stored in a charged capacitor by all three equivalent routes, and convert it into an energy per cubic metre of the field that holds it.
Predict what happens to the charge, the potential difference, the field and the stored energy when a dielectric slab is inserted, in both the case where the charge is fixed and the case where the voltage is fixed.
Explain the dielectric constant in terms of the charge induced on the faces of the slab, and compute that induced charge and the energy density inside the material.
Syllabus coverage
Capacitance
What capacitance means and what it does not depend on; the capacitance of a parallel plate arrangement from Gauss's law; coaxial cylinders, concentric spheres and the isolated sphere; and networks of capacitors in series and in parallel.
The week line names three topics and carries no chapter numbers, so no chapter number is quoted anywhere in this section.
covered
Dielectrics
The dielectric constant and what it multiplies; the two standard insertion problems, one at fixed charge and one at fixed voltage; and the maximum ; the molecular picture, the induced surface charge, and the form Gauss's law takes inside the material.
covered
Electric Energy Storage
The work needed to charge a capacitor and the three equivalent expressions for it; the energy per unit volume of an electric field, in vacuum and inside a dielectric; and where the energy goes when a slab is inserted or the plates are pulled apart.
covered
The attractive force between the plates of a charged capacitor
The force one plate feels in the field of the other, equal to one half the charge times the field.
Not named on the week line. It is kept because it gives a genuinely independent check on the energy calculations, and it is flagged here so that nobody revises it as examinable material on the strength of this page alone.
off_syllabus
How a capacitor fills and empties over time
The time dependence of the charge on a capacitor when it is connected through a resistance.
Deferred. The whole question is about current through a resistance, and both belong to later weeks. Everything on this page is a statement about a final resting state, never about how long it took to get there.
deferred
Recall first
The field just outside a charged conductor
At the surface of a conductor in equilibrium the field is perpendicular to the surface and has magnitude $E = \sigma/\varepsilon_0$, where $\sigma$ is the local surface charge density. This came from a pillbox argument with Gauss's law.
It is the whole first half of the parallel plate derivation. The field in the gap is read straight off the charge on the plate.
Gauss's law
The flux of the electric field out of any closed surface equals the charge enclosed divided by $\varepsilon_0$: $\oint\vec E\cdot d\vec A = Q_{\rm enc}/\varepsilon_0$. With enough symmetry it gives the field in one line.
Every capacitance on this page starts by putting charge on the conductors and asking Gauss's law for the field between them.
Potential difference from the field
$V_b - V_a = -\int_a^b \vec E\cdot d\vec l$, and for a uniform field along a straight path of length $d$ this is just $Ed$ in magnitude.
It is the second half of every derivation here: having got the field, you integrate it across the gap to get the voltage the definition of capacitance needs.
The field and potential of a point charge
$E = kQ/r^{2}$ pointing away from a positive charge, and $V = kQ/r$ with the zero taken at infinity.
The and the isolated sphere are built from these two, and the isolated sphere result is nothing more than the second one rearranged.
The field of a long line of charge
A long straight line carrying charge $\lambda$ per unit length makes a radial field $E = \lambda/(2\pi\varepsilon_0 r)$ at distance $r$ from it.
It is the field inside a coaxial cable, and the logarithm in the cable's capacitance comes from integrating it.
A conductor in equilibrium is all at one potential
The field inside the metal is zero, so no work is done moving a charge about inside it, so every point of one connected piece of conductor is at the same potential.
It is why it makes sense to speak of the potential difference between two plates at all, and it is the reason the middle of a series chain carries equal and opposite charges.
Try it yourself first (3 questions)
1§06.1 — what a charged capacitor actually carries●○○○○
Before anything new: one sentence from the previous sections, and it catches most people. A pair of metal plates faces each other across a gap. One carries $+8.0\ \mathrm{nC}$ and the other carries $-8.0\ \mathrm{nC}$.
Given
one plate carries $+8.0\ \mathrm{nC}$
the other plate carries $-8.0\ \mathrm{nC}$
nothing else is nearby
Find
(a) True or false: the pair as a whole carries a charge of $16\ \mathrm{nC}$. Say in one sentence what it does carry.
Hint 1/4
The question is about a sum, not about a magnitude. Add the two numbers as signed quantities and see what you get.
Hint 2/4
Charge is conserved and it adds algebraically. Two equal and opposite charges on two pieces of metal add to zero.
Hint 3/4
Here the two numbers are $+8.0\ \mathrm{nC}$ and $-8.0\ \mathrm{nC}$, so the sum is $+8.0 - 8.0$.
Hint 4/4
False. The total is zero; what the arrangement carries is separated charge, not net charge.
charge is a signed scalar, so the two contributions cancel exactly rather than reinforcing
Answer $$\boxed{\,Q_{\rm net} = 0,\ \text{while the stored (separated) charge is } 8.0\ \mathrm{nC}\,}$$
Check
Independent test: hang the whole arrangement from a thread near a charged rod. A net charge would be pulled or pushed as a whole; a neutral pair is not, and that is exactly what happens.
2§06.2 — the field between two oppositely charged plates●●○○○
The trap in this one is a factor of two that half the class carries forward into every capacitance they ever compute. Two large parallel plates carry uniform surface charge densities $+\sigma$ and $-\sigma$.
Given
surface charge density $+\sigma$ on one plate
surface charge density $-\sigma$ on the other
the plates are large and close together
Find
(a) What is the magnitude of the field at a point in the gap between them?
Hint 1/4
Two sheets are present, and the question is what each of them contributes at a point that lies between them.
Hint 2/4
One infinite sheet of charge makes a field $\sigma/(2\varepsilon_0)$ on each side of itself, pointing away from it if it is positive.
Hint 3/4
Between these two plates the positive sheet pushes away from itself and the negative sheet pulls towards itself, and both of those point the same way.
Hint 4/4
The two contributions add rather than cancel, so the gap field is $\sigma/\varepsilon_0$.
Show solutionTake the two sheets one at a time
$$E_+ = \frac{\sigma}{2\varepsilon_0}\ \text{away from the positive sheet}$$
this is the single sheet result from the Gauss's law section, and it does not depend on distance
$$E_- = \frac{\sigma}{2\varepsilon_0}\ \text{towards the negative sheet}$$
same size, and in the gap this direction coincides with the first one
Independent check from a different law: a pillbox drawn half inside the metal of one plate encloses charge $\sigma A$ and has flux only through its outer face, giving $EA = \sigma A/\varepsilon_0$ at once. Two different arguments, one answer.
3§06.3 — volts across a uniform gap●○○○○
One line of arithmetic from the potential section, because every capacitance in this section ends with it. A uniform field of $2.40\times10^{5}\ \mathrm{V/m}$ fills the gap between two plates $0.500\ \mathrm{mm}$ apart.
Given
uniform field $E = 2.40\times10^{5}\ \mathrm{V/m}$
gap $d = 0.500\ \mathrm{mm} = 5.00\times10^{-4}\ \mathrm{m}$
the field points from one plate straight across to the other
Find
(a) What is the potential difference between the plates?
Hint 1/4
The field is the same at every point of the path, so no integral is needed and the path length is the only geometry involved.
Hint 2/4
For a uniform field along a straight path, $V = Ed$, with $d$ measured along the field.
Hint 3/4
Here $E = 2.40\times10^{5}\ \mathrm{V/m}$ and $d = 5.00\times10^{-4}\ \mathrm{m}$.
Hint 4/4
The potential difference is $120\ \mathrm{V}$.
Show solutionMultiply, and watch the units
$$V = Ed = (2.40\times10^{5}\ \mathrm{V/m})(5.00\times10^{-4}\ \mathrm{m}) = 120\ \mathrm{V}$$
volts per metre times metres is volts, which is the unit check and the reason V/m and N/C are the same unit
Answer $$\boxed{\,V = 120\ \mathrm{V}\,}$$
Check
Order of magnitude check: this is a field a sixth of the way to breaking down dry air across half a millimetre, so a value of a hundred or so volts is what one expects, and a value in the millivolts or the megavolts would signal a slipped power of ten.
Notation
symbol
reads as
means
watch out
$C$
C
capacitance, in ; the charge one conductor carries for each volt of potential difference between the pair
it is fixed by the geometry and the material in the gap, so it does not change when the charge or the voltage changes; also, it is not the coulomb, which is the unit of the quantity in the numerator
$Q$
Q
the magnitude of the charge on either conductor of the pair, in coulombs
the capacitor as a whole is neutral; the charge that appears in every formula on this page is the charge on one plate
$F$
farad
the SI unit of capacitance, one coulomb per volt
it is an enormous unit, so real values arrive as microfarads or picofarads; a bare pair of plates you could hold in your hand is tens of picofarads
$K$
K
the dielectric constant of an insulating material, a pure number at least equal to one
it multiplies the capacitance and divides the field, and getting those two the wrong way round is the standard error in this material
$\varepsilon = K\varepsilon_0$
epsilon
the , used to shorten formulas that would otherwise carry K and epsilon nought separately
in vacuum K is exactly one and this reduces to the permittivity of free space from the earlier sections
$u$
u
energy density, the energy stored per cubic metre of field, in joules per cubic metre
lower case u is a density and upper case U is a total energy; the two differ by a volume and are easy to swap in a hurry
$\sigma_{\rm ind}$
sigma induced
the surface charge density that appears on the faces of a dielectric slab when it sits in a field
it is bound charge and cannot be drained off the slab, unlike the on a metal plate
$C_{\rm eq}$
C equivalent
the single capacitance that would take the same charge at the same voltage as a whole network
it is a stand-in for the network as seen from outside; the individual charges and voltages inside still have to be recovered one by one
Conventions used here
What Q and V mean for a capacitor
$Q$ is the magnitude of the charge on either conductor, not the sum of the two. A charged capacitor carries $+Q$ on one plate and $-Q$ on the other, so its total charge is zero and it is never described as holding charge $2Q$. $V$ is the magnitude of the potential difference between the two conductors, taken as a positive number throughout, with the plate carrying $+Q$ understood to be the one at the higher potential.
Which direction d is measured in
Between parallel plates the field points from the positive plate to the negative one, and $d$ is the separation measured along that direction, so $V = Ed$ comes out positive. In a coaxial or spherical arrangement the inner conductor is the one taken to be positive unless the question says otherwise, so the field points outward and the potential falls as the radius grows.
What a source of fixed voltage is allowed to mean here
A battery appears in several problems, and it is used for one property only: it holds a fixed potential difference between the two conductors it is attached to. Nothing about current, resistance or how long anything takes is used or needed on this page, and every result quoted is a statement about the final resting state. Connected means the voltage is the fixed quantity; disconnected means the charge is.
Constants used throughout
$k = 1/(4\pi\varepsilon_0) = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$ and $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$, the same two numbers as in the previous sections, together with the elementary charge $e = 1.602\times10^{-19}\ \mathrm{C}$. The field at which dry air breaks down is taken as about $3\times10^{6}\ \mathrm{V/m}$ and is used only for order of magnitude remarks.
How many digits an answer keeps
Every answer is rounded to three significant figures at the very end and not before, matching the three figures the data carries. Intermediate lines keep four so that the rounding does not creep into the last digit.
How a capacitance is quoted
A capacitance is a positive number of farads with a prefix that keeps it readable: picofarads for anything built from bare plates, microfarads for anything with a dielectric rolled up inside it. $1\ \mathrm{\mu F} = 10^{-6}\ \mathrm{F}$ and $1\ \mathrm{pF} = 10^{-12}\ \mathrm{F}$, and the two are six orders of magnitude apart, which is the single most common place an answer loses its digits.
What the dielectric constant is called
The dielectric constant is written $K$ here, it is a pure number with no units, and it is never smaller than one. Some books write the same quantity as $\kappa$ and some absorb it into a permittivity $\varepsilon = K\varepsilon_0$; all three say the same thing, and $\varepsilon$ is used on this page only where it shortens a formula.
6.1Capacitance: the charge a pair of conductors takes per volt
A pair of conductors takes charge in proportion to the voltage across them, and that constant of proportionality is the capacitance.
Four sections told us what a given arrangement of charge does. The question now runs the other way: how much charge will a piece of hardware accept?
Solvable with what we have
Find the field between two plates, given the charge on them.
Find the potential difference across a gap, given the field.
Find the speed a proton gains crossing a stated potential difference.
Find the force between two charges you have been handed.
Not solvable yet
Say how much charge sits on a pair of plates held at $12.0\ \mathrm{V}$.
Say how much energy waits in a camera flash before it fires.
Say which of two devices takes more charge at the same voltage.
Say what happens to any of that when a plastic sheet is slid into the gap.
The obvious move is Coulomb's law. But every expression it produces begins with $Q$, and $Q$ is what nobody has given us. Writing $E = \sigma/\varepsilon_0$ and then $V = Ed$ gives $V = Qd/(\varepsilon_0 A)$: one equation, two unknowns.
Why it fails
It fails because it is one equation short, not because the physics is wrong. The missing fact comes from the hardware: for a fixed pair of conductors the ratio of charge to voltage is itself fixed. Define that ratio first, then compute it from the geometry.
DefinitionDefinition 6.1: capacitance and the farad
Conditions
Two conductors, carrying equal and opposite charges $+Q$ and $-Q$
Electrostatic equilibrium, so each conductor is at a single potential
$V$ is the magnitude of the potential difference between the two conductors
The capacitance is how many coulombs the pair takes for every volt across it, so one farad means one coulomb per volt. Because the hardware fixes the ratio, one number works both ways: charge is capacitance times voltage, voltage is charge over capacitance.
Proof
What needs proving is not the formula but the constancy: why is $Q/V$ the same at every voltage?
Suppose a charge $Q$ produces a field pattern $\vec E(\vec r)$. Now scale every charge by $\lambda$.
By superposition the field everywhere scales by $\lambda$, since each contribution is proportional to its own source charge.
The potential difference $V = \int \vec E\cdot d\vec l$ between them therefore scales by $\lambda$ too.
The new ratio is $\lambda Q/\lambda V = Q/V$: unchanged. So $C$ depends on the shape and the material and on nothing else, which is what makes it worth naming.
Measured charge against measured voltage for one fixed pair of plates. The points lie on a straight line through the origin, and $\textcolor{#1f6feb}{\text{its slope}}$ is the capacitance: $36.0\ \mathrm{nC}$ of extra charge for $18.0\ \mathrm{V}$ of extra voltage is $2.00\ \mathrm{nF}$. Raising the voltage slides the point up the line; it does not tilt the line.
Looks like this, but is not
A capacitor stores charge, so a charged one must be carrying a lot of it and a flat one none. It sounds like a bucket of water.
The bucket picture gets the arithmetic wrong. A charged capacitor holds as much charge as an uncharged one: zero. Some has merely moved from one plate to the other, so the plates read $+Q$ and $-Q$. Every $Q$ here is the amount moved, which is why a capacitor stays neutral and is still dangerous.
V (volts)
Q on one plate (nC)
Q/V (nF)
0
0
not defined
5.00
10.0
2.00
10.0
20.0
2.00
15.0
30.0
2.00
20.0
40.0
2.00
The first two columns change by a factor of four down the table and the third does not change at all. That constancy is the experimental content of the definition, and it is why one number is enough to describe the device. The first row has no ratio because zero over zero is not a number.
A pair of plates that takes 24.0 nC at 12.0 V
A pair of conductors is measured twice. At a potential difference of $12.0\ \mathrm{V}$ one of them carries $24.0\ \mathrm{nC}$. Find the capacitance, then find the charge the same pair takes at $30.0\ \mathrm{V}$, and say what the capacitance is at that second voltage.
using C as a multiplier rather than as a quotient is the whole reason for defining it; the hardware has not been touched, so the same number is still valid
raising the voltage moves the operating point along the straight line in the figure; it does not change the slope, and the slope is what the word capacitance names
Independent check by a ratio that never uses $C$: charge should scale with voltage, and $60.0/24.0 = 2.50$ while $30.0/12.0 = 2.50$, which is the statement that both points sit on one line through the origin. Order of magnitude: nanocoulombs per tens of volts is nanofarads, the scale of a bare pair of plates.
Two divisions and one multiplication, and not a single piece of information about the shape of the plates.
One measured pair of numbers fixes the device at every other voltage. That is why data sheets quote a capacitance and not a table.
Reading a component marked 470 microfarads, 25 V
A capacitor from a drawer is marked $470\ \mathrm{\mu F}$ and $25\ \mathrm{V}$. Find the charge on one plate when it is held at its marked voltage, find how many surplus electrons that is on the negative plate, and say what the second number on the label means.
charge comes in whole elementary units, so dividing by e counts them; the four figure numerator is kept until the final rounding
Say what the voltage marking is
$$25\ \mathrm{V}\ \text{is a limit, not a capacitance}$$
it is the largest potential difference the insulation inside can stand before it conducts and destroys the component, so it caps V rather than describing the ratio Q over V
Answer $$\boxed{\,Q = 1.18\times10^{-2}\ \mathrm{C},\qquad N = 7.33\times10^{16}\ \text{electrons}\,}$$
Check
Independent cross-check against the previous example, on entirely different hardware: the capacitances are in the ratio $2.35\times10^{5}$ and the voltages in the ratio $2.08$, so the charges should be in the ratio $4.9\times10^{5}$. Dividing the two charges directly gives $4.9\times10^{5}$, which agrees.
A hundredth of a coulomb is about a million times the ten or so nanocoulombs of a winter doorknob spark, which is why large capacitors are treated with respect. The two numbers on the case are the two you need: how much per volt, and how many volts you are allowed.
Checkpoint
§06.1 — one measurement fixes the device●○○○○
Thirty seconds. A capacitor is found to carry $8.00\ \mathrm{\mu C}$ on one plate when the potential difference across it is $4.00\ \mathrm{V}$.
Given
$Q = 8.00\ \mathrm{\mu C}$ at $V = 4.00\ \mathrm{V}$
the same capacitor is then taken to $V = 10.0\ \mathrm{V}$
Find
(a) Find its capacitance and the charge it then carries.
Hint 1/4
Only one relation connects the three symbols in this question, and it has already been given to you twice over in the first measurement.
Hint 2/4
$C = Q/V$, and the same $C$ multiplied by a new voltage gives the new charge.
Hint 3/4
The measured pair is $8.00\ \mathrm{\mu C}$ with $4.00\ \mathrm{V}$, and the new voltage is $10.0\ \mathrm{V}$.
Hint 4/4
It is $2.00\ \mathrm{\mu F}$, and it then carries $20.0\ \mathrm{\mu C}$.
Independent check without computing $C$ at all: the voltage was multiplied by $10.0/4.00 = 2.50$, so the charge must be too, and $8.00 \times 2.50 = 20.0\ \mathrm{\mu C}$.
⚠ Calling the total charge on a capacitor 2Q
the words store and hold suggest a container that fills up, and both plates plainly have charge on them
wrong$$Q_{\rm total} = (+Q) + (Q) = 2Q$$
right$$Q_{\rm total} = (+Q) + (-Q) = 0,\qquad \text{the } Q \text{ in } C = Q/V \text{ is one plate's magnitude}$$
⚠ Believing the capacitance grows as the capacitor is charged
$C$ sits next to $Q$ in the formula, and everything else in the formula does change when the charge changes
wrong$$C \propto Q$$
right$$C = \frac{Q}{V} = \text{constant, fixed by geometry and by the material in the gap}$$
⚠ Losing six orders of magnitude between microfarads and picofarads
both prefixes are read as small and the difference between them is never visible in the algebra, only in the final number
6.2The parallel plate capacitor, from Gauss's law in two lines
Facing area over gap, times the permittivity of free space, and the charge cancels out of the ratio as promised.
The definition names a number but does not produce one; to produce one we have to do the calculation it forbids us from skipping, and the flat plate pair is the case where that calculation takes two lines.
RuleResult 6.2: the capacitance of a parallel plate pair
Conditions
Two flat parallel conductors with facing (overlap) area $A$ and separation $d$
$d$ much smaller than the width of the plates, so that the edge fringing can be ignored
Vacuum or air in the gap, with nothing else nearby to distort the field
$$\boxed{\,C = \frac{\varepsilon_0 A}{d}\,}$$
The capacitance of a flat pair is the permittivity of free space multiplied by the facing area and divided by the gap. Wider plates take more charge at the same voltage, and a wider gap costs more volts for the same charge, so the area goes upstairs and the separation goes downstairs.
Proof
Put $+Q$ on one plate and $-Q$ on the other. In equilibrium the charge sits on the facing surfaces with density $\sigma = Q/A$.
A pillbox drawn with one face inside the metal and one in the gap encloses $\sigma A_{\rm box}$ and has flux only through its outer face, so $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$, uniform across the gap.
The field is uniform, so the potential difference is the field times the distance: $V = Ed = Qd/(\varepsilon_0 A)$.
Form the defining ratio: $C = Q/V = Q\varepsilon_0 A/(Qd) = \varepsilon_0 A/d$.
Notice what just happened to $Q$. It cancelled, exactly as the constancy argument of the previous block said it had to. If it had not cancelled, one of the two steps above would contain an error.
The two line derivation drawn once. The $\textcolor{#1f6feb}{\text{plates}}$ carry $\pm Q$, the dashed pillbox turns that into the $\textcolor{#cf222e}{\text{field}}$ $\sigma/\varepsilon_0$, and the gap $d$ turns the field into volts. The dashed curls at the rims are the fringing the formula throws away, and throwing it away is only honest when the gap is much smaller than the plates.
Looks like this, but is not
Two square plates of area $1.00\ \mathrm{m^2}$ are held $50.0\ \mathrm{cm}$ apart, so the formula gives $C = (8.85\times10^{-12})(1.00)/(0.500) = 17.7\ \mathrm{pF}$. Every symbol has been given a value and the units come out right.
The arithmetic is fine and the answer is wrong, because the second condition in the box has been broken. With plates a metre across and half a metre apart, there is no region where the field is anything like uniform: the fringing that the figure draws as a small curl at the rim is now most of the picture, and the real capacitance is larger than the formula says. The formula is a small gap approximation, and the small gap condition is checked before it is used, not after. As a rule of thumb, the plates should be at least ten times wider than the gap.
gap d (mm)
C (pF)
C times d (F m)
0.100
885
8.85 e-14
0.200
443
8.85 e-14
0.500
177
8.85 e-14
1.00
88.5
8.85 e-14
Doubling the gap halves the capacitance exactly, and the third column says the same thing more sharply: the product stays at 8.85 times ten to the minus fourteen farad metres, which is the permittivity times the area. If your own answer does not keep that product fixed, the error is a power of ten.
Four square centimetres of plate, a tenth of a millimetre apart
Two flat plates each of area $4.00\ \mathrm{cm^2}$ face each other across a gap of $0.100\ \mathrm{mm}$ of air. Find the capacitance, then the charge on one plate when $50.0\ \mathrm{V}$ is put across the pair, then the field in the gap. Check that the air holds.
both lengths are converted to metres before anything is multiplied, because square centimetres carry a factor of ten thousand that is invisible once the numbers are on the page
going through V and d is one step; going through the charge would need sigma and epsilon nought and would give the same thing, so the shorter road is taken
Test the gap against breakdown
$$\frac{5.00\times10^{5}}{3\times10^{6}} = 0.17$$
dry air conducts at about three million volts per metre, so this gap is running at a sixth of its limit and the arrangement is safe
Independent check on the field by the other route, through the charge instead of the voltage: $\sigma = Q/A = 4.425\times10^{-6}\ \mathrm{C/m^2}$, and $E = \sigma/\varepsilon_0 = 5.00\times10^{5}\ \mathrm{V/m}$. Two roads, one number.
One conversion of area, one of length, and three arithmetic lines. The conversion is where the marks are lost.
Thirty five picofarads from a plate the size of a postage stamp is the scale to remember. Anything quoted in microfarads has an enormous rolled up area or something in the gap, usually both.
How large would a one farad pair of bare plates have to be?
Suppose you insist on building a $1.00\ \mathrm{F}$ capacitor out of two flat plates with air between them and a gap of $1.00\ \mathrm{mm}$. What area would each plate need, and how long would its side be if it were square?
a square root turns an area nobody can picture into a length everybody can; 113 square kilometres is a square about ten and a half kilometres on a side
Answer $$\boxed{\,A = 1.13\times10^{8}\ \mathrm{m^2},\qquad L \approx 10.6\ \mathrm{km}\,}$$
Check
Independent check by going backwards: $\varepsilon_0A/d$ with this area and gap is $(8.85\times10^{-12})(1.13\times10^{11}) = 1.00\ \mathrm{F}$, so the area does return one farad.
The farad is not a unit anyone reaches by making the plates bigger. Real ones get there by a very thin gap and by something in that gap which multiplies the answer, both of which are later blocks here.
Checkpoint
§06.2 — halving the gap with the charge held fixed●●○○○
Thirty seconds, and the second half is the part worth arguing. A parallel plate capacitor has $C = 20.0\ \mathrm{pF}$ with its plates $2.00\ \mathrm{mm}$ apart. It is charged, disconnected so that the charge on it cannot change, and then the plates are pushed to $1.00\ \mathrm{mm}$ apart.
Given
$C = 20.0\ \mathrm{pF}$ at $d = 2.00\ \mathrm{mm}$
the charge $Q$ on the plates is held fixed
the new separation is $d = 1.00\ \mathrm{mm}$
Find
(a) What is the new capacitance?
(b) Does the field in the gap change? Give the reason in one sentence.
Hint 1/4
The first part only needs to know how $C$ depends on $d$. The second part is asking which of the three quantities in the field is actually being held still.
Hint 2/4
$C = \varepsilon_0 A/d$, so $C$ and $d$ are inversely proportional; and the field between plates is $E = \sigma/\varepsilon_0$ with $\sigma = Q/A$.
Hint 3/4
Here $A$ is untouched, $Q$ is held fixed, and $d$ is halved from $2.00\ \mathrm{mm}$ to $1.00\ \mathrm{mm}$.
Hint 4/4
The capacitance doubles to $40.0\ \mathrm{pF}$, and the field does not change at all.
Show solutionUse the proportionality rather than the full formula
Independent consistency check through the definition: $Q = CV$ must hold before and after, and the capacitance doubled while the voltage halved, so their product is the same. That is exactly the statement that the charge was held fixed, which is what the question said.
⚠ Using the single sheet field in the gap between two plates
the result $\sigma/(2\varepsilon_0)$ is the one memorised from the Gauss's law section, and it is correct there
wrong$$E_{\rm gap} = \frac{\sigma}{2\varepsilon_0} \Rightarrow C = \frac{2\varepsilon_0 A}{d}$$
right$$E_{\rm gap} = \frac{\sigma}{\varepsilon_0} \Rightarrow C = \frac{\varepsilon_0 A}{d}$$
⚠ Leaving the area in square centimetres
lengths in centimetres are converted by reflex, but an area carries the conversion factor squared and the reflex does not
⚠ Using the whole plate area when the plates only partly overlap
the formula says A and each plate has an area, so the number is taken from whichever plate is mentioned first
wrong$$A = A_{\rm larger\ plate}$$
right$$A = A_{\rm overlap},\qquad \text{the facing region where the field actually is}$$
6.3Cables, shells, and one lonely sphere
The same three steps that gave the flat pair also give a coaxial cable and a pair of shells, and the second of those has a famous limit.
The plate calculation had three moves in it: charge the conductors, get the field with Gauss's law, integrate the field across the gap. Nothing in those three moves is about flatness, so they run just as well on a cable and on a pair of spheres.
RuleResult 6.3: three more capacitances from the same recipe
Conditions
The inner conductor carries $+Q$ and the outer carries $-Q$
For the cable, the length $L$ is large compared with $R_b$ so that the ends can be ignored
For a cable, the capacitance per metre depends only on the ratio of the two radii, through a logarithm, which is why doubling both radii changes nothing. For two shells it is set by the two radii and by the gap between them. For one sphere on its own it is simply the radius divided by the Coulomb constant, so a metre sized sphere manages about a tenth of a nanofarad.
Proof
Cable. Put charge $\lambda = Q/L$ per metre on the inner wire. A Gaussian cylinder of radius $r$ inside the gap encloses $\lambda \ell$, so $E = \lambda/(2\pi\varepsilon_0 r)$, radial.
Integrate outward from $R_a$ to $R_b$: $V = \int E\,dr = \dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{R_b}{R_a}$. Divide $Q = \lambda L$ by it and the charge cancels again.
Shells. A Gaussian sphere of radius $r$ in the gap encloses $Q$, so $E = kQ/r^{2}$, and integrating from $R_a$ to $R_b$ gives $V = kQ\left(\frac{1}{R_a}-\frac{1}{R_b}\right)$.
Lone sphere. Let $R_b\to\infty$ in the shell result. Then $R_aR_b/(R_b-R_a)\to R_a$, and $C = 4\pi\varepsilon_0 R_a$. The far conductor has not disappeared; it has just moved to infinity, which is where the zero of potential already lives.
A cable in cross-section. The $\textcolor{#1f6feb}{\text{inner wire and the outer sheath}}$ carry $\pm\lambda$ per metre, the $\textcolor{#cf222e}{\text{field}}$ between them is radial and dies as $1/r$, and the dashed Gaussian cylinder is the surface that produced that statement. Because the field varies across the gap, the voltage is an integral rather than a product, and integrating $1/r$ is where the logarithm comes from.
Looks like this, but is not
A single metal sphere has only one conductor, so it cannot have a capacitance; capacitance is a property of a pair. The definition does say pair, and there is visibly only one ball.
The second conductor is present, it is just very far away. When we say the sphere is at $V = kQ/R$, that number is measured against infinity, and infinity is where the returning charge $-Q$ has gone. Setting $R_b\to\infty$ in the two shell result is exactly that statement, and it leaves $C = 4\pi\varepsilon_0 R$ behind. The pair is still a pair; one member of it is the rest of the universe.
Rb / Ra
ln of the ratio
C per metre (pF)
2.00
0.693
80.2
3.00
1.099
50.6
5.00
1.609
34.6
10.0
2.303
24.1
A factor of five in the radius ratio buys only a factor of 3.3 in the capacitance, because the logarithm is a very slow function. That is why cable capacitance per metre is stuck in the tens of picofarads whatever the geometry, and why a cable maker changes what is in the gap instead.
The capacitance of a metre of coaxial cable
A cable has an inner wire of radius $0.500\ \mathrm{mm}$ inside a sheath of inner radius $2.50\ \mathrm{mm}$, with air between them. Find the capacitance of a one metre length, and then say what happens if both radii are doubled.
the ratio is what the logarithm sees, and doubling both radii leaves the ratio alone; a fatter cable of the same shape has the same capacitance per metre
Answer $$\boxed{\,C = 34.6\ \mathrm{pF}\ \text{per metre, and unchanged when both radii are doubled}\,}$$
Check
Independent check by a crude parallel plate stand-in, which should be in the right region without being exact: the mean circumference is $2\pi(1.50\times10^{-3}) = 9.42\times10^{-3}\ \mathrm{m}$, giving an area of $9.42\times10^{-3}\ \mathrm{m^2}$ per metre across a gap of $2.00\times10^{-3}\ \mathrm{m}$, so $\varepsilon_0A/d = 4.2\times10^{-11}\ \mathrm{F}$. That is $42\ \mathrm{pF}$ against $34.6\ \mathrm{pF}$: the same order, and larger, as it should be, because the flat estimate ignores the way the field weakens outward.
One ratio, one logarithm, one division. The logarithm is the only transcendental function in the whole section.
Because the answer sits inside a logarithm of a ratio, cable capacitance is very hard to change: a radius ratio of ten instead of two cuts it by barely a factor of three.
The Earth, treated as one isolated conductor
Treat the Earth as an isolated conducting sphere of radius $6.37\times10^{6}\ \mathrm{m}$. Find its capacitance, and find how much charge it would take to raise the whole planet to $1.00\ \mathrm{V}$ above infinity.
Given
$R = 6.37\times10^{6}\ \mathrm{m}$
$k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$
nothing else nearby, so the isolated sphere result applies
Find
the capacitance of the Earth and the charge needed for one volt
SolutionUse the shortest form of the isolated sphere result
$$C = 4\pi\varepsilon_0 R = \frac{R}{k} = \frac{6.37\times10^{6}}{8.99\times10^{9}}$$
writing it as R over k saves carrying four pi and epsilon nought separately, and the two forms are the same number
Independent check by going back through the potential of a point charge, which is where the formula came from: a charge of $7.09\times10^{-4}\ \mathrm{C}$ at a distance of $6.37\times10^{6}\ \mathrm{m}$ gives $V = kQ/R = (8.99\times10^{9})(7.09\times10^{-4})/(6.37\times10^{6}) = 1.00\ \mathrm{V}$, as required.
The whole planet manages seven hundred microfarads, and a component the size of your thumb beats it. Real capacitors get there with a folded area, a tiny gap, and a third trick the later blocks are about.
Checkpoint
§06.3 — a thin gap between two shells●●○○○
Thirty seconds with a calculator. A spherical capacitor has an inner shell of radius $5.00\ \mathrm{cm}$ inside an outer shell of inner radius $5.50\ \mathrm{cm}$, with vacuum between them.
Given
$R_a = 5.00\ \mathrm{cm} = 0.0500\ \mathrm{m}$
$R_b = 5.50\ \mathrm{cm} = 0.0550\ \mathrm{m}$
vacuum in the gap
Find
(a) Find the capacitance.
Hint 1/4
Two radii are given and the formula wants both of them plus their difference, so work out the difference first and keep it in metres.
Hint 2/4
$C = \dfrac{1}{k}\dfrac{R_aR_b}{R_b - R_a}$, which is the same as $4\pi\varepsilon_0R_aR_b/(R_b-R_a)$.
Hint 3/4
Here $R_aR_b = (0.0500)(0.0550)$ and $R_b - R_a = 5.00\times10^{-3}\ \mathrm{m}$.
using k rather than four pi epsilon nought keeps this to a single division
Answer $$\boxed{\,C = 61.2\ \mathrm{pF}\,}$$
Check
Independent check by pretending the gap is flat, which is legitimate because it is only a tenth of the radius: the mean area is $4\pi(0.0525)^{2} = 3.46\times10^{-2}\ \mathrm{m^2}$, and $\varepsilon_0A/d$ with $d = 5.00\times10^{-3}\ \mathrm{m}$ comes to $6.13\times10^{-11}\ \mathrm{F}$. The two agree to three figures, which is the sense in which a thin spherical gap is a parallel plate capacitor rolled into a ball.
⚠ Turning the ratio inside the logarithm into a difference
every other formula on the page has a gap width in it, so the eye expects a subtraction here too
Side by side the capacitances add; end to end the reciprocals add, and the result is smaller than the smallest.
Real hardware almost never arrives as one capacitor, and the two ways of joining a pair of them are settled by asking a single question: which quantity are the two forced to share?
RuleResult 6.4: equivalent capacitance of two standard arrangements
Conditions
Side by side (parallel): both capacitors are connected between the same two conductors, so both carry the same $V$
End to end (series): the piece of metal between the two is isolated and started out neutral, so both carry the same $Q$
Side by side, both feel the whole voltage and each takes its own charge, so the charges add and the capacitances add with them. End to end, the same charge is forced onto each and each demands its own share of the voltage, so the reciprocals add. That always gives something below the smallest term, so a chain is worse than any capacitor in it.
Proof
Side by side. Both are connected between the same two conductors, so both experience the same $V$. Then $Q_{\rm tot} = Q_1 + Q_2 = C_1V + C_2V = (C_1+C_2)V$, and dividing by $V$ gives $C_{\rm par} = C_1+C_2$.
End to end. Look at the piece of metal joining the two capacitors: one plate of the first, one plate of the second, and the wire between them. It touches nothing else, and it was neutral before anything was charged.
So if $-Q$ is drawn onto its left half, $+Q$ must be left on its right half. Charge conservation on an isolated island forces the same $Q$ onto every capacitor in the chain, whatever their sizes.
The voltages then add along the chain: $V = V_1 + V_2 = Q/C_1 + Q/C_2$.
Divide by $Q$: $V/Q = 1/C_1 + 1/C_2$, and the left side is $1/C_{\rm ser}$ by the definition. The last step is the one people forget: what has been computed is the reciprocal, and it still has to be inverted.
The two arrangements, and the reason each one behaves as it does. Side by side, both $\textcolor{#1f6feb}{\text{capacitors}}$ hang between the same pair of conductors, so both see the same $\textcolor{#8250df}{\text{potential difference}}$. End to end, the dashed $\textcolor{#8250df}{\text{island}}$ of metal between them is isolated and was neutral to begin with, so whatever is pulled off one side of it appears on the other, and both capacitors are stuck with the same charge.
Looks like this, but is not
Two capacitors end to end have four plates instead of two, and more plates should mean more capacitance. Side by side it worked that way, after all.
Count what actually got added. Side by side, the extra plates are extra area at the same voltage, so extra charge and a bigger $C$. End to end, they come with an extra gap, and gaps cost volts: the same charge now crosses two gaps, so the voltage doubles and $Q/V$ halves. Two $10.0\ \mathrm{\mu F}$ capacitors end to end give $5.00\ \mathrm{\mu F}$.
C1
C2
side by side
end to end
2.00
2.00
4.00
1.00
2.00
6.00
8.00
1.50
2.00
100
102
1.96
2.00
10000
10002
2.00
Read the last column downwards. However enormous the second capacitor becomes, the chain never gets past 2.00, the smaller one; it only creeps up to it. The third column runs away with the larger member instead. A chain is controlled by its smallest capacitor, a side by side group by its largest.
4.00 and 12.0 microfarads, joined both ways across 20.0 V
Two capacitors, $C_1 = 4.00\ \mathrm{\mu F}$ and $C_2 = 12.0\ \mathrm{\mu F}$, are put across a fixed $20.0\ \mathrm{V}$, first side by side and then end to end. In each case find the equivalent capacitance and the charge and voltage on each capacitor.
each voltage comes from that capacitor's own capacitance; the smaller capacitor takes the larger share, which is the opposite of most people's first guess
Two independent checks, one for each arrangement. Parallel: the total charge $80.0 + 240 = 320\ \mathrm{\mu C}$ must equal $C_{\rm par}V = (16.0)(20.0) = 320\ \mathrm{\mu C}$, and it does. Series: the two voltages must add up to the applied one, and $15.0 + 5.00 = 20.0\ \mathrm{V}$, which they do. Neither check reuses the line it is checking.
Two additions, one reciprocal sum, and four small divisions. The reciprocal sum is the only place a mark is usually lost.
The end to end answer, $3.00\ \mathrm{\mu F}$, is below $4.00\ \mathrm{\mu F}$, the smaller of the pair. That is not an accident of these numbers, and it is a one second sanity test on every chain answer you write.
A three capacitor network reduced and then unpicked
A capacitor $C_1 = 6.00\ \mathrm{\mu F}$ is joined end to end with a side by side pair $C_2 = 3.00\ \mathrm{\mu F}$ and $C_3 = 9.00\ \mathrm{\mu F}$. A potential difference of $24.0\ \mathrm{V}$ is applied across the whole thing. Find the equivalent capacitance, and then the charge and the voltage on each of the three.
Given
$C_1 = 6.00\ \mathrm{\mu F}$ end to end with the pair
$C_2 = 3.00\ \mathrm{\mu F}$ side by side with $C_3 = 9.00\ \mathrm{\mu F}$
$24.0\ \mathrm{V}$ across the whole arrangement
Find
the equivalent capacitance and the charge and voltage on each capacitor
the side by side pair is the only group whose members share a known quantity from the start, so it is the only group that can be collapsed without more information
Two independent checks. The voltages along the chain must add to the applied one: $16.0 + 8.00 = 24.0\ \mathrm{V}$. The charges in the side by side pair must add to the charge that came down the chain: $24.0 + 72.0 = 96.0\ \mathrm{\mu C}$. Both are arithmetic that was not used in getting the answers.
Two collapses going in, three expansions coming out. Every network problem has this shape, however many capacitors are in it.
The order is always the same: collapse inwards to one number, get the total charge, then expand outwards carrying at each stage the quantity the arrangement makes common.
Checkpoint
§06.4 — two identical capacitors, two arrangements●○○○○
Thirty seconds, no calculator. Two identical capacitors, each of $10.0\ \mathrm{\mu F}$, are available.
Given
two capacitors, each $C = 10.0\ \mathrm{\mu F}$
they can be joined side by side or end to end
Find
(a) Find the equivalent capacitance of each arrangement, and the factor between them.
Hint 1/4
Do not reach for the formulas yet. Ask which quantity each arrangement forces the two to share, and the answers follow from that.
Hint 2/4
Side by side the capacitances add; end to end the reciprocals add and the result must then be inverted.
Hint 3/4
Both capacitances are $10.0\ \mathrm{\mu F}$, so the sum is $20.0$ and the reciprocal sum is $2/10.0$.
Hint 4/4
Side by side gives $20.0\ \mathrm{\mu F}$ and end to end gives $5.00\ \mathrm{\mu F}$, a factor of four apart.
Independent check through the parallel plate picture rather than through the formulas: side by side is two plates of area $A$ next to each other, that is area $2A$ at the same gap, so $2C$. End to end is the same area with twice the total gap, so $C/2$. Both agree with the numbers above.
⚠ Adding capacitors end to end the way resistances are added
the two arrangements have the same picture in every book, and the rule that goes with each picture gets swapped
Charge does not all cross the full voltage, so the stored energy is half the charge times the voltage, not the whole product.
A charged capacitor will happily fire a flash tube, so energy went into it while it was being charged; the question is how much, and the answer is not the one the definition tempts you into writing.
RuleResult 6.5: stored energy, and the energy density of a field
Conditions
The capacitor starts uncharged and ends with charge $Q$ and potential difference $V$
Nothing is lost on the way, so the work done is what is stored
The density formula as written is for vacuum; a dielectric multiplies it by $K$
The energy stored is half the charge times the voltage, and the same number can be written with either the charge or the voltage eliminated. The last expression says the energy can be thought of as spread through the gap at a rate of half the permittivity times the square of the field per cubic metre, whether or not there is a capacitor there at all.
Proof
Charge the capacitor a slice at a time. When it already carries $q$, the potential difference across it is $q/C$, so moving the next slice $dq$ across costs $dW = (q/C)\,dq$.
Add up the slices: $W = \int_0^{Q} \frac{q}{C}\,dq = \frac{Q^{2}}{2C}$. Substituting $Q = CV$ gives the other two forms.
The one half is not decoration. The first slice crosses no voltage at all and the last crosses the full $V$, so the average is $V/2$, and the figure shows that average as the area of a triangle rather than of a rectangle.
Now re-attribute it. For a flat pair, $U = \tfrac12 CV^{2} = \tfrac12\left(\frac{\varepsilon_0A}{d}\right)(Ed)^{2} = \tfrac12\varepsilon_0E^{2}(Ad)$.
The bracket $Ad$ is the volume of the gap, so the energy per unit volume is $u = \tfrac12\varepsilon_0E^{2}$. The derivation used a capacitor, but the result mentions only the field, and it is used wherever a field is.
Why the answer carries a half. As charge is moved across, the $\textcolor{#cf222e}{\text{voltage the next slice has to climb}}$ rises in proportion to the charge already moved. The work done is the $\textcolor{#8250df}{\text{area under that line}}$, a triangle of area $\tfrac12 QV$, and for the $2.00\ \mathrm{\mu F}$ capacitor drawn here that is $400\ \mathrm{\mu J}$. Multiplying $Q$ by the final $V$ would count the whole rectangle and be wrong by a factor of two.
Looks like this, but is not
$U = \tfrac12 CV^{2}$, so a larger capacitor always stores more energy. The capacitance is sitting right there in the numerator.
Only at the same voltage. At the same charge the right form is $U = Q^{2}/2C$, and that one falls as $C$ rises. Two capacitors carrying $60.0\ \mathrm{\mu C}$ each, one of $4.00\ \mathrm{\mu F}$ and one of $12.0\ \mathrm{\mu F}$, hold $450\ \mathrm{\mu J}$ and $150\ \mathrm{\mu J}$ respectively: the bigger capacitor holds three times less. The three expressions are the same number, but they are not the same statement about what happens when something is changed, and choosing the wrong one is how the dielectric problems in the next block go wrong.
V (volts)
Q (μC)
U (μJ)
5.00
10.0
25.0
10.0
20.0
100
20.0
40.0
400
40.0
80.0
1600
Doubling the voltage doubles the charge and quadruples the energy, all the way down the table. That asymmetry is why a capacitor's voltage rating decides how much energy it can hold: doubling the working voltage is worth four times doubling the capacitance.
The three joules behind a camera flash
A flash unit stores its energy in a $150\ \mathrm{\mu F}$ capacitor charged to $200\ \mathrm{V}$. Find the charge on it and the energy stored, and find the average power delivered if that energy leaves in $2.00\ \mathrm{ms}$.
Independent check by the form that was not used: $U = Q^{2}/2C$, and here that is $(9.00\times10^{-4})/(3.00\times10^{-4}) = 3.00\ \mathrm{J}$. Same answer through a different pair of quantities. Physical check: a kilowatt and a half for two milliseconds is the right order for a bright flash, and if the answer had come out in watts or in megawatts a power of ten would have slipped.
One multiplication, one square, one division, and a decision about which of the three energy forms to use.
The capacitor did not create the kilowatt and a half. It collected three joules slowly and gave them back in two milliseconds, and that ratio of times is the whole point.
The same energy counted twice, once as a total and once per cubic metre
A parallel plate capacitor has plates of area $2.00\times10^{-2}\ \mathrm{m^2}$ separated by $1.00\ \mathrm{mm}$ of air, held at $100\ \mathrm{V}$. Find the stored energy from the capacitance, then find the field, the energy density and the volume of the gap, and get the same energy that way.
The agreement of the two routes is itself the check, and it is a genuine one because they share no intermediate quantity: the first goes through $C$ and never mentions the field, the second goes through $E$ and never mentions the capacitance. A unit check on the second closes it: $\mathrm{C^{2}\,N^{-1}m^{-2}}$ times $\mathrm{V^{2}m^{-2}}$ gives joules per cubic metre.
Six lines instead of three, to get an answer that three lines already gave. The point is not the answer but the licence to speak of energy sitting in the field.
Since the energy can be written using only the field, it can be attributed to the field, and then it can be asked for in places where no capacitor exists. One consequence, which is not on this week's syllabus line but gives a useful independent check later: differentiating $U = Q^{2}d/(2\varepsilon_0A)$ with respect to $d$ at fixed charge gives the attraction between the plates, $F = Q^{2}/(2\varepsilon_0A) = \tfrac12 QE$. The half is there because each plate sits in the field of the other one only.
Checkpoint
§06.5 — tripling the voltage, then tripling the charge●●○○○
Thirty seconds, no calculator, and the second half is not a repeat of the first. A capacitor is charged so that it stores $50.0\ \mathrm{\mu J}$.
Given
stored energy at the start is $50.0\ \mathrm{\mu J}$
the capacitance itself is not altered in either part
Find
(a) The voltage across it is then tripled. What is the new stored energy?
(b) Starting again from $50.0\ \mathrm{\mu J}$, the charge on it is tripled instead. What is the new stored energy?
Hint 1/4
Both parts are about how the energy depends on one quantity while the capacitance stands still. Write down which of the three energy forms mentions only that quantity and the capacitance.
Hint 2/4
$U = \tfrac12 CV^{2}$ has the voltage squared in it, and $U = Q^{2}/2C$ has the charge squared in it.
Hint 3/4
In both parts $C$ is unchanged and the quantity in question is multiplied by three, so the energy is multiplied by three squared.
⚠ Quoting the energy density where the total energy was asked for
lower case u and upper case U are one keystroke apart and both are called energy in conversation
wrong$$U = \tfrac12\varepsilon_0E^{2}$$
right$$U = \tfrac12\varepsilon_0E^{2}\times(\text{volume}),\qquad u = \tfrac12\varepsilon_0E^{2}$$
6.6A slab in the gap: what the dielectric constant multiplies
An insulating slab in the gap multiplies the capacitance by a number K, and divides the field by the same number.
Every capacitance so far has assumed vacuum in the gap, and no real component is built that way; filling it with an insulator is the largest of the tricks that make a farad possible.
RuleResult 6.6: the dielectric constant
Conditions
The slab completely fills the gap between the conductors
$K$ is a property of the material, a pure number never less than one
The field stays below the dielectric strength of the material, or it stops being an insulator
$$\boxed{\;C = K C_0 = \frac{K\varepsilon_0 A}{d} = \frac{\varepsilon A}{d},\qquad \varepsilon = K\varepsilon_0,\qquad K \ge 1\;}$$
Filling the gap with an insulator multiplies the capacitance by a number belonging to the material. Everything else follows from that plus one question: what is held fixed. With the charge fixed, the voltage and the field fall by $K$; with the voltage fixed, they stand still and the charge rises.
Proof
Charge a capacitor to $Q$, disconnect it so $Q$ cannot change, and measure $V_0$. Slide the slab in and measure again: the voltage has fallen to $V_0/K$. That measurement defines $K$.
The charge did not change and the voltage fell by a factor $K$, so $C = Q/V$ rose by a factor $K$: $C = KC_0$.
The gap $d$ did not change either, and $V = Ed$, so the field also fell by $K$: $E = E_0/K$. Where it went is the next block.
$C$ belongs to the hardware, not to how it is being used, so $K$ stays put whatever is done afterwards. If the slab goes in while the voltage rather than the charge is held fixed, $C$ still rises by $K$ and the extra charge $Q = KC_0V$ is drawn onto the plates instead.
For a flat pair this reads $C = K\varepsilon_0 A/d$, often shortened to $\varepsilon A/d$ with $\varepsilon = K\varepsilon_0$. Vacuum is the case $K = 1$, so nothing earlier is contradicted.
A slab of $\textcolor{#1a7f37}{K = 3.00}$ filling the right hand half of a gap. Two of every three $\textcolor{#cf222e}{\text{field lines}}$ stop on the bound charge at the slab's face, so the field inside the slab is a third of the field outside it. Below, the same story told as a $\textcolor{#8250df}{\text{potential}}$: the drop is steep where the field is strong and shallow where it is weak, and three quarters of the whole voltage is used up in the empty half.
Looks like this, but is not
A dielectric is an insulator, so putting one into the gap must make it harder to get charge onto the plates and therefore reduce the capacitance. Insulators block things; that is what the word means.
The slab is never in the way of the charge. The charge lives on the metal plates and arrives by the route it always did; the slab sits between them and never touches the supply. What it does is weaken the field, so the same charge costs fewer volts and $Q/V$ goes up. No material has $K$ below one, so a slab can never reduce a capacitance.
material
K, roughly
strength (MV/m), roughly
vacuum
1 exactly
no limit
air
1.0006
3
paper
3.7
15
mica
7
150
water
80
not usable
Two things. Air sits within a thousandth of vacuum, which is why every air gap here has been treated as vacuum without apology. And the two columns do not rise together: water has the largest K and is useless because anything dissolved in it conducts, while mica has a middling K and an enormous breakdown field. A dielectric is chosen on both, and usually the second decides.
Why a paper capacitor beats an air one by a factor of ninety
A capacitor is built from plates of area $6.00\ \mathrm{cm^2}$ separated by $0.0500\ \mathrm{mm}$ of paper, for which take $K = 3.7$ and a dielectric strength of $15\times10^{6}\ \mathrm{V/m}$. Find its capacitance, the largest voltage it can take, and the largest energy it can hold. Then compare with the same plates and the same gap filled with air, for which $K = 1.0006$ and the strength is $3\times10^{6}\ \mathrm{V/m}$.
Independent check on the energy ratio without touching either energy: the paper wins a factor $3.7/1.0006 = 3.70$ on capacitance and a factor $15/3 = 5$ on voltage, and the energy carries the voltage squared, so the ratio should be $3.70\times25 = 92.4$. Dividing the two energies directly gives $92.4$ as well.
Four short calculations done twice. The only new physical idea is that dielectric strength is a field and has to be multiplied by the gap.
Most of the ninety comes from the breakdown field, not from $K$. A dielectric is worth having twice over: it multiplies the capacitance, and it lets you put a far bigger field across the gap before anything sparks.
A slab that fills only half the gap
A parallel plate capacitor of $C_0 = 40.0\ \mathrm{pF}$ has a slab of dielectric constant $K = 3.00$ slid in so that it fills exactly half the gap, lying flat against one plate and leaving vacuum in the other half. Find the new capacitance, and find what fraction of the total voltage appears across the empty half.
Given
$C_0 = 40.0\ \mathrm{pF}$ with the gap empty
the slab has $K = 3.00$ and thickness $d/2$
the slab lies flat against one plate, so the two regions are stacked across the gap
Find
the new capacitance, and the share of the voltage taken by the empty half
SolutionTurn the stack into two capacitors
$$\text{imagine a very thin metal foil laid on the face of the slab}$$
it changes nothing, because the field is perpendicular to it and it carries no net charge, but it lets the gap be read as two capacitors end to end
Independent check on the general formula at its two limits, neither of which was used to derive it. Put $K = 1$: the factor becomes $2/2 = 1$, so $C = C_0$, correct, because a slab of vacuum is nothing at all. Let $K\to\infty$: the factor tends to $2$, so $C\to 2C_0$, also correct, because an infinitely good dielectric behaves like a conductor filling half the gap, and that halves the effective separation. A formula that gets both ends right is very unlikely to be wrong in the middle.
One imaginary foil, one , one voltage divide. The imaginary foil is the whole trick.
Half a slab is worth much less than half a slab's worth, because the two halves form a chain and a chain is dominated by its smallest member, here the empty half. For the full factor $K$ the material has to fill the gap.
Checkpoint
§06.6 — sliding a slab in with the supply still attached●●○○○
Thirty seconds, and the whole question is which quantity is nailed down. A capacitor stays connected to a source that holds it at $12.0\ \mathrm{V}$. A slab with $K = 2.50$ is then slid in until it fills the gap completely.
Given
the potential difference is held at $12.0\ \mathrm{V}$ throughout
the slab has $K = 2.50$ and fills the gap
the plate area and the separation are not touched
Find
(a) Does the charge on the plates rise, fall or stay put, and by what factor?
(b) What happens to the field in the gap?
Hint 1/4
Write down which quantity the question has nailed down, and then look at each of $Q = CV$ and $E = V/d$ to see which symbols in them are free to move.
Hint 2/4
$C = KC_0$ always, and with $V$ held fixed $Q = CV$ must follow $C$; meanwhile $E = V/d$ mentions neither $C$ nor $K$.
Hint 3/4
Here $V = 12.0\ \mathrm{V}$ is fixed, $d$ is fixed, and $K = 2.50$.
Hint 4/4
The charge rises by a factor of $2.50$, and the field does not change at all.
Show solutionThe capacitance first, since it never depends on the circumstances
$$C = KC_0 = 2.50\,C_0$$
this line is true in every insertion problem and is the only line that is
Now use the quantity that was pinned
$$Q = CV = 2.50\,C_0V = 2.50\,Q_0$$
V is fixed by the source, so the whole change in C shows up in the charge
$$E = \frac{V}{d}\ \text{with both } V \text{ and } d \text{ fixed} \Rightarrow E = E_0$$
neither symbol in this expression moved, so nothing about the field can have moved either
Independent check by the bound charge instead: the field inside a dielectric is set by the net surface charge, free minus induced. Here the free charge rose by 2.50 and the induced charge is $\sigma(1 - 1/2.50) = 0.600\sigma$, so the net is $0.400 \times 2.50 = 1.00$ times the original. The field is unchanged, as found.
⚠ Dividing the capacitance by K instead of multiplying
the field really does get divided by K, and the two statements sit one line apart
wrong$$C = \frac{C_0}{K}$$
right$$C = KC_0,\qquad E = \frac{E_0}{K}$$
⚠ Saying the voltage falls by K when the source is still attached
the sentence the voltage falls by K is how the constant is defined, and the condition attached to it (the charge is fixed) travels less well than the sentence does
⚠ Treating a partly filled gap as if it were fully filled
the slab is visibly there, so the factor K is applied whole
wrong$$\text{slab of thickness } d/2: \ C = KC_0$$
right$$\text{slab of thickness } d/2: \ C = \frac{2K}{K+1}C_0$$
6.7Where the missing field went: bound charge on the faces
The molecules line up, their ends cancel everywhere except at the two faces, and the charge left there is what weakens the field.
The previous block found the field inside the slab smaller by a factor $K$ and left the reason open; the reason is a second layer of charge, and it can be computed exactly.
TheoremResult 6.7: bound charge, and Gauss's law with a dielectric present
Conditions
The slab fills the gap and the field inside it is uniform
$\sigma$ is the free charge density on the metal plate, $\sigma_{\rm ind}$ the bound charge density on the slab face
The material responds in proportion to the field, which is true for every material in this course
The field inside the slab is made by the free charge on the metal minus the bound charge on the slab's face, and that bound charge is the fraction one minus one over K of the free charge. The last line saves work: carry a K inside the flux integral and only the free charge ever needs counting.
Proof
Each molecule in the slab is, or becomes, a small dipole: water molecules already are and simply turn, paraffin molecules get stretched. Either way they end up pointing the same way.
Deep inside the slab the positive end of one molecule sits beside the negative end of the next and the two cancel. Only at the faces is anything left: $-\sigma_{\rm ind}$ towards the positive plate, $+\sigma_{\rm ind}$ on the other. That charge is bound.
A pillbox in the slab now encloses two sheets, so $E = (\sigma - \sigma_{\rm ind})/\varepsilon_0$.
But the previous block measured $E = E_0/K = \sigma/(K\varepsilon_0)$. Equating, $\sigma - \sigma_{\rm ind} = \sigma/K$, so $\sigma_{\rm ind} = \sigma(1 - 1/K)$.
Since $K \ge 1$ the bracket lies between $0$ and $1$, so the bound charge is always smaller than the free charge: it weakens the field, never cancels or reverses it. Hiding it inside the constant gives the last line of the box.
Why a face ends up charged. With no field the $\textcolor{#1a7f37}{\text{molecular dipoles}}$ point in every direction and every exposed end has a neighbour of the opposite sign, so the faces carry nothing. Switch the $\textcolor{#cf222e}{\text{field}}$ on and they line up: now the leftmost column of negative ends and the rightmost column of positive ends have nobody to cancel against, and those two rows are the bound charge that weakens the field inside.
Looks like this, but is not
If the bound charge can grow with $K$, then a good enough dielectric cancels the field completely, so a metal is just a dielectric with $K$ infinite. The algebra does say $\sigma_{\rm ind}\to\sigma$ as $K\to\infty$.
Half of that is a useful memory hook and half is wrong about the material. The limit is right: $K = \infty$ in $C = 2KC_0/(K+1)$ gives $2C_0$, which is what a metal slab of that thickness gives. But bound charge is tethered to its molecule and shifts only a fraction of a molecular diameter, so $\sigma_{\rm ind}$ stays below $\sigma$. In a metal the charge travels the whole body and can be removed from it. Same limit, different mechanism.
K
1/K
fraction cancelled
1.0006
0.9994
0.0006
2.00
0.500
0.500
3.70
0.270
0.730
7.00
0.143
0.857
80.0
0.0125
0.988
The last column climbs quickly and then crawls. K from 2 to 7 raises the cancelled fraction from a half to six sevenths, but the rest of the way to one costs another factor of ten in K. Even water leaves one and a quarter per cent uncancelled. Nothing made of tethered charges reaches the end of that column; only a metal does.
How much charge appears on the face of the slab
The plates of area $4.00\ \mathrm{cm^2}$ from earlier carry $1.77\ \mathrm{nC}$ each, with a gap of $0.100\ \mathrm{mm}$ that has been filled with a slab of $K = 3.40$ after the supply was disconnected. Find the free surface charge density, the bound charge density on each face of the slab, the total bound charge, and the field inside the slab.
Independent check on the field by the other road, which never mentions the bound charge: with the slab out, $E_0 = \sigma/\varepsilon_0 = (4.425\times10^{-6})/(8.85\times10^{-12}) = 5.00\times10^{5}\ \mathrm{V/m}$, and the previous block says the slab divides that by $K$, giving $5.00\times10^{5}/3.40 = 1.47\times10^{5}\ \mathrm{V/m}$. The two agree, which is the arithmetic content of the claim that the bound charge is where the missing field went.
Four divisions and one subtraction, and the only thing that has to be remembered is that the bracket carries one over K and not K.
The bound charge is $1.25\ \mathrm{nC}$ against a free charge of $1.77\ \mathrm{nC}$: seventy per cent neutralised, and only the remaining $0.52\ \mathrm{nC}$ left to make a field. Thirty per cent is exactly $1/3.40$.
Energy density inside the slab, and the total it adds up to
For the same arrangement, with $E = 1.47\times10^{5}\ \mathrm{V/m}$ inside a slab of $K = 3.40$ filling a gap of $1.00\times10^{-4}\ \mathrm{m}$ between plates of $4.00\times10^{-4}\ \mathrm{m^2}$, find the energy density and hence the total stored energy, and compare it with the $44.3\ \mathrm{nJ}$ the same capacitor held before the slab went in.
Given
$E = 1.47\times10^{5}\ \mathrm{V/m}$ inside the slab, $K = 3.40$
gap volume is $A d = (4.00\times10^{-4})(1.00\times10^{-4})\ \mathrm{m^3}$
before the slab was inserted the stored energy was $44.3\ \mathrm{nJ}$
Find
the energy density in the slab, the total energy, and the ratio to the energy before
Independent check by the capacitor formula rather than by the field: with the slab in, $C = 3.40\times35.4\ \mathrm{pF} = 120\ \mathrm{pF}$, and with $Q = 1.77\ \mathrm{nC}$ held fixed, $U = Q^{2}/2C = (1.77\times10^{-9})^{2}/(2\times1.204\times10^{-10}) = 1.30\times10^{-8}\ \mathrm{J}$. The field route and the capacitor route agree to three figures.
Energy went missing and did not vanish. The field acting on the induced charge pulled the slab in, and that work is exactly the $31.2\ \mathrm{nJ}$ difference. Hold a slab at the mouth of a charged capacitor and it really is tugged in.
Checkpoint
§06.7 — what fraction of the plate charge gets cancelled●●○○○
Thirty seconds. A slab with $K = 5.00$ is slid into a gap where the field was $2.00\times10^{5}\ \mathrm{V/m}$, with the charge on the plates held fixed.
Given
$K = 5.00$, the slab fills the gap
the field before insertion was $E_0 = 2.00\times10^{5}\ \mathrm{V/m}$
the plate charge is held fixed
Find
(a) What is the field inside the slab?
(b) What fraction of the plate charge is cancelled by the bound charge on the face?
Hint 1/4
The two parts are two readings of the same number. One asks what the constant does to the field, the other asks what it does to the charge that produced the change in the field.
Hint 2/4
$E = E_0/K$ and $\sigma_{\rm ind}/\sigma = 1 - 1/K$.
Hint 3/4
Here $E_0 = 2.00\times10^{5}\ \mathrm{V/m}$ and $K = 5.00$, so $1/K = 0.200$.
Hint 4/4
The field is $4.00\times10^{4}\ \mathrm{V/m}$ and four fifths of the plate charge is cancelled.
Independent check by a limiting case: at $K = 1$ the formula gives no cancellation and no reduction, which is correct for a gap with nothing in it, and as $K$ grows the fraction approaches but never reaches one, which is correct for a material whose charges are tethered to their own molecules.
⚠ Letting the bound charge exceed the free charge
the formula is often half remembered as sigma times K rather than sigma times one minus one over K
it is charge and it is on a surface, so it looks like the free charge on a conductor, which can indeed be drained
wrong$$\text{ground the slab} \Rightarrow \sigma_{\rm ind} = 0$$
right$$\sigma_{\rm ind}\ \text{is bound to its own molecules and stays; only free charge can be drained}$$
⚠ Using the vacuum energy density inside the slab
the density formula was met first without a K in it, and the K is easy to leave behind when the field is the quantity in hand
wrong$$u = \tfrac12\varepsilon_0E^{2}\ \text{inside a dielectric}$$
right$$u = \tfrac12 K\varepsilon_0E^{2}\ \text{inside a dielectric of constant } K$$
Reducing a network of capacitors, and coming back out again
Any question with more than one capacitor in it. Six steps, three in and three out, and the order never changes.
Redraw before calculating
Two capacitors are side by side only if both of their terminals are joined to the same two conductors, and end to end only if the metal between them touches nothing else. Being drawn next to each other on the page means nothing. Redraw until every group is one or the other.
Collapse the innermost group first
Add the capacitances for a side by side group, add the reciprocals and then invert for a chain. Write the collapsed value onto the picture; it will be needed again on the way out.
Repeat until one number is left
That number is $C_{\rm eq}$, and it is the whole network as seen from outside. Nothing inside the network has been solved yet.
Total charge, from the applied voltage
$Q_{\rm tot} = C_{\rm eq}V$. This is the charge that entered the network, and it is the quantity that gets carried back inwards.
Expand outwards, carrying the common quantity
At each stage ask what the grouping forces the members to share: a chain shares the charge, a side by side group shares the voltage. Give every member that shared quantity, then use $C = Q/V$ on each member individually to get the other one.
Check with two sums that were not used
The voltages along any chain must add up to the voltage across that chain, and the charges in any side by side group must add up to the charge that came into it. Both are arithmetic you did not use, so both are real checks.
Where it goes wrong
Forgetting the final inversion, so the reciprocal sum is reported as the capacitance.
Splitting the voltage equally along a chain of unequal capacitors.
Calling two capacitors side by side because the drawing puts them next to each other.
Something about the capacitor changed: what was held fixed?
Any question in which a gap is altered, a slab is inserted or removed, or plates are moved. This is the single most examined shape in this material, and almost all of the marks turn on step 2.
Name the change
Gap, area, or material. Write down which one, because the next steps treat them differently.
Ask whether the source is still attached
Attached means the potential difference is fixed and the charge is free to move on or off the plates. Disconnected means the charge is fixed and the voltage is free. Everything downstream depends on this one sentence, so write it out.
Recompute the capacitance
$C = K\varepsilon_0A/d$ with the new geometry and the new material. This step is the same whatever the answer to step 2 was, because the capacitance belongs to the hardware and not to how it is being used.
Push the fixed quantity through the definition
If $V$ is fixed, the new charge is $Q = CV$. If $Q$ is fixed, the new voltage is $V = Q/C$. Only one of these two lines is ever correct in a given problem.
Field next
For a flat pair, $E = V/d$ is the shortest route if you now know $V$. If you know $Q$ instead, $E = Q/(K\varepsilon_0A)$ gets there without a detour through the voltage.
Energy last, with the matching form
Use $U = Q^{2}/2C$ when the charge was the fixed quantity and $U = \tfrac12 CV^{2}$ when the voltage was. Using the other one gives a change in the wrong direction, and it is a change of the same size, which makes it hard to spot.
Say in words where any energy went
If the energy fell, something did work on the surroundings, usually by pulling a slab in. If it rose, the source paid for it. An answer with energy appearing from nowhere is not finished.
Where it goes wrong
Getting the new capacitance right and then using the wrong fixed quantity, which spoils everything after it.
Using $\tfrac12 CV^{2}$ in a disconnected problem, which reports the energy as rising by $K$ when it actually falls by $K$.
Reporting an energy change without saying where the energy came from or went.
Getting a capacitance out of a new shape
When the geometry is not one of the three on the formula card. The same five steps produced all three of those.
Charge the conductors
Put $+Q$ on the inner or upper conductor and $-Q$ on the other. $Q$ is a symbol here, not a number; it is going to cancel.
Choose the Gaussian surface the symmetry allows
A sphere for spherical symmetry, a coaxial cylinder for a cable, a pillbox for a plate. If no symmetry fits, the capacitance has to be measured or computed numerically instead.
Field between the conductors
Gauss's law gives $E$ at every point of the gap, proportional to $Q$. Nothing outside the gap matters, because the field is zero inside both conductors.
Integrate across the gap
$V = \int E\,dl$ along any path from the positive conductor to the negative one, which for these symmetries is a straight radial line. Still proportional to $Q$.
Divide, and check that Q disappears
$C = Q/V$. If $Q$ survives into the answer, one of steps 3 and 4 has an error in it, because the constancy argument says it cannot survive. Then test a limit you already know.
Where it goes wrong
Integrating along a path that does not end on the other conductor.
Including the region inside the metal in the integral, where the field is zero anyway.
Leaving $Q$ in the final expression and not noticing.
The slab goes in with the supply disconnected
A capacitor of $C_0 = 35.4\ \mathrm{pF}$ is charged to $50.0\ \mathrm{V}$ and then disconnected, so the charge on it can no longer change. A slab of $K = 3.40$ is slid in until it fills the gap. Find the capacitance, the charge, the voltage, the field and the stored energy afterwards.
Given
$C_0 = 35.4\ \mathrm{pF}$, charged to $V_0 = 50.0\ \mathrm{V}$, then disconnected
the slab has $K = 3.40$ and fills the gap
gap $d = 0.100\ \mathrm{mm}$, unchanged throughout
Find
the five quantities after insertion, with the charge held fixed
SolutionState what is nailed down before anything else
the energy fell by a factor K, and the missing 31.2 nJ was spent pulling the slab into the gap
Answer $$\boxed{\;C = 120\ \mathrm{pF},\ Q = 1.77\ \mathrm{nC}\ \text{(fixed)},\ V = 14.7\ \mathrm{V},\ E = 1.47\times10^{5}\ \mathrm{V/m},\ U = 13.0\ \mathrm{nJ}\;}$$
Check
Independent check on the energy through the field rather than through the capacitor: $u = \tfrac12K\varepsilon_0E^{2} = 0.325\ \mathrm{J/m^{3}}$ over a gap volume of $4.00\times10^{-8}\ \mathrm{m^{3}}$ gives $1.30\times10^{-8}\ \mathrm{J}$, which is the same number by a road that never uses $C$.
Everything except the charge moved, and everything that moved went down by the same factor $K$.
The same slab goes in with the supply still attached
The same capacitor, $C_0 = 35.4\ \mathrm{pF}$, is held at $50.0\ \mathrm{V}$ by a source that stays connected while the same slab of $K = 3.40$ is slid in. Find the capacitance, the charge, the voltage, the field and the stored energy afterwards.
Given
$C_0 = 35.4\ \mathrm{pF}$, held at $V = 50.0\ \mathrm{V}$ throughout
the slab has $K = 3.40$ and fills the gap
gap $d = 0.100\ \mathrm{mm}$, unchanged throughout
Find
the five quantities after insertion, with the voltage held fixed
SolutionState what is nailed down before anything else
$$V = 50.0\ \mathrm{V}\ \text{, fixed by the source}$$
connected means the source will move whatever charge it takes to keep this number where it is
Capacitance, identical to the other case
$$C = KC_0 = 120\ \mathrm{pF}$$
the hardware is the same hardware, so this line cannot depend on whether a source is attached to it
the voltage is the fixed quantity here, so this is the form with nothing hidden moving in it
$$\frac{U}{U_0} = \frac{150}{44.3} = 3.40$$
the energy rose by a factor K instead of falling by it, and the source paid for the increase
Answer $$\boxed{\;C = 120\ \mathrm{pF},\ Q = 6.02\ \mathrm{nC},\ V = 50.0\ \mathrm{V}\ \text{(fixed)},\ E = 5.00\times10^{5}\ \mathrm{V/m},\ U = 150\ \mathrm{nJ}\;}$$
Check
Independent check on the charge by the bound charge argument: with the field unchanged, the net surface density must be unchanged, so $\sigma(1 - 1/K)$ worth of new free charge must have arrived to be cancelled. That means $\sigma_{\rm new} = K\sigma_{\rm old}$, so $Q = 3.40\times1.77 = 6.02\ \mathrm{nC}$, as computed.
Everything except the voltage moved, and everything that moved went up.
Identical hardware, identical slab, identical $K$, and the capacitance comes out identical at $120\ \mathrm{pF}$ in both. Every other answer differs: the voltage and the field fall by $3.40$ in one and stand still in the other, and the stored energy falls by $3.40$ in one while rising by $3.40$ in the other, a factor of $11.6$ apart.
How to tell them apart
Read the problem for one word before you write anything: is the source still attached? Attached pins the voltage, so the charge and the energy rise. Disconnected pins the charge, so the voltage, the field and the energy fall. The capacitance is the only quantity that does not care, and that is precisely why starting your solution from the capacitance leaves you with no way of telling the two cases apart.
Scaffolding comes off
The common skeleton
Say what the arrangement is, and name the quantity that is nailed down: the charge if nothing is attached, the potential difference if a source is.
Work out every capacitance in the problem from the geometry and the material, before and after any change. This step never depends on the previous one.
If there is more than one capacitor, reduce the network to a single equivalent capacitance.
Push the nailed down quantity through $C = Q/V$ to get the other one for the arrangement as a whole.
Expand back out to the individual capacitors, carrying at each stage the quantity that the grouping makes common.
Get the fields and the energies last, choosing the energy form that is built from whatever was nailed down.
Check: voltages along a chain must add up, charges in a group must add up, and any energy that appeared or disappeared must have somewhere to have come from.
1 · fully worked
Pulling the plates apart on a disconnected capacitor
A parallel plate capacitor has plates of area $8.00\ \mathrm{cm^2}$ separated by $0.250\ \mathrm{mm}$ of air. It is charged by a $24.0\ \mathrm{V}$ source, then disconnected, and then the plates are pulled apart to $0.500\ \mathrm{mm}$. Find the capacitance, charge, voltage, field and stored energy before and after, and find the work that had to be done to pull the plates apart.
energy does not appear from nowhere; the plates attract each other, so pulling them apart is work done against that attraction, and it lands in the store
Independent check on the work by computing the force instead of the energy. Each plate sits in the field of the other one only, which is half the total, so the attraction is $F = \tfrac12 QE = 3.26\times10^{-5}\ \mathrm{N}$. The field is constant during the move, so the work is $F\Delta d = 8.16\times10^{-9}\ \mathrm{J}$, matching the energy difference exactly.
Two capacitances, one charge, one voltage, two fields, two energies and a subtraction. The whole problem is the first line: the charge is what is fixed.
The field did not change while the plates were separated, and that is the reason the force was constant and the work came out as a simple product. Any question that pulls plates apart at fixed charge has this property; any question that does it with the source attached does not.
2 · you write the reasoning
Easier than rung 1: no geometry, no change, and only three lines. Two capacitors, $C_1 = 3.00\ \mathrm{\mu F}$ and $C_2 = 6.00\ \mathrm{\mu F}$, are placed side by side across a fixed $12.0\ \mathrm{V}$. The three lines below are all correct. Before opening the model reasons, say in your own words why each one is allowed, paying particular attention to what the third line is checking.
reasoning
The two capacitances may be added only because both capacitors hang between the same pair of conductors, so both of them span the same potential difference. That is what side by side means, and it is the assumption doing the work here, not the arithmetic. If the two were end to end the same picture would give a very different number, $2.00\ \mathrm{\mu F}$, so the classification has to be settled before this line is written.
reasoning
Each capacitor is treated on its own with its own capacitance and the shared voltage of $12.0\ \mathrm{V}$. Nothing is divided between them, because nothing is being shared out: the source holds both at twelve volts and each takes whatever charge its own capacitance asks for. The larger capacitor takes the larger charge, in exact proportion, which is the opposite of what happens in a chain.
reasoning
This line is not a new result; it is a check. The total charge that entered the arrangement should equal the charge the equivalent capacitor would have taken, and it does. Because the check uses the first line and the second line and combines them in a way neither of them assumed, it would catch an error in either of them. A check that reuses the same multiplication would catch nothing.
3 · find the buried error
Harder than rung 2: two capacitors end to end, and a dielectric goes into one of them. $C_1 = 2.00\ \mathrm{\mu F}$ and $C_2 = 3.00\ \mathrm{\mu F}$ are joined end to end across a fixed $25.0\ \mathrm{V}$, and then a slab with $K = 2.00$ is slid in so that it completely fills $C_2$. A student works out the charge and voltage on $C_1$ afterwards and gets $12.2\ \mathrm{\mu C}$ and $6.11\ \mathrm{V}$. Exactly two of the four steps below are faulty. Find them.
the two buried errors (2)
⚠ step 1
The dielectric constant multiplies the capacitance, it does not divide it: $C_2' = KC_2 = (2.00)(3.00) = 6.00\ \mathrm{\mu F}$, not $1.50\ \mathrm{\mu F}$. The stated reason is the source of the error: $K$ divides the field and multiplies the capacitance, and those are two different statements that live one line apart in every textbook.
The two facts are learnt together in the same breath, they involve the same constant, and one of them really does divide. A student who remembers only the phrase the dielectric weakens things will apply the division to whatever quantity is in front of them. The wrong line is also self-consistent with a plausible sentence, so nothing about it looks odd on the page.
right
Anchor the direction on the definition rather than on a word. $C = Q/V$; with the charge fixed, inserting a slab makes $V$ smaller, so $Q/V$ must get bigger. A slab can never reduce a capacitance, because there is no material with $K$ below one. With the correction, $C_2' = 6.00\ \mathrm{\mu F}$.
⚠ step 3
In a chain the charge does not divide at all: every capacitor carries the same charge as the equivalent one. So $Q_1 = Q_2 = Q_{\rm eq}$, and with the corrected first step that is $Q = (1.50)(25.0) = 37.5\ \mathrm{\mu C}$ on both. Nothing is ever multiplied by a fraction of a capacitance sum in a series problem.
Dividing a total in proportion to the parts is a habit that works in a side by side group, where the charge really does split as $C_i/\sum C$. Carrying that habit across to a chain is the standard slip, and the fraction chosen even looks reasonable. It also produces a smaller number than the equivalent charge, which feels right because sharing normally makes things smaller.
right
Go back to the island argument: the metal between the two capacitors is isolated and started neutral, so whatever is drawn off one side of it must appear on the other. That forces one single charge onto the whole chain. Correctly: $C_{\rm eq} = 1.50\ \mathrm{\mu F}$, $Q = 37.5\ \mathrm{\mu C}$ on both, $V_1 = 18.75\ \mathrm{V}$ and $V_2 = 6.25\ \mathrm{V}$, which add to the applied $25.0\ \mathrm{V}$ as they must.
4 · the bare problem
§06.6 — a slab into one member of a chain●●●●○
No scaffolding this time. Two capacitors, $C_1 = 5.00\ \mathrm{\mu F}$ and $C_2 = 20.0\ \mathrm{\mu F}$, are joined end to end across a source that holds $60.0\ \mathrm{V}$ and stays connected. A slab with $K = 4.00$ is then slid in so that it completely fills $C_1$.
Given
$C_1 = 5.00\ \mathrm{\mu F}$ and $C_2 = 20.0\ \mathrm{\mu F}$, joined end to end
the source holds $60.0\ \mathrm{V}$ across the pair and stays connected
a slab with $K = 4.00$ fills $C_1$ completely
Find
(a) Find the equivalent capacitance before and after the slab is inserted.
(b) Find the charge on the chain after insertion.
(c) Find the voltage across each capacitor after insertion.
Hint 1/4
Three things are going on at once: a dielectric, a chain, and a fixed voltage. Deal with them in that order, and notice that only one capacitance changes.
Hint 2/4
$C_1' = KC_1$; then $1/C_{\rm eq} = 1/C_1' + 1/C_2$; then $Q = C_{\rm eq}V$ with the same $Q$ on both; then $V_i = Q/C_i$ for each.
Hint 3/4
Here $K = 4.00$ acts only on the $5.00\ \mathrm{\mu F}$, the other capacitor stays at $20.0\ \mathrm{\mu F}$, and the applied voltage is fixed at $60.0\ \mathrm{V}$.
Hint 4/4
The equivalent goes from $4.00$ to $10.0\ \mathrm{\mu F}$, the charge becomes $600\ \mathrm{\mu C}$, and the two voltages come out equal at $30.0\ \mathrm{V}$ each.
connected means V is the fixed quantity, so this is the line that is allowed; the charge rose from 240 to 600 microcoulombs and the source supplied the difference
Two independent checks. The voltages must add to the applied one: $30.0 + 30.0 = 60.0\ \mathrm{V}$. And before the slab went in the same method gives $Q = (4.00)(60.0) = 240\ \mathrm{\mu C}$ with $V_1 = 48.0\ \mathrm{V}$ and $V_2 = 12.0\ \mathrm{V}$, which also add to $60.0\ \mathrm{V}$; the slab moved $18.0\ \mathrm{V}$ of that from the first capacitor to the second, closing the $36.0\ \mathrm{V}$ gap between them completely, which is the qualitative effect one expects from making the first capacitor four times larger.
A dielectric in one member of a chain does two things at once: it raises the equivalent capacitance, and it shifts voltage off the capacitor it fills and onto the others. The second effect is the one that matters in practice, because it is what decides which capacitor in a chain breaks down first.
Full exam-style question
Full marks on an air gap capacitor that is then filled with plasticexam format
A parallel plate capacitor has plates of area $1.20\times10^{-2}\ \mathrm{m^2}$ separated by $1.50\ \mathrm{mm}$ of air. It is connected to a $300\ \mathrm{V}$ source, then disconnected. A slab of dielectric with $K = 2.60$ is then slid in until it completely fills the gap. (a) Find the capacitance and the charge before the slab is inserted, and check that the air does not break down. (b) Find the field and the stored energy before insertion. (c) Find the capacitance, potential difference, field and stored energy after insertion. (d) Find the bound charge that appears on each face of the slab. (e) Account for the energy that has gone missing.
Given
$A = 1.20\times10^{-2}\ \mathrm{m^2}$, $d = 1.50\times10^{-3}\ \mathrm{m}$, air in the gap
charged to $300\ \mathrm{V}$ and then disconnected
slab with $K = 2.60$ fills the gap; air breaks down at about $3\times10^{6}\ \mathrm{V/m}$
Find
capacitance, charge, field, potential difference and energy before and after, the bound charge, and the energy balance
Solution(a) Capacitance and charge, and the safety check
the bracket is 0.6154, so about sixty per cent of the plate charge is neutralised and the remaining forty per cent makes the reduced field
(e) The energy balance
$$\Delta U = 3.19 - 1.23 = 1.96\ \mathrm{\mu J}\ \text{has left the capacitor}$$
energy does not disappear, so this number has to be accounted for and not merely reported
$$W_{\rm on\ slab} = 1.96\ \mathrm{\mu J}$$
the field at the mouth of the gap is not uniform, and it pulls on the induced charge of the slab, so the field does work dragging the slab in; that work is where the missing energy went
Answer $$\boxed{\;C_0 = 70.8\ \mathrm{pF},\ Q = 21.2\ \mathrm{nC},\ U_0 = 3.19\ \mathrm{\mu J};\quad C = 184\ \mathrm{pF},\ V = 115\ \mathrm{V},\ E = 7.69\times10^{4}\ \mathrm{V/m},\ U = 1.23\ \mathrm{\mu J};\quad Q_{\rm ind} = 13.1\ \mathrm{nC}\;}$$
Check
Independent check on the energy before insertion by the field route, which shares no line with the capacitor route: $u_0 = \tfrac12\varepsilon_0E_0^{2} = 0.177\ \mathrm{J/m^{3}}$, and the gap volume is $Ad = 1.80\times10^{-5}\ \mathrm{m^{3}}$, giving $U_0 = 3.19\times10^{-6}\ \mathrm{J}$. Second check, on the whole after-insertion block: every quantity except $C$ and $Q$ should have moved by exactly $2.60$, and $300/115.4 = 2.60$, $2.00\times10^{5}/7.69\times10^{4} = 2.60$, $3.186/1.225 = 2.60$. Three independent ratios, one number.
Five parts, eleven lines, and one decision in the first line of part (c) that determines whether parts (c), (d) and (e) score anything at all.
Notice how much of the work part (e) does for one mark. An examiner asking where the energy went is checking whether you know that the capacitor is a physical object being pulled at, not a symbol in a formula. Any answer that stops at the numbers has answered a smaller question than the one asked.
Practice
A · concept 4 questions
1§06.1 — what a charged capacitor is carrying●●○○○
One mark, and it separates reading the word capacitor from understanding it. A capacitor is charged until one plate carries $+15.0\ \mathrm{nC}$.
Given
one plate carries $+15.0\ \mathrm{nC}$
the capacitor was neutral before it was charged
nothing has been added to or removed from the capacitor as a whole
Find
(a) True or false: the capacitor now carries a net charge of $15.0\ \mathrm{nC}$. Give your reason in one sentence.
Hint 1/4
Ask what happened to the charge that left the plate which is now positive.
Hint 2/4
Charge is conserved, and charging a capacitor moves charge from one plate to the other rather than adding any.
Hint 3/4
Here $+15.0\ \mathrm{nC}$ is on one plate, so $-15.0\ \mathrm{nC}$ must be on the other.
Hint 4/4
False: the net charge is still zero, and the $15.0\ \mathrm{nC}$ is the amount that has been separated.
Show solutionConservation, applied to the whole object
Independent test: bring the whole charged capacitor near a suspended charged ball. A net charge of 15 nC would deflect it; a neutral pair with its field trapped in the gap does essentially nothing, which is what is observed.
2§06.5 — pulling the plates apart on an isolated capacitor●●●○○
A capacitor is charged, then disconnected so that no charge can enter or leave it, and then its plates are slowly pulled twice as far apart. Four quantities are listed below and exactly one of them is the same at the end as it was at the beginning.
Given
the capacitor is disconnected before the plates are moved
the separation is doubled and the plate area is unchanged
nothing is put into the gap
Find
(a) Which quantity is unchanged?
Hint 1/4
Take each quantity in turn and write down the expression it is built from, then ask whether any symbol in that expression moved.
Hint 2/4
$C = \varepsilon_0A/d$, $Q$ is fixed by the disconnection, $V = Q/C$, $E = \sigma/\varepsilon_0$ and $U = Q^{2}/2C$.
Hint 3/4
Here $A$ and $Q$ are fixed and $d$ is doubled, so look for the expression that mentions neither $d$ nor $C$.
Hint 4/4
The field is unchanged; the capacitance halves, and the voltage and the energy both double.
Show solutionWrite each one in terms of what is fixed
$$C = \frac{\varepsilon_0 A}{d} \to \tfrac12 C$$
d is in the denominator and it doubled
$$V = \frac{Q}{C} \to 2V$$
the charge is fixed and the capacitance halved
$$E = \frac{Q}{\varepsilon_0 A} \to E$$
neither symbol on the right hand side changed, so this is the answer
$$U = \frac{Q^{2}}{2C} \to 2U$$
the charge is fixed and the capacitance halved, so the energy doubles
Answer $$\boxed{\,\text{the field between the plates is unchanged}\,}$$
Check
Independent check by the other definition of the field: $E = V/d$, and both the voltage and the separation doubled, so their ratio is untouched. Two different expressions for the same quantity, both giving no change.
3§06.4 — how small a chain can get●●○○○
A quick one that is worth turning into a reflex, because it catches wrong answers on much longer questions. Two capacitors of different sizes are joined end to end.
Given
two capacitors $C_1$ and $C_2$, joined end to end
the two values are different from each other
no dielectric is inserted and nothing is moved
Find
(a) True or false: the equivalent capacitance is always smaller than either of the two. Justify it in one sentence.
Hint 1/4
Look at the shape of the formula rather than at any particular numbers, and ask what adding a positive quantity to $1/C_1$ must do.
Hint 2/4
$1/C_{\rm eq} = 1/C_1 + 1/C_2$, and both terms on the right are positive.
Hint 3/4
Adding the positive term $1/C_2$ makes $1/C_{\rm eq}$ larger than $1/C_1$ on its own.
Hint 4/4
True: a larger reciprocal means a smaller capacitance, so the chain beats neither member.
Show solutionArgue on the reciprocal, where the inequality is obvious
Independent numerical spot check at a lopsided pair: $C_1 = 1.00\ \mathrm{\mu F}$ with $C_2 = 1000\ \mathrm{\mu F}$ gives $C_{\rm eq} = 0.999\ \mathrm{\mu F}$, just under the smaller one, which is the limiting behaviour the argument predicts.
4§06.6 — a slab inserted with the source still attached●●●○○
A capacitor remains connected to a source that holds a fixed potential difference across it while a dielectric slab of constant $K$ is slid in until it fills the gap completely.
Given
the source stays connected throughout, so the potential difference is fixed
the slab has dielectric constant $K$ and fills the gap
the plate area and the separation are not touched
Find
(a) What happens to the energy stored in the capacitor?
Hint 1/4
Decide first which quantity is nailed down by the connection, then choose the energy expression that is built out of that quantity.
Hint 2/4
With $V$ fixed the safe form is $U = \tfrac12 CV^{2}$, and the capacitance becomes $KC_0$.
Hint 3/4
Here $V$ does not move and $C$ is multiplied by $K$, so only one factor in the expression changes.
Hint 4/4
The energy rises by a factor of $K$, and the source supplied the extra.
Show solutionChoose the form built from the fixed quantity
$$U = \tfrac12 CV^{2},\qquad V\ \text{fixed}$$
the charge is free to move here, so any form containing Q would hide a second change
$$C \to KC_0 \Rightarrow U \to KU_0$$
one factor changed by K and the rest of the expression stood still
Answer $$\boxed{\,U \to K\,U_0\,}$$
Check
Independent check by counting what the source did: it pushed an extra charge $\Delta Q = (K-1)C_0V$ across a fixed $V$, doing work $(K-1)C_0V^{2}$. Half of that, $\tfrac12(K-1)C_0V^{2}$, is exactly the rise in stored energy from $\tfrac12C_0V^{2}$ to $\tfrac12KC_0V^{2}$; the other half went into pulling the slab in.
B · computation 8 questions
1§06.1 — one capacitor, three questions●●○○○
A warm up that uses the definition in all three directions. A capacitor is found to carry $4.50\ \mathrm{\mu C}$ when the potential difference across it is $9.00\ \mathrm{V}$.
Given
$Q = 4.50\ \mathrm{\mu C}$ at $V = 9.00\ \mathrm{V}$
the same capacitor is used throughout
Find
(a) Find its capacitance.
(b) Find the charge it carries at $20.0\ \mathrm{V}$.
(c) Find the voltage at which it carries $1.00\ \mathrm{\mu C}$.
Hint 1/4
All three parts are the same relation read in different directions; identify which of the three symbols is unknown in each part.
Hint 2/4
$C = Q/V$, $Q = CV$ and $V = Q/C$ are one statement written three ways.
Hint 3/4
The measured pair is $4.50\ \mathrm{\mu C}$ with $9.00\ \mathrm{V}$; the two later parts supply $20.0\ \mathrm{V}$ and $1.00\ \mathrm{\mu C}$.
Hint 4/4
The capacitance is $0.500\ \mathrm{\mu F}$, the charge is $10.0\ \mathrm{\mu C}$, and the voltage is $2.00\ \mathrm{V}$.
Independent check without using $C$ at all: charge and voltage must stay in proportion. $20.0/9.00 = 2.22$ and $10.0/4.50 = 2.22$; $1.00/4.50 = 0.222$ and $2.00/9.00 = 0.222$. Both pairs of ratios match.
2§06.2 — a flat pair from its dimensions●●●○○
Straight application of the geometry, with a design question at the end. Two flat plates of area $25.0\ \mathrm{cm^2}$ face each other across $0.400\ \mathrm{mm}$ of air.
Given
$A = 25.0\ \mathrm{cm^2}$ for each plate, fully overlapping
(b) Find the potential difference when one plate carries $0.500\ \mathrm{nC}$.
(c) Find the plate area that would double the capacitance at the same separation.
Hint 1/4
Before anything else, convert the area and the separation into SI units, because both carry powers of ten that are invisible once they are inside the formula.
Hint 2/4
$C = \varepsilon_0A/d$, and then $V = Q/C$ from the definition.
Hint 3/4
Here $A = 2.50\times10^{-3}\ \mathrm{m^2}$, $d = 4.00\times10^{-4}\ \mathrm{m}$ and $Q = 5.00\times10^{-10}\ \mathrm{C}$.
Hint 4/4
The capacitance is $55.3\ \mathrm{pF}$, the voltage is $9.04\ \mathrm{V}$, and doubling the capacitance needs $50.0\ \mathrm{cm^2}$.
Independent check on the voltage by the field route: $\sigma = Q/A = 2.00\times10^{-7}\ \mathrm{C/m^2}$, so $E = \sigma/\varepsilon_0 = 2.26\times10^{4}\ \mathrm{V/m}$ and $V = Ed = 9.04\ \mathrm{V}$. Same number, no capacitance used.
3§06.3 — twelve metres of coaxial cable●●●○○
A length of coaxial cable has an inner conductor of radius $1.00\ \mathrm{mm}$ inside a sheath of inner radius $4.00\ \mathrm{mm}$, with air between them. The run is $12.0\ \mathrm{m}$ long.
Given
$R_a = 1.00\ \mathrm{mm}$, $R_b = 4.00\ \mathrm{mm}$, air between
$L = 12.0\ \mathrm{m}$
dry air breaks down at about $3\times10^{6}\ \mathrm{V/m}$
Find
(a) Find the capacitance of the whole run.
(b) Find the charge on the inner conductor at $500\ \mathrm{V}$.
(c) Find the field at the surface of the inner conductor at that voltage, and say whether the air is safe.
Hint 1/4
Two of the three parts are about the cable as a single capacitor, and the last one needs the field at a particular radius, which is where the field is largest.
Hint 2/4
$C = 2\pi\varepsilon_0L/\ln(R_b/R_a)$, then $Q = CV$, and the field at radius $r$ is $E = \lambda/(2\pi\varepsilon_0 r)$ with $\lambda = Q/L$.
Hint 3/4
Here the radius ratio is $4.00$, the length is $12.0\ \mathrm{m}$ and the voltage is $500\ \mathrm{V}$; the field is wanted at $r = 1.00\times10^{-3}\ \mathrm{m}$.
Hint 4/4
The capacitance is $481\ \mathrm{pF}$, the charge is $241\ \mathrm{nC}$, and the field is $3.61\times10^{5}\ \mathrm{V/m}$, about an eighth of the breakdown value.
Show solutionCapacitance per metre, then the whole run
Independent check on the field without going through the charge: for a cable the voltage is $E(R_a)R_a\ln(R_b/R_a)$, so $E(R_a) = V/[R_a\ln(R_b/R_a)] = 3.61\times10^{5}\ \mathrm{V/m}$. Same number from the voltage rather than from the charge.
4§06.4 — three capacitors, one applied voltage●●●○○
A standard network. $C_1 = 8.00\ \mathrm{\mu F}$ is joined end to end with a side by side pair made of $C_2 = 4.00\ \mathrm{\mu F}$ and $C_3 = 12.0\ \mathrm{\mu F}$, and $36.0\ \mathrm{V}$ is applied across the whole arrangement.
Given
$C_1 = 8.00\ \mathrm{\mu F}$ end to end with the pair
$C_2 = 4.00\ \mathrm{\mu F}$ side by side with $C_3 = 12.0\ \mathrm{\mu F}$
$36.0\ \mathrm{V}$ across the whole arrangement
Find
(a) Find the equivalent capacitance.
(b) Find the charge and the voltage on $C_1$.
(c) Find the charge on $C_2$ and on $C_3$.
Hint 1/4
Collapse the group that already shares a known quantity, then collapse what is left, then come back out carrying whichever quantity each grouping makes common.
Hint 2/4
Side by side the capacitances add; end to end the reciprocals add and the same charge sits on every member.
Hint 3/4
Here the pair is $4.00 + 12.0$, and that result is then in a chain with $8.00$, across $36.0\ \mathrm{V}$.
Hint 4/4
The equivalent is $5.33\ \mathrm{\mu F}$, the chain carries $192\ \mathrm{\mu C}$, and the pair splits it as $48.0$ and $144\ \mathrm{\mu C}$.
Show solutionCollapse the pair
$$C_{23} = 4.00+12.0 = 16.0\ \mathrm{\mu F}$$
the pair is the only group whose shared quantity is known from the start
Two independent checks that were not used above. The voltages along the chain must add to the applied one: $24.0 + 12.0 = 36.0\ \mathrm{V}$. The charges in the pair must add to the chain charge: $48.0 + 144 = 192\ \mathrm{\mu C}$.
5§06.5 — energy in a large capacitor, and a design target●●●○○
A $25.0\ \mathrm{\mu F}$ capacitor is charged to $120\ \mathrm{V}$.
the square root is why the answer is only 1.67 times the original voltage even though the energy target is 2.78 times the original energy
Answer $$\boxed{\,Q = 3.00\times10^{-3}\ \mathrm{C},\qquad U = 0.180\ \mathrm{J},\qquad V = 200\ \mathrm{V}\,}$$
Check
Independent check on part (b) using the charge form, which uses a different pair of quantities: $U = \tfrac12QV = \tfrac12(3.00\times10^{-3})(120) = 0.180\ \mathrm{J}$. And a ratio check on part (c): the energy went up by $0.500/0.180 = 2.78$, so the voltage should go up by $\sqrt{2.78} = 1.67$, and $200/120 = 1.67$.
6§06.6 — a slab into a disconnected capacitor●●●○○
A $90.0\ \mathrm{pF}$ capacitor is charged to $40.0\ \mathrm{V}$ and then disconnected. A slab with $K = 3.00$ is then slid in until it fills the gap.
Given
$C_0 = 90.0\ \mathrm{pF}$, charged to $40.0\ \mathrm{V}$
the capacitor is disconnected before the slab is inserted
the slab has $K = 3.00$ and fills the gap
Find
(a) Find the charge on the plates, before and after insertion.
(b) Find the capacitance and the potential difference after insertion.
(c) Find the stored energy before and after insertion.
(d) Find the bound charge on each face of the slab.
Hint 1/4
Write down in one sentence which quantity the disconnection nails down, and then work out every other quantity from it.
Hint 2/4
$C = KC_0$ always; with $Q$ fixed, $V = Q/C$ and $U = Q^{2}/2C$; and $Q_{\rm ind} = Q(1 - 1/K)$.
Hint 3/4
Here $Q$ was set by $C_0 = 90.0\ \mathrm{pF}$ at $40.0\ \mathrm{V}$, and $K = 3.00$.
Hint 4/4
The charge stays at $3.60\ \mathrm{nC}$, the capacitance becomes $270\ \mathrm{pF}$, the voltage falls to $13.3\ \mathrm{V}$, the energy falls from $72.0$ to $24.0\ \mathrm{nJ}$, and $2.40\ \mathrm{nC}$ appears on each face.
Show solutionThe fixed quantity, computed while both C and V are known
Independent check on the final energy by the voltage form applied to the final state, which is legitimate because both $C$ and $V$ are now known: $\tfrac12CV^{2} = \tfrac12(2.70\times10^{-10})(13.33)^{2} = 2.40\times10^{-8}\ \mathrm{J}$. The two agree, which confirms that the fall by a factor of three was not an arithmetic slip.
7§06.7 — a component designed around its dielectric●●●○○
A capacitor is built from plates of area $30.0\ \mathrm{cm^2}$ separated by $0.200\ \mathrm{mm}$ of a plastic with $K = 5.00$ and a dielectric strength of $20\times10^{6}\ \mathrm{V/m}$.
Here $A/d = 15.0\ \mathrm{m}$, $K = 5.00$, $E_{\max} = 20\times10^{6}\ \mathrm{V/m}$ and $d = 2.00\times10^{-4}\ \mathrm{m}$.
Hint 4/4
The capacitance is $664\ \mathrm{pF}$, the limit is $4.00\ \mathrm{kV}$, the energy is $5.31\ \mathrm{mJ}$ and the bound charge is $2.12\ \mathrm{\mu C}$.
Independent check on the energy by the charge form: $U = \tfrac12QV = 5.31\times10^{-3}\ \mathrm{J}$. Also a comparison: the same plates with an air gap manage only $133\ \mathrm{pF}$ at $600\ \mathrm{V}$, that is $23.9\ \mathrm{\mu J}$, so the plastic wins by $5.00\times(20/3)^{2} = 222$, and the two energies do stand in that ratio.
8§06.3 — a lone metal sphere as a capacitor●●●○○
An isolated conducting sphere of radius $25.0\ \mathrm{cm}$ hangs in dry air, far from everything else.
Given
$R = 25.0\ \mathrm{cm} = 0.250\ \mathrm{m}$
$k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$
dry air breaks down at about $3\times10^{6}\ \mathrm{V/m}$
Find
(a) Find its capacitance.
(b) Find the charge needed to raise it to $15.0\ \mathrm{kV}$.
(c) Find the highest potential it can be raised to before the surrounding air breaks down.
Hint 1/4
A single sphere counts as a capacitor with its partner at infinity, and the field at its surface is the quantity the last part is really about.
Hint 2/4
$C = R/k$, then $Q = CV$; and for a sphere the surface field is $E = kQ/R^{2} = V/R$.
Hint 3/4
Here $R = 0.250\ \mathrm{m}$, the target potential is $1.50\times10^{4}\ \mathrm{V}$, and the limiting field is $3\times10^{6}\ \mathrm{V/m}$.
Hint 4/4
The capacitance is $27.8\ \mathrm{pF}$, the charge is $417\ \mathrm{nC}$, and the limit is about $750\ \mathrm{kV}$.
Independent check on part (b) through the potential of a point charge, which never mentions capacitance: $V = kQ/R = 1.50\times10^{4}\ \mathrm{V}$, as required. Check on part (c): at $15.0\ \mathrm{kV}$ the surface field is $6.00\times10^{4}\ \mathrm{V/m}$, a fiftieth of breakdown, and the answer to (c) is indeed fifty times $15.0\ \mathrm{kV}$.
C · exam level 5 questions
1§06.6 — a slab inserted with the supply left connected●●●●○
Exam length, five parts, and the last one is where the marks separate. A parallel plate capacitor has plates of area $4.00\times10^{-2}\ \mathrm{m^2}$ separated by $2.00\ \mathrm{mm}$ of air, and is connected to a source holding $500\ \mathrm{V}$. With the source still connected, a slab of $K = 4.50$ is slid in until it fills the gap.
Given
$A = 4.00\times10^{-2}\ \mathrm{m^2}$, $d = 2.00\times10^{-3}\ \mathrm{m}$, air at first
the source holds $500\ \mathrm{V}$ and stays connected throughout
the slab has $K = 4.50$ and fills the gap completely
Find
(a) Find the capacitance, charge and stored energy before insertion.
(b) Find the capacitance, charge and stored energy after insertion.
(c) Find the field in the gap before and after.
(d) Find the bound charge on each face of the slab.
(e) Find the work done by the source during the insertion, and account for all of it.
Hint 1/4
Start by writing one sentence saying which quantity the connection nails down, because parts (b) to (e) all depend on it.
Hint 2/4
$C = K C_0$; with $V$ fixed use $Q = CV$ and $U = \tfrac12CV^{2}$; the field is $E = V/d$; $Q_{\rm ind} = Q(1 - 1/K)$; and the source does work $V\,\Delta Q$.
Hint 3/4
Here $V = 500\ \mathrm{V}$ is fixed, $d = 2.00\times10^{-3}\ \mathrm{m}$ is fixed, $A/d = 20.0\ \mathrm{m}$ and $K = 4.50$.
Hint 4/4
The capacitance goes from $177$ to $797\ \mathrm{pF}$, the charge from $88.5$ to $398\ \mathrm{nC}$, the energy from $22.1$ to $99.6\ \mathrm{\mu J}$, the field stays at $2.50\times10^{5}\ \mathrm{V/m}$, the bound charge is $310\ \mathrm{nC}$, and the source does $155\ \mathrm{\mu J}$ of work, exactly half of which ends up stored.
the remainder, and it went into dragging the slab into the gap against nothing at all, which is why in practice it ends up as heat and sound when the slab arrives
Independent check on the field claim through the charges rather than through $V/d$: the net surface density $(Q - Q_{\rm ind})/A$ comes to $2.21\times10^{-6}\ \mathrm{C/m^2}$, and dividing by $\varepsilon_0$ gives $2.50\times10^{5}\ \mathrm{V/m}$, the field it was before. Second check, algebraic: $W_{\rm source} = (K-1)C_0V^{2}$ while $\Delta U = \tfrac12(K-1)C_0V^{2}$, so the ratio is exactly two for any $K$.
Two general lessons hide in part (e). A source charging a capacitor always delivers twice what the capacitor keeps, and a dielectric being drawn into a gap always takes the other half.
2§06.4 — a network, and then a slab into one branch of it●●●●○
Exam length. $C_1 = 6.00\ \mathrm{\mu F}$ and $C_2 = 3.00\ \mathrm{\mu F}$ are joined side by side, and that pair is joined end to end with $C_3 = 9.00\ \mathrm{\mu F}$. A source holding $40.0\ \mathrm{V}$ is connected across the whole arrangement and stays connected.
Given
$C_1 = 6.00\ \mathrm{\mu F}$ side by side with $C_2 = 3.00\ \mathrm{\mu F}$
that pair is end to end with $C_3 = 9.00\ \mathrm{\mu F}$
$40.0\ \mathrm{V}$ across the whole arrangement, source left connected
Find
(a) Find the equivalent capacitance, and the charge and voltage on each of the three capacitors.
(b) A slab with $K = 2.00$ is now slid in so that it fills $C_3$. Find the new equivalent capacitance and the new charge and voltage on each capacitor.
(c) Find the energy stored in the whole arrangement before and after the slab is inserted.
Hint 1/4
Do the network exactly as before, and treat the slab as a change to one capacitance and nothing else. The voltage across the whole arrangement is pinned by the source in both parts.
Hint 2/4
Side by side the capacitances add, end to end the reciprocals add, the chain shares its charge, the pair shares its voltage, and $C_3' = KC_3$.
Hint 3/4
Here the pair is $6.00 + 3.00$, the chain is that against $9.00$ then against $18.0$, and the applied voltage is $40.0\ \mathrm{V}$ in both parts.
Hint 4/4
The equivalent goes from $4.50$ to $6.00\ \mathrm{\mu F}$, the chain charge from $180$ to $240\ \mathrm{\mu C}$, and the stored energy from $3.60$ to $4.80\ \mathrm{mJ}$.
Three independent checks, none of which was used in the working. Voltages along the chain must add to $40.0\ \mathrm{V}$: before, $20.0 + 20.0$; after, $13.33 + 26.67$. Charges in the pair must add to the chain charge: before, $120 + 60.0 = 180$; after, $160 + 80.0 = 240\ \mathrm{\mu C}$. And summing the individual energies after insertion gives $\tfrac12(18.0)(13.33)^{2} + \tfrac12(6.00)(26.67)^{2} + \tfrac12(3.00)(26.67)^{2}$ microjoules, that is $1600 + 2133 + 1067 = 4800\ \mathrm{\mu J}$, matching part (c).
3§06.6 — a slab that fills only part of the gap●●●●○
Exam length, and the geometry is the part that has to be got right first. A parallel plate capacitor has plates of area $1.00\times10^{-2}\ \mathrm{m^2}$ separated by $1.00\ \mathrm{mm}$. A slab of dielectric with $K = 4.00$ and thickness $0.600\ \mathrm{mm}$ is laid flat against one plate, leaving $0.400\ \mathrm{mm}$ of air. The arrangement is then held at $200\ \mathrm{V}$.
Given
$A = 1.00\times10^{-2}\ \mathrm{m^2}$, total gap $d = 1.00\times10^{-3}\ \mathrm{m}$
the potential difference across the pair is $200\ \mathrm{V}$
Find
(a) Find the capacitance, and compare it with the empty gap value and with the fully filled value.
(b) Find the charge on the plates.
(c) Find the field in the air layer and in the slab.
(d) Find the potential difference across each layer, and check that they add up.
(e) Say whether the air layer is in danger of breaking down.
Hint 1/4
The two layers are stacked across the gap, so the same charge crosses both of them; that single sentence turns the problem into one you have already solved.
Hint 2/4
Treat the two layers as two capacitors end to end, each with its own thickness and its own $K$, then combine with the reciprocal rule; afterwards $E = \sigma/(K\varepsilon_0)$ in each layer.
Hint 3/4
Here the air layer is $4.00\times10^{-4}\ \mathrm{m}$ thick with $K = 1$ and the slab is $6.00\times10^{-4}\ \mathrm{m}$ thick with $K = 4.00$, over an area of $1.00\times10^{-2}\ \mathrm{m^2}$ at $200\ \mathrm{V}$.
Hint 4/4
The capacitance is $161\ \mathrm{pF}$, the charge is $32.2\ \mathrm{nC}$, the fields are $3.64\times10^{5}$ and $9.09\times10^{4}\ \mathrm{V/m}$, and the voltages are $145$ and $54.5\ \mathrm{V}$.
Independent check on the capacitance from the voltages rather than from the layer formula: $C = Q/V = 1.61\times10^{-10}\ \mathrm{F}$, the definition applied to the finished answer. Second check, a limit: with no air layer the chain would be the slab alone at full thickness, $354\ \mathrm{pF}$, and with no slab it would be $88.5\ \mathrm{pF}$; the answer sits between them, near the small end, as a chain always is.
Any thin air gap left inside a capacitor is dangerous out of all proportion to its thickness, because the field crowds into it. Real components are impregnated to drive the air out for exactly this reason.
4§06.5 — pulling plates apart with the source left connected●●●●●
Exam length, and it is deliberately the mirror image of the worked ladder problem, in which the capacitor was disconnected first. A parallel plate capacitor has plates of area $6.00\times10^{-3}\ \mathrm{m^2}$ separated by $0.500\ \mathrm{mm}$ of air and is held at $100\ \mathrm{V}$ by a source that stays connected. The plates are then slowly pulled apart to $1.00\ \mathrm{mm}$.
Given
$A = 6.00\times10^{-3}\ \mathrm{m^2}$, air in the gap
$d$ goes from $5.00\times10^{-4}\ \mathrm{m}$ to $1.00\times10^{-3}\ \mathrm{m}$
the source holds $100\ \mathrm{V}$ throughout and is never disconnected
Find
(a) Find the capacitance, charge and stored energy before and after.
(b) Say how much charge went back into the source, and how much energy went with it.
(c) Find the work the external agent had to do on the plates, from energy conservation.
(d) Confirm that work by integrating the force between the plates.
Hint 1/4
Everything turns on the fact that the voltage, not the charge, is the fixed quantity here, so charge leaves the plates as they separate.
Hint 2/4
$C = \varepsilon_0A/d$, $Q = CV$, $U = \tfrac12CV^{2}$; the source does work $V\Delta Q$, which is negative if charge returns to it; and $W_{\rm ext} = \Delta U - W_{\rm source}$.
Hint 3/4
Here $A/d$ goes from $12.0$ to $6.00\ \mathrm{m}$ at a fixed $100\ \mathrm{V}$.
Hint 4/4
The capacitance and the charge both halve, the energy halves from $0.531$ to $0.266\ \mathrm{\mu J}$, $5.31\ \mathrm{nC}$ and $0.531\ \mathrm{\mu J}$ go back to the source, and the agent does $0.266\ \mathrm{\mu J}$ of work.
positive, as it must be: the plates attract each other, so separating them is work done against that attraction, and here that work goes into the source rather than into the capacitor
(d) The same number from the force
$$F = \tfrac12 QE = \tfrac12(C V)\left(\frac{V}{d}\right) = \frac{\varepsilon_0 AV^{2}}{2d^{2}}$$
each plate sits in the field of the other only, which is half the total field, and that is where the leading half comes from
The agreement between parts (c) and (d) is the check, and it is a real one: (c) uses only energy bookkeeping and the source, (d) uses only the force between the plates and an integral, with no mention of the source. A cheaper second check: in the disconnected version of this move the field stayed constant and the work was a simple product; here the field falls as the plates separate, so the work is smaller than a constant force estimate.
Same hardware, same movement, and the stored energy goes one way when the source is attached and the other way when it is not. The agent does positive work in both cases; what changes is where that work ends up.
5§06.6 — where the source's energy goes when a slab goes in●●●●○
A parallel plate capacitor of capacitance $C_0$ is held at a fixed potential difference $V$ by a source that stays connected. A slab of dielectric constant $K$ is then slid in until it fills the gap. The source has to push extra charge onto the plates while this happens, and it does work $W$ doing so. Exam question: how much of that $W$ ends up as extra energy stored in the capacitor?
Given
the source holds $V$ fixed and stays connected throughout
the capacitance goes from $C_0$ to $KC_0$
$W$ is the work the source does moving the extra charge
Find
(a) What fraction of $W$ shows up as extra stored energy?
Hint 1/4
Two different energies are moving at once here: the one the source hands over, and the one the capacitor keeps. Write an expression for each before you try to compare them.
Hint 2/4
The source does $W = V\,\Delta Q$ with no factor of a half, because every coulomb crosses the full fixed voltage; the capacitor's own store is $U = \tfrac12 CV^{2}$.
Hint 3/4
Here $V$ is fixed, the capacitance goes from $C_0$ to $KC_0$, so $\Delta Q = (K-1)C_0V$ and the store goes from $\tfrac12 C_0V^{2}$ to $\tfrac12 KC_0V^{2}$.
Hint 4/4
The ratio is exactly one half, for every $K$; the missing half is the work the field does dragging the slab into the gap.
the charge is what moves and the voltage is what is held, so this is the only form allowed here; there is no factor of a half because every coulomb crosses the full fixed voltage
What the capacitor keeps
$$\Delta U = \tfrac12(KC_0)V^{2} - \tfrac12 C_0V^{2} = \tfrac12(K-1)C_0V^{2}$$
the voltage form of the energy is the safe one here for the same reason: V is the pinned symbol, so nothing on the right hand side is guesswork
the whole block cancels, which is why no property of the slab and no property of the plates survives into the answer
Answer $$\boxed{\;\Delta U = \tfrac12 W\ \text{for every}\ K\;}$$
Check
Independent check with numbers instead of symbols, using the first question in this set: there $W = 155\ \mathrm{\mu J}$ and the store went from $22.1$ to $99.6\ \mathrm{\mu J}$, a rise of $77.4\ \mathrm{\mu J}$, and $77.4/155 = 0.499$. That run had $K = 4.50$, so a second value of $K$ would be needed to see the cancellation, and $K = 2.00$ gives $W = C_0V^{2}$ against $\Delta U = \tfrac12 C_0V^{2}$: the same half.
This is the same one half that appears when a source charges an empty capacitor from scratch, and for the same reason: the capacitor's store carries a factor of a half that the source's bookkeeping does not. Whenever a question says the supply stayed connected, expect half the supplied energy to have gone somewhere other than the store, and be ready to say where.
D · interleaved 4 questions
1§06.2 — two charged sheets, and what they add up to●●●○○
This set is deliberately mixed, so decide for yourself which tool each question wants before reaching for one. Two large parallel conducting plates, each of area $5.00\times10^{-2}\ \mathrm{m^2}$, face each other across $0.300\ \mathrm{mm}$ of air. The facing surfaces carry uniform charge densities of $+4.00\times10^{-6}$ and $-4.00\times10^{-6}\ \mathrm{C/m^2}$.
Given
$A = 5.00\times10^{-2}\ \mathrm{m^2}$ for each plate
$d = 3.00\times10^{-4}\ \mathrm{m}$, air in the gap
$\sigma = \pm4.00\times10^{-6}\ \mathrm{C/m^2}$ on the facing surfaces
Find
(a) Find the field in the gap.
(b) Find the potential difference between the plates.
(c) Find the charge on one plate, and hence the capacitance of the arrangement.
(d) Check your capacitance against the geometry formula.
Hint 1/4
Nothing here is labelled as a capacitance problem, and the first two parts are not; identify what each part is asking for before choosing a formula.
Hint 2/4
A pillbox at a conductor surface gives $E = \sigma/\varepsilon_0$; a uniform field gives $V = Ed$; the definition gives $C = Q/V$; and the geometry gives $C = \varepsilon_0A/d$.
Hint 3/4
Here $\sigma = 4.00\times10^{-6}\ \mathrm{C/m^2}$, $d = 3.00\times10^{-4}\ \mathrm{m}$ and $A = 5.00\times10^{-2}\ \mathrm{m^2}$.
Hint 4/4
The field is $4.52\times10^{5}\ \mathrm{V/m}$, the voltage is $136\ \mathrm{V}$, the charge is $200\ \mathrm{nC}$, and the capacitance is $1.48\ \mathrm{nF}$ by both routes.
the surface density has vanished from this route entirely, which is the point of the check
Answer $$\boxed{\,E = 4.52\times10^{5}\ \mathrm{V/m},\ V = 136\ \mathrm{V},\ Q = 200\ \mathrm{nC},\ C = 1.48\ \mathrm{nF}\,}$$
Check
The agreement between (c) and (d) is itself the independent check, and it is genuine: the first route goes through the charge and the field and the voltage, and the second uses only the plate dimensions. A safety check as well: $4.52\times10^{5}\ \mathrm{V/m}$ is fifteen per cent of the breakdown field of air, so the arrangement is physically possible.
A question can describe a capacitor without ever using the word. Whenever two conductors carry equal and opposite charge and you are asked for a voltage, there is a capacitance in the problem whether or not anybody names it.
2§06.5 — a proton let go inside a charged capacitor●●●●○
Decide which tool this one wants before reaching for a formula. A $40.0\ \mathrm{pF}$ capacitor with its plates $0.800\ \mathrm{mm}$ apart is charged to $250\ \mathrm{V}$ and then disconnected. A proton is released from rest right at the surface of the positive plate.
Given
$C = 40.0\ \mathrm{pF}$, charged to $250\ \mathrm{V}$, then disconnected
proton: charge $1.602\times10^{-19}\ \mathrm{C}$, mass $1.67\times10^{-27}\ \mathrm{kg}$
Find
(a) Find the field between the plates.
(b) Find the speed of the proton when it reaches the other plate.
(c) Find the energy stored in the capacitor.
(d) Find how many protons would have to make that crossing to use up the whole store.
Hint 1/4
Two of these parts are about one particle and two are about the whole capacitor, and the last part connects them by comparing two energies.
Hint 2/4
$E = V/d$; energy conservation gives $qV = \tfrac12mv^{2}$; the store is $U = \tfrac12CV^{2}$; and the count is the store divided by the energy of one crossing.
Hint 3/4
Here $V = 250\ \mathrm{V}$, $d = 8.00\times10^{-4}\ \mathrm{m}$, $C = 4.00\times10^{-11}\ \mathrm{F}$, and one proton crossing gains $qV$.
Hint 4/4
The field is $3.13\times10^{5}\ \mathrm{V/m}$, the speed is $2.19\times10^{5}\ \mathrm{m/s}$, the store is $1.25\ \mathrm{\mu J}$, and it would take $3.12\times10^{10}$ crossings.
thirty billion protons, which is a vanishing amount of matter and a reminder of how large a joule is on the atomic scale
Answer $$\boxed{\,E = 3.13\times10^{5}\ \mathrm{V/m},\ v = 2.19\times10^{5}\ \mathrm{m/s},\ U = 1.25\ \mathrm{\mu J},\ N = 3.12\times10^{10}\,}$$
Check
Independent check on the speed by the mechanics route, which shares nothing with the energy route: the acceleration is $a = qE/m$, which comes to $3.00\times10^{13}\ \mathrm{m/s^{2}}$, and then $v^{2} = 2ad = 4.80\times10^{10}\ \mathrm{m^{2}/s^{2}}$, giving $v = 2.19\times10^{5}\ \mathrm{m/s}$. Same answer from force and kinematics as from energy.
The charge of the capacitor, $10.0\ \mathrm{nC}$, is about $6\times10^{10}$ elementary charges, and the count in part (d) came out at half that. That is not a coincidence: half, because the store is $\tfrac12QV$ and one crossing costs $eV$.
3§06.3 — two spheres joined by a long thin wire●●●●○
Mixed set, so work out what is being asked before choosing a method. A conducting sphere of radius $6.00\ \mathrm{cm}$ carries $12.0\ \mathrm{nC}$ and hangs far from everything. A second, uncharged conducting sphere of radius $3.00\ \mathrm{cm}$ hangs far away from it. The two are then joined by a long thin wire and left to settle.
they are far apart, so each behaves as an isolated sphere even after joining
Find
(a) Find the capacitance of each sphere on its own.
(b) Find the common potential after joining, and the charge that ends up on each sphere.
(c) Find the field at the surface of each sphere afterwards, and say which is larger.
(d) Find the total stored energy before and after joining.
Hint 1/4
Once joined, the two spheres are one conductor, so they must share something; decide what that is before writing anything down.
Hint 2/4
$C = R/k$ for an isolated sphere; joined conductors share a potential, so $V = Q_{\rm tot}/(C_1+C_2)$ and $Q_i = C_iV$; the surface field of a sphere is $E = V/R$; and $U = \tfrac12QV$.
Hint 3/4
Here $R_1 = 0.0600\ \mathrm{m}$, $R_2 = 0.0300\ \mathrm{m}$ and $Q_{\rm tot} = 12.0\ \mathrm{nC}$ throughout.
Hint 4/4
The capacitances are $6.67$ and $3.34\ \mathrm{pF}$, the common potential is $1.20\ \mathrm{kV}$, the charges split as $8.00$ and $4.00\ \mathrm{nC}$, the smaller sphere has twice the surface field, and the energy falls from $10.8$ to $7.19\ \mathrm{\mu J}$.
Independent check on the final energy by adding the two spheres separately instead of treating them as one: $Q_1^{2}/2C_1 + Q_2^{2}/2C_2 = 4.79\times10^{-6} + 2.40\times10^{-6} = 7.19\times10^{-6}\ \mathrm{J}$, matching. Check on the charges: they must add back to $12.0\ \mathrm{nC}$, and $8.00 + 4.00$ does.
Charge does not divide equally between joined conductors; potential does. And the arrangement that results always stores less energy than the one it came from, which is why this process only ever runs one way.
4§06.5 — the energy of a charged sphere, counted twice●●●●●
The last one in the mixed set, and it asks you to get the same number from two places that look unrelated. A conducting sphere of radius $10.0\ \mathrm{cm}$ carries $5.00\ \mathrm{nC}$ and hangs far from everything else.
(a) Find its capacitance and hence the stored energy.
(b) Find the energy density just outside its surface.
(c) By integrating the energy density over all the space outside the sphere, show that the total comes to $kQ^{2}/2R$, and check that this agrees with part (a).
Hint 1/4
Two different ways of saying where the energy is: one attributes it to the charge on the sphere, the other to the field filling the space around it. They must agree.
Hint 2/4
$C = R/k$ and $U = Q^{2}/2C$; the density is $u = \tfrac12\varepsilon_0E^{2}$ with $E = kQ/r^{2}$; and a shell of radius $r$ has volume $4\pi r^{2}\,dr$.
Hint 3/4
Here $R = 0.100\ \mathrm{m}$ and $Q = 5.00\times10^{-9}\ \mathrm{C}$, and the integral runs from $r = R$ to infinity because the field inside the metal is zero.
Hint 4/4
The capacitance is $11.1\ \mathrm{pF}$, the energy is $1.12\ \mathrm{\mu J}$, the surface density is $8.94\times10^{-5}\ \mathrm{J/m^{3}}$, and the integral returns the same $1.12\ \mathrm{\mu J}$.
The agreement of (a) and (c) is the check, and it is as independent as checks get: one route is a capacitance and a charge, the other is an integral of a field over infinite space, and they share only the value of $k$. An algebraic version of the same check: $Q^{2}/2C$ with $C = R/k$ is $kQ^{2}/2R$ identically, for any $R$ and any $Q$.
Energy can be booked to the charges or to the field and the total is the same either way. The field picture is the one that survives into later work, because it keeps working in places where there is no capacitor to point at.
Mistake ledger (25 entries)
⚠ Calling the total charge on a capacitor 2Q
the words store and hold suggest a container that fills up, and both plates plainly have charge on them
wrong$$Q_{\rm total} = (+Q) + (Q) = 2Q$$
right$$Q_{\rm total} = (+Q) + (-Q) = 0,\qquad \text{the } Q \text{ in } C = Q/V \text{ is one plate's magnitude}$$
⚠ Believing the capacitance grows as the capacitor is charged
$C$ sits next to $Q$ in the formula, and everything else in the formula does change when the charge changes
wrong$$C \propto Q$$
right$$C = \frac{Q}{V} = \text{constant, fixed by geometry and by the material in the gap}$$
⚠ Losing six orders of magnitude between microfarads and picofarads
both prefixes are read as small and the difference between them is never visible in the algebra, only in the final number
⚠ Quoting the energy density where the total energy was asked for
lower case u and upper case U are one keystroke apart and both are called energy in conversation
wrong$$U = \tfrac12\varepsilon_0E^{2}$$
right$$U = \tfrac12\varepsilon_0E^{2}\times(\text{volume}),\qquad u = \tfrac12\varepsilon_0E^{2}$$
⚠ Dividing the capacitance by K instead of multiplying
the field really does get divided by K, and the two statements sit one line apart
wrong$$C = \frac{C_0}{K}$$
right$$C = KC_0,\qquad E = \frac{E_0}{K}$$
⚠ Saying the voltage falls by K when the source is still attached
the sentence the voltage falls by K is how the constant is defined, and the condition attached to it (the charge is fixed) travels less well than the sentence does
it is charge and it is on a surface, so it looks like the free charge on a conductor, which can indeed be drained
wrong$$\text{ground the slab} \Rightarrow \sigma_{\rm ind} = 0$$
right$$\sigma_{\rm ind}\ \text{is bound to its own molecules and stays; only free charge can be drained}$$
⚠ Using the vacuum energy density inside the slab
the density formula was met first without a K in it, and the K is easy to leave behind when the field is the quantity in hand
wrong$$u = \tfrac12\varepsilon_0E^{2}\ \text{inside a dielectric}$$
right$$u = \tfrac12 K\varepsilon_0E^{2}\ \text{inside a dielectric of constant } K$$
⚠ Applying the full factor K to a gap the slab does not fill
The two layers are stacked across the gap, so they share their charge and combine as a chain, and a chain is dominated by its smallest member, which is the layer with nothing in it.
wrong$$\text{slab of thickness } d/2:\ C = KC_0$$
right$$C = \frac{2K}{K+1}C_0,\qquad \text{from two capacitors end to end}$$
⚠ Sharing the charge out between capacitors in a chain
The proportional split is the rule for a side by side group, where the voltage is shared. In a chain the isolated island between the capacitors forces one single charge onto all of them, and nothing is divided at all.
wrong$$Q_i = Q_{\rm eq}\frac{C_i}{\sum_j C_j}$$
right$$Q_i = Q_{\rm eq}\ \text{for every capacitor in a chain}$$
⚠ Using a constant force for the work of separating plates at fixed voltage
At fixed charge the field between the plates does not change as they separate, so the force is constant and the work is a product. At fixed voltage the field falls as the gap grows, so the force falls with it and the work is an integral.
wrong$$W = F\,\Delta d\ \text{with}\ F\ \text{constant, when } V \text{ is held fixed}$$
right$$W = \int F\,dd = \int \frac{\varepsilon_0AV^{2}}{2d^{2}}\,dd,\qquad F\ \text{constant only when } Q \text{ is fixed}$$
⚠ Forgetting that the source delivers twice what the capacitor keeps
Every coulomb the source moves crosses the full fixed voltage, while the capacitor's own store carries a factor of one half because its voltage climbed from zero as it filled. The missing half goes into whatever mechanical change was happening, or into heat.
wrong$$W_{\rm source} = \Delta U$$
right$$W_{\rm source} = V\,\Delta Q = 2\,\Delta U\ \text{when } V \text{ is held fixed}$$
Formula card
Definition of capacitance
$$C = \frac{Q}{V}$$
two conductors carrying $+Q$ and $-Q$; $V$ is the potential difference between them; $Q$ is one plate's magnitude
Parallel plate capacitor
$$C = \frac{\varepsilon_0 A}{d}$$
flat parallel plates, overlap area $A$, gap $d$ much smaller than the plate width, vacuum or air between
Coaxial cable
$$C = \frac{2\pi\varepsilon_0 L}{\ln(R_b/R_a)}$$
length $L$ long compared with $R_b$; only the ratio of the radii matters
Concentric spheres, and the lone sphere
$$C = 4\pi\varepsilon_0\frac{R_aR_b}{R_b-R_a}\ \xrightarrow{\;R_b\to\infty\;}\ C = 4\pi\varepsilon_0R = \frac{R}{k}$$
$R_a$ inner and $R_b$ outer radius; the limit puts the second conductor at infinity
the first case is a disconnected capacitor, the second one still attached to its source
Slab filling only part of the gap
$$C = \frac{2K}{K+1}C_0\quad\text{for a slab of thickness } d/2$$
the layers are stacked across the gap, so they form a chain; the general case is a reciprocal sum over the layers
Bound charge on the faces of a dielectric
$$\sigma_{\rm ind} = \sigma\left(1-\frac{1}{K}\right),\qquad E = \frac{\sigma-\sigma_{\rm ind}}{\varepsilon_0}$$
uniform field in the slab; $\sigma$ is the free charge on the metal and $\sigma_{\rm ind}$ the bound charge on the slab face
Gauss's law with a dielectric present
$$\oint K\vec E\cdot d\vec A = \frac{Q_{\rm free}}{\varepsilon_0}$$
$K$ constant over the surface; only the free charge on the conductors is counted
Dielectric strength and the working voltage
$$V_{\max} = E_{\max}\,d$$
$E_{\max}$ is a property of the material, and $d$ the thickness it has to hold off
Attraction between the plates
$$F = \tfrac12 QE = \frac{Q^{2}}{2\varepsilon_0 A}$$
each plate sits in the field of the other only, which is half the total, hence the leading half; not named on the week line
Check yourself
Close the page and write out from memory: the definition of capacitance and what it does not depend on; the flat pair formula and the condition it needs; the cable, the two shells and the lone sphere; both combination rules and which quantity each makes common; all three forms of the stored energy and the reason for the half; the energy density, in vacuum and in a dielectric; what $K$ multiplies and what it divides; what happens to charge, voltage, field and energy when a slab goes in, connected and disconnected; and the bound charge formula. Then open the formula card and mark only what you missed.
Take one measured pair of numbers for a capacitor and produce its charge at any other voltage, and say what the marked voltage on a component means?
c-capacitance
Derive $C = \varepsilon_0A/d$ from a pillbox and an integral in two lines, and say what is wrong with applying it to plates a metre across held half a metre apart?
c-parallel-plate
Get the capacitance of a coaxial cable per metre from its two radii, and explain why an isolated sphere has a capacitance at all when there is only one conductor in sight?
c-other-geometries
Reduce a three capacitor network to one number and then recover the charge and voltage on each capacitor, with the two sum checks at the end?
c-combinations
Get the stored energy by all three routes, say why the answer carries a half, and turn it into an energy per cubic metre of the gap?
c-energy
Take the same capacitor and the same slab through both insertion problems, one with the source attached and one without, and get every quantity right in both?
c-dielectrics
Compute the bound charge on a slab face from the plate charge and the dielectric constant, and use it to explain why the field inside the slab is smaller by exactly that constant?
c-dielectric-molecular
Glossary (19 terms)
capacitancesığa
The charge one conductor of a pair carries for each volt of potential difference between them, measured in farads. It is fixed by the shape of the two conductors and by the material between them, and does not depend on how much charge is actually on them.
capacitorkondansatör
Any pair of conductors used to hold separated charge, with equal and opposite amounts on the two of them. The device as a whole stays electrically neutral, whatever it is holding.
faradfarad
The SI unit of capacitance, equal to one coulomb per volt. It is an enormous unit: a lone metal sphere would have to be nine million kilometres across to reach one farad, so practical values are quoted in microfarads and picofarads.
Two flat conductors facing each other across a small gap, whose capacitance is the permittivity times the overlap area divided by the gap. The formula assumes the gap is much smaller than the plates.
The part of the field that bulges outwards near the rims of a pair of plates instead of running straight across the gap. Every flat plate formula on this page ignores it, which is only fair when the gap is much smaller than the plates.
silindirik kondansatör
A pair of coaxial conducting cylinders, as in a cable. Its capacitance per unit length depends only on the ratio of the two radii, through a logarithm, so doubling both radii changes nothing.
spherical capacitorküresel kondansatör
A pair of concentric conducting shells. Its capacitance depends on both radii and on the gap between them, and it becomes the isolated sphere result when the outer shell is moved to infinity.
equivalent capacitanceeşdeğer sığa
The single capacitance that would take the same charge at the same potential difference as a whole network. It stands in for the network seen from outside, and the individual charges and voltages inside still have to be recovered one at a time.
paralel bağlama
Capacitors joined so that both terminals of each are attached to the same two conductors, so all of them span the same potential difference and their charges add. The capacitances add directly.
series combinationseri bağlama
Capacitors joined end to end so that the metal between each neighbouring pair is isolated. That island started neutral and must stay so, which forces the same charge onto every capacitor in the chain, and the reciprocals of the capacitances add.
energy densityenerji yoğunluğu
The energy stored per cubic metre of an electric field, equal to half the permittivity times the square of the field. It lets the energy of an arrangement be attributed to the field itself rather than to the charges that made it.
dielectricdielektrik
An insulating material placed in the gap of a capacitor. It never carries the charge that the plates carry; what it does is weaken the field between them, so that the same charge costs fewer volts.
dielectric constantdielektrik sabiti
The pure number, never less than one, by which filling the gap with a given material multiplies the capacitance. The same number divides the field inside the material when the charge on the plates is held fixed.
dielectric strengthdielektrik dayanımı
The largest field a material can stand before it stops insulating and begins to conduct. It is a field and not a voltage, so it has to be multiplied by the thickness before it says anything about a component's rating.
bound chargebağlı yük
The layer of charge that appears on the faces of a dielectric because its molecules have lined up with the field. Each contribution belongs to its own molecule and cannot be drained away, and the total is always smaller than the charge on the plates.
free chargeserbest yük
Charge that can travel through a conductor and be added to or removed from it, as opposed to the bound charge tethered inside a dielectric. In a capacitor it is the charge on the metal plates.
polar molekül
A molecule that is already a small dipole before any field is applied, because its charge is not distributed symmetrically. Water is the standard example, and materials made of such molecules have much larger dielectric constants than those that are not.
permittivity of a materialelektriksel geçirgenlik
The product of the dielectric constant and the permittivity of free space, written as a single symbol so that formulas for a filled gap look like the vacuum ones. In vacuum the dielectric constant is one and the two coincide.
working voltageçalışma gerilimi
The largest potential difference a component is allowed to have across it, printed on the case beside the capacitance. It is set by the dielectric strength of the material inside multiplied by its thickness, and exceeding it destroys the component.
What comes next
§07 · Electric Current and Resistance
Everything on this page has been a statement about a final resting state: where the charge ends up, what the field is once it has settled, how much energy is sitting in the gap. Not one line has said how long any of it took. The next section takes the lid off that, and asks what is happening while the charge is actually moving.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook. Its treatment of capacitance, dielectrics and the energy stored in an electric field covers the same ground as this section, and its end of chapter problems run a level harder than the ones here, which is the right next step once this set feels comfortable.
Course syllabus, week 6 line and assessment table The scope of this section comes from the week line, which reads Capacitance, Dielectrics, Electric Energy Storage, and from nowhere else. The line names three topics and carries no chapter numbers, so no chapter number is quoted anywhere on this page. The assessment weights quoted on the card are the published ones and nothing finer than them is claimed.
SI values of the electric constants used here The permittivity of free space is 8.85 times ten to the minus twelfth in SI units and the Coulomb constant 8.99 times ten to the ninth, the same two numbers as in the previous sections. The elementary charge is 1.602 times ten to the minus nineteenth coulombs and the proton mass 1.67 times ten to the minus twenty seventh kilograms. Dry air is taken to break down at about three times ten to the sixth volts per metre.
Representative dielectric constants and dielectric strengths The values in the tables here are round representative figures, used only to show the range and the trade-off between the two columns. For a named material in a calculation the number comes from the textbook table, and every worked example states the value it is using.