6 concepts16 worked examples28 exercises4 exam-level6 figures
What are you here for?
01Electric charge and the electric field: charge, Coulomb's law, and the field of point charges
Run a plastic ruler along your sleeve four or five times and hold it a centimetre above the paper punchings on the desk. They jump up and stick to it. Those punchings came out of a fresh stack a minute ago, nobody rubbed them, and nothing was added to them.
By the end of this section you can say why an untouched scrap of paper is pulled towards the ruler, put a number in newtons on that pull, and compute the force and the field produced by any set of whose coordinates you are given.
In 60 seconds
Matter carries a second kind of quantity besides mass, it comes in two signs and in whole multiples of one small unit, and the force between two lumps of it falls off as the square of the distance. Everything else this term is built on that one force.
$$Q = n e,\qquad e = 1.602\times 10^{-19}\ \mathrm{C}$$
counting electrons transferred, or checking whether a quoted charge is physically possible
Coulomb's law, magnitude
$$F = k\,\frac{|q_1||q_2|}{r^{2}},\qquad k = 8.99\times 10^{9}\ \mathrm{N\cdot m^{2}/C^{2}}$$
two objects small compared with their separation; the sign of each charge decides direction, not size
you want a property of the empty point itself, independent of what you later put there
Field of a single point charge
$$E = k\,\frac{|Q|}{r^{2}}$$
one source charge; direction is away from a positive source and towards a negative one
Three most common mistakes
Feeding signed charges into the formula and then also arguing the direction from the picture. The sign gets used twice and the arrow ends up backwards. Put magnitudes in the formula and read the direction off the diagram, once.
Adding the magnitudes of two forces or two fields that are not parallel. Two pulls of 1.80 N and 0.22 N give 1.94 N here and 1.58 N in a different geometry; only components can tell you which.
Leaving micro or nano in the numbers. A charge written as 3.0 in a formula that expects coulombs is out by a factor of a million, and the answer looks reasonable enough to hand in.
The published weights for this course are Midterm 1 20%, Midterm 2 20%, quizzes 10%, homework 5%, final 25% and laboratory 20%. This is the first week, so nothing here has been examined yet; what can be said is that Coulomb's law and the field of point charges are the tools the rest of the weeks are written in, and an error made here reappears in every later answer.
How much time do you have?
10 minutes
The two formulas that carry the whole week, plus the sign rule that decides which way the arrow points.
The 60 second card · Coulomb's law and how strong the pull is · The electric field and what it means · Formula card
45 minutes
Everything that turns into a number: counting electrons, the pair force, the component sum for three charges, the field at a point, and the place on a line where the field vanishes.
The 60 second card · Charge and its two signs · Coulomb's law and how strong the pull is · Three charges at once: the vector sum · The electric field and what it means · The field of several charges and the · The faded ladder · Practice set B (the computations)
full read
Adds the part the formulas do not contain: why a neutral scrap of paper moves at all, what a does that an does not, and why physics bothers to define a field instead of just listing forces.
The opening question · Charge and its two signs · Why a neutral scrap of paper still moves · Coulomb's law and how strong the pull is · Three charges at once: the vector sum · The electric field and what it means · The field of several charges and the null point · The method boxes · The contrast pair · The faded ladder · The exam style example · Practice sets A to D · The self audit
By the end of this section
Count the electrons behind a given net charge, and decide from a quoted value whether a charge is physically possible at all.
Explain why a charged object attracts a neutral one, and trace what ends up where when a conductor is charged by contact or by induction.
Compute the magnitude and direction of the Coulomb force between two point charges, and predict how it changes when a charge or the separation is scaled.
Combine the forces exerted by several point charges on one of them, by resolving each pair force into components and summing column by column.
Convert between the force on a and the field at its location, and compute the field of a single point charge with its direction.
Locate the point on a line where the net field of two charges vanishes, and evaluate the net field of a group of point charges at an arbitrary point.
Syllabus coverage
Two signs of charge, conservation, quantisation and the , conductors and insulators, and by induction, of a neutral object
Two concepts share this token: what charge is and how it is counted, then what happens when it is free to move inside a material and what happens when it is not.
covered
Electric Field
Coulomb's law and the superposition of pair forces, the field as force per unit test charge, the field of a point charge, and the net field of several point charges
Four concepts share this token. Coulomb's law comes first because the field is defined from it; the last concept is the vector sum that every later week reuses.
covered
Comparison with Newtonian gravitation
The ratio of the electric to the gravitational force between two protons, used once as a scale check
The week line does not name gravitation. It is used here for exactly one purpose: the two laws have identical shape, so the ratio of the two forces is a pure number that does not depend on separation, and computing it once tells you why gravity never appears again in this half of the course. Nothing about gravitation is assumed beyond the force law itself, which is restated in the recall list.
off_syllabus
Field lines and continuous charge distributions
Drawing the field as continuous lines, and charge smeared along a rod, a ring or a sheet
Deferred to the next section, which the syllabus gives its own week for. Everything here is done with arrows at chosen points and with charges treated one at a time, so nothing in this section depends on the deferred material.
deferred
Recall first
A vector is handled through its components
A vector of magnitude $F$ making an angle $\theta$ with the $x$ axis has components $F_x = F\cos\theta$ and $F_y = F\sin\theta$. To add vectors, add the $x$ components, add the $y$ components separately, and rebuild with $F = \sqrt{F_x^{2}+F_y^{2}}$ and $\tan\theta = F_y/F_x$.
Every problem in this section with more than two charges is a component sum. It is the single most common place marks are lost, and the loss is always the same one: adding magnitudes.
Distance between two points
The distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $r=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$, and the unit vector pointing from the first to the second is that displacement divided by $r$.
Coulomb's law needs the distance between the two charges, not the distance from the origin and not one of the coordinate differences. Half the wrong answers in two dimensional problems start here.
Newton's second law
$\vec{F}_{\text{net}} = m\vec{a}$: the acceleration of a body is the net force on it divided by its mass, and it points the same way as that net force.
Used only to turn a computed force into something you can picture, such as the acceleration of an electron, and to compare an electric pull with a weight.
Newton's third law
If body A pushes on body B with a force $\vec{F}$, then B pushes on A with $-\vec{F}$: same size, opposite direction, and the two forces act on different bodies.
It survives untouched here. A tiny charge and a huge one pull on each other equally hard, and that is unintuitive enough that this section gives it a question of its own twice, once in the pretest and once in the concept practice, instead of a passing remark.
Newton's law of universal gravitation
Two masses attract along the line joining them with $F = G\,m_1m_2/r^{2}$, where $G = 6.674\times 10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$.
Needed for one calculation only, the ratio of the electric to the gravitational force between two protons. It is quoted here so that the comparison does not lean on anything you have to go and look up.
An inverse square quantity
If $y = C/r^{2}$, then doubling $r$ divides $y$ by four and tripling $r$ divides it by nine. Multiplying $r$ by a factor $s$ divides $y$ by $s^{2}$.
Both laws in this section are of this shape, and several questions here are answered by scaling alone, with no number ever put into a formula: the pretest opener, the check after Coulomb's law, and the first of the exam level problems.
Try it yourself first (3 questions)
1§01.0 — how an inverse square force scales●○○○○
Before anything electrical, a check on the shape of the law. Two small lead spheres hang side by side and attract each other gravitationally with a force of 8.0 N when their centres are a distance $d$ apart. Somebody moves them so the centres are $2d$ apart, changing nothing else.
Given
Force at separation $d$ is $8.0\ \mathrm{N}$
New separation is $2d$
Masses unchanged
Find
(a) What is the new force?
Hint 1/4
Nothing needs to be computed. The question is only what factor multiplies the force when the separation is multiplied by two.
Hint 2/4
For $F = C/r^{2}$, replacing $r$ by $sr$ divides $F$ by $s^{2}$.
Hint 3/4
Substitute the data from the question, $F=8.0\ \mathrm{N}$ and $s=2$: the new force is $8.0/2^{2}$.
Hint 4/4
So the new force is $2.0\ \mathrm{N}$.
Show solution
Taking the ratio of the two force expressions is shorter than computing either one, because $G$ and both masses are unknown and would cancel anyway.
the ratio is applied to the one force that was measured
Answer $$\boxed{F(2d) = 2.0\ \mathrm{N}}$$
Check
Independent route: the force at $2d$ should be smaller than at $d$ and larger than at $3d$, where the same rule gives $8.0/9 = 0.89\ \mathrm{N}$. The value 2.0 N sits between them, and the three values 8.0, 2.0, 0.89 fall off faster than the separations 1, 2, 3 rise, as an inverse square must.
In a scaling question the shape of the law does all the work: an inverse square divides by the square of whatever factor the separation grew by. Coulomb's law has exactly this shape, so every ratio argument used here carries over to it untouched.
2§01.0 — resolving a force into components●○○○○
A rope pulls a crate across a floor with a steady force of $5.0\ \mathrm{N}$, and the rope is tilted $37^{\circ}$ above the horizontal. The whole of this section is component arithmetic, so this is the tool being checked, not the crate.
Given
$F = 5.0\ \mathrm{N}$
$\theta = 37^{\circ}$ above the $+x$ axis
Find
(a) Find the $x$ and $y$ components of the pull.
Hint 1/4
You are asked to split one arrow into two arrows along the axes, not to find anything about the crate.
Hint 2/4
The component along the axis the angle is measured from carries the cosine, the other carries the sine.
Hint 3/4
Substitute the data from the question, $F=5.0\ \mathrm{N}$ and $\theta=37^{\circ}$: $F_x = 5.0\cos 37^{\circ}$ and $F_y = 5.0\sin 37^{\circ}$.
Hint 4/4
So $F_x = 4.0\ \mathrm{N}$ and $F_y = 3.0\ \mathrm{N}$.
Show solution
The angle is measured from the $x$ axis, so the cosine belongs to $x$; had it been measured from the vertical, the two would swap and the answer would be different.
Rebuild the magnitude from the two components: $\sqrt{4.0^{2}+3.0^{2}} = 5.0$, which is the force we started from, so nothing was lost or double counted.
Which trigonometric function goes with which axis is decided by what the angle is measured from, never by habit: the cosine belongs to the axis the angle opens from. Settle that before writing a single component, because every vector sum in this section starts at this step.
3§01.0 — the third law with very unequal partners●●○○○
This one is deliberately built around the intuition most people bring in and that this section will keep testing. A lorry of mass $8000\ \mathrm{kg}$ collides with a stationary bicycle of mass $12\ \mathrm{kg}$.
Given
Lorry mass $8000\ \mathrm{kg}$
Bicycle mass $12\ \mathrm{kg}$
They are in contact during the collision
Find
(a) True or false: during contact the lorry pushes on the bicycle harder than the bicycle pushes on the lorry.
Hint 1/4
Separate two different questions: how hard each body is pushed, and how much each body is changed by being pushed.
Hint 2/4
The third law pairs two forces of equal size and opposite direction, acting on different bodies; mass does not appear in it.
Hint 3/4
Substitute the data from the question, $8000\ \mathrm{kg}$ and $12\ \mathrm{kg}$: the forces are equal, so the accelerations are in the ratio $8000:12$, about 670 to 1.
Hint 4/4
So the statement is false: the pushes are equal and only the responses to them differ.
Show solution
Argue from the accelerations rather than from the law itself, because the law is exactly what the intuition refuses to believe.
same force divided by very different masses, which is the asymmetry the eye actually sees
Answer $$\boxed{\text{False: equal forces, accelerations in the ratio }670:1}$$
Check
Independent check on the direction of the conclusion: if the forces were not equal there would be a net force on the pair as a whole coming from nowhere, and the two bodies together would start accelerating without anything outside pushing them.
Equal forces do not mean equal effects. Whenever an argument sounds like "the bigger one pushes harder", the intuition is really about acceleration, and the repair is always to divide the same force by two different masses.
Notation
symbol
reads as
means
watch out
$q,\ Q$
q, or capital Q
an electric charge in coulombs; lower case is the usual choice for a small charge being acted on, capital for a source charge doing the acting
the letter carries the sign with it, so $q=-3.0\ \mathrm{\mu C}$ is a complete statement and writing a second minus in front of it is a double negative
$|q|$
the magnitude of q
the size of the charge with the sign stripped off, which is what goes into Coulomb's law
if a minus sign survives into the formula, the number that comes out is not a magnitude and the direction has already been decided twice
$e$
e, the elementary charge
$1.602\times 10^{-19}\ \mathrm{C}$, the size of the charge on one proton and on one electron
$e$ is a positive number by definition; an electron carries $-e$, and writing the electron charge as $e$ loses the whole sign of the problem
$r$
r
the distance between the two charges being considered, always centre to centre
not the distance from the origin, and in two dimensions not one of the coordinate differences either
$\hat{r}$
r hat
a unit vector pointing from the source charge towards the point of interest, obtained by dividing the displacement by its own length
it has length one and no units; the metres in it have already cancelled
$\vec{E}$
E vector, the electric field
force per unit charge at a point, in newtons per coulomb, defined whether or not anything is sitting there
$E$ without the arrow is the magnitude only; a field quoted with no direction is half an answer
$k,\ \varepsilon_0$
k, and epsilon nought
$k = 8.99\times 10^{9}\ \mathrm{N\cdot m^{2}/C^{2}}$ is the , and $\varepsilon_0 = 8.85\times 10^{-12}$ in SI units is the same constant repackaged as $k = 1/(4\pi\varepsilon_0)$
one or the other, never both in the same expression; textbooks and lecturers differ and both forms are correct
$\mathrm{\mu C},\ \mathrm{nC}$
microcoulomb, nanocoulomb
$10^{-6}\ \mathrm{C}$ and $10^{-9}\ \mathrm{C}$, the sizes real laboratory charges actually come in
convert to coulombs before the formula, never after; a forgotten micro is a factor of a million and the answer still looks plausible
Conventions used here
Signs go in the picture, magnitudes go in the formula.
Throughout this section, Coulomb's law is evaluated with $|q_1|$ and $|q_2|$, so the number it returns is always positive and is a magnitude. The direction of the force is then decided separately, in one sentence, from the two signs: like signs push apart along the line joining the charges, unlike signs pull together along it.
The alternative, keeping the signs inside the formula and reading the direction off the resulting sign, works in one dimension and quietly breaks in two, because there the sign of a product says nothing about which way an arrow points.
Axes and angles.
The $x$ axis runs to the right and the $y$ axis upwards, and a positive angle is measured anticlockwise from the $+x$ axis. When an answer is quoted as an angle below the axis, the word below is written out rather than being hidden in a minus sign.
Half of the two dimensional answers in this section are a magnitude plus an angle, and a reader who has to guess the reference direction cannot check the answer.
Which constants are used.
These notes use $k = 8.99\times 10^{9}\ \mathrm{N\cdot m^{2}/C^{2}}$, equivalently $\varepsilon_0 = 8.85\times 10^{-12}\ \mathrm{C^{2}/(N\cdot m^{2})}$, and $e = 1.602\times 10^{-19}\ \mathrm{C}$ everywhere. Where a weight is compared with an electric force, $g = 9.80\ \mathrm{m/s^{2}}$.
A constant that changes value between examples changes the last digit of every answer, and a reader has no way of telling which version a given number came from.
How many digits an answer keeps.
Results are quoted to three significant figures, or fewer when the data deserve fewer. Intermediate values carry a guard digit and are never rounded on the way through; rounding happens once, at the end.
Rounding twice moves the last digit by more than the data justify, and it makes a correct method look like an arithmetic error.
What a point charge is allowed to mean.
Every charged object in this section is treated as a point: its size is negligible compared with the distances between the objects. Where a real object is described as a small sphere or a scrap of paper, that idealisation is being made and it is stated in the problem.
Coulomb's law as written applies to points. Saying so in advance means that when an extended object does turn up later, the difference is a genuine complication and not a hidden assumption you were never told about.
What a test charge is allowed to mean.
A test charge is small enough that its own field does not move the charges producing the field being measured. It is written $q_0$ and its numerical value never appears in a final answer for the field.
Otherwise the definition of the field would be circular, because the probe would be changing the very thing it is probing, and two different probes would report two different fields at the same point.
1.1Charge: two signs, conserved, and counted in whole numbers
Names the quantity every force in this course acts on, and fixes the smallest amount of it that can exist.
Start with what actually moved when the ruler crossed the sleeve.
Solvable with what we have
Rub the ruler and watch the paper punchings jump up to it.
State the rule: like charges repel, unlike charges attract.
Notice that the effect dies away on a humid day.
Not solvable yet
Say what sign the punching carries.
Say how much charge the ruler picked up.
Say whether a charge of any size at all is allowed.
Apply the known rule. The punching is attracted, attraction means opposite signs, so if the ruler is negative the punching must be positive:
That is a definite prediction: the punchings were charged before the ruler arrived.
Why it fails
They were not. They came from a sealed pack and touched nothing but each other. The rule was quoted outside the situation it describes: it is about two objects that are already charged, and says nothing about a neutral one. Fixing that needs the next concept.
RuleRule: the three facts about electric charge
Conditions
the conservation statement needs an isolated system, so draw the boundary before using it
the counting number $n$ is any whole number, including zero
neutral means equal amounts of both signs, not an absence of charge
$$\boxed{\begin{aligned}&\text{like signs repel},\quad \text{unlike signs attract}\\[3pt]&\textstyle\sum Q \text{ of an isolated system is constant}\\[3pt]&Q = n\,e,\qquad e = 1.602\times 10^{-19}\ \mathrm{C}\end{aligned}}$$
There are two kinds of charge and they behave oppositely towards each other. Charging never manufactures charge, it only moves charge that already existed, so whatever one object gains another has lost. And charge comes in identical grains: a measured charge is always a whole number of them.
Rubbing does not make charge, it relocates it. The two boxes in the lower row carry equal and opposite amounts, and together they carry exactly what they carried before the rubbing started.
Looks like this, but is not
This is quantisation working: a sphere measured at $-4.806\times 10^{-19}\ \mathrm{C}$. Divide by $e$ and you get exactly $-3$, so it carries three more electrons than protons and the value is allowed.
This looks equally harmless and cannot happen: a sphere carrying $-1.0\times 10^{-19}\ \mathrm{C}$. Divided by $e$ that is $-0.624$ of an electron, and there is no such object. The number is small, tidy and impossible, which is why the test is worth doing: only the division warns you.
situation
typical net charge
electrons behind it
one electron
$-1.602\times 10^{-19}\ \mathrm{C}$
1
a rubbed plastic rod
about $10^{-7}\ \mathrm{C}$
about $10^{12}$
a charged laboratory sphere
about $10^{-6}\ \mathrm{C}$
about $10^{13}$
a lightning stroke
about $10\ \mathrm{C}$
about $10^{20}$
The interesting column is the third one. Even the smallest charge you can produce by rubbing is a million million grains, which is why charge feels like a smooth quantity in the laboratory, and why nobody discovered that it is not until they could work with single particles.
How many electrons make up a charge of one microcoulomb
A metal ball carries a net charge of $-1.0\ \mathrm{\mu C}$. Find the number of electrons it has in excess of its protons.
Work with magnitudes and read the sign off in words at the end. Carrying the minus through the division only risks reporting a negative count of objects, which is not a thing.
Independent sense check rather than the same division again: a copper ball a centimetre across contains of order $10^{24}$ conduction electrons. Six million million sounds enormous, but against $10^{24}$ it is about one electron in $10^{11}$, which is why the ball's mass and chemistry are completely unchanged by charging it.
One unit conversion and one division. The conversion is where the marks go.
A microcoulomb is a large charge in the laboratory and a colossal number of electrons. Keep that pairing in mind: whenever a problem hands you microcoulombs, you are in the world of visible sparks and centimetre separations, not of single particles.
Where the charge went when the rod was rubbed
Rubbing a plastic rod with a wool cloth moves $1.0\times 10^{12}$ electrons from the cloth to the rod. Both were neutral to begin with, and no charge escapes to the air or the bench. Find the final charge of each, and of the pair together.
Given
$N = 1.0\times 10^{12}$ electrons transferred, cloth to rod
Both objects initially neutral
Nothing else exchanges charge with them
Find
$Q_{\text{rod}}$, $Q_{\text{cloth}}$, and their sum.
Solution
Compute one object and get the other from conservation rather than from a second count. That is not laziness: it makes the check at the end independent of the arithmetic, because the two numbers were not obtained the same way.
The object that gained electrons
$$Q_{\text{rod}} = -N e = -(1.0\times 10^{12})(1.602\times 10^{-19}\ \mathrm{C})$$
each arriving electron adds $-e$, and the rod started from zero, so the total is the count times one electron charge
Check the size against the previous example instead of redoing the multiplication. There, $6.2\times 10^{12}$ electrons made one microcoulomb. Here the count is about six times smaller, so the charge should be about a sixth of a microcoulomb, and $1.6\times 10^{-7}\ \mathrm{C}$ is exactly that.
The pattern generalises: in any charging problem, find the object whose electrons you can count, then get every other object from the fact that the total has not moved.
Checkpoint
§01.1 — which charges can actually exist●●○○○
Thirty seconds on quantisation, in the form that actually catches people out: four values are read off a very sensitive instrument, and three of them are impossible for a reason that has nothing to do with the instrument.
Given
$e = 1.602\times 10^{-19}\ \mathrm{C}$
Find
(a) Which of these could be the net charge of an object?
Hint 1/4
The question is not which number is smallest or neatest. It is which one is a whole number of grains.
Hint 2/4
A charge is possible exactly when $Q/e$ is a whole number, with $e$ the elementary charge.
Hint 3/4
Substitute the data, with $e = 1.602\times 10^{-19}\ \mathrm{C}$: the four quotients are $-5.00$, $-1.50$, $+3.50$ and $+0.624$.
Hint 4/4
So only $-8.01\times 10^{-19}\ \mathrm{C}$ is possible, and it is five excess electrons.
Show solution
Divide every candidate by $e$ rather than looking for a pattern in the digits, because half integer values are designed to look exactly as plausible as whole ones.
Independent check on the winner: multiply back, $5\times 1.602\times 10^{-19} = 8.01\times 10^{-19}$, and the digits reproduce the quoted value exactly rather than approximately, which is the signature of a genuine multiple.
To test whether a charge can exist, divide it by $e$ and ask whether the count comes out whole; the look of the digits never decides it. Any quantity built out of indivisible units is checked the same way.
⚠ Saying that rubbing creates charge
the rod was inert before and attracts things afterwards, so something obviously appeared, and calling that something new charge is the natural description
Explains the one demonstration everybody has seen, and the difference between a material that lets charge run and one that does not.
The rule about two charged objects was fine. The scrap of paper is not two charged objects, so the question is what a single charged object does to a neutral neighbour.
RuleRule: a charged object always attracts a neutral one
Conditions
the neutral object must be free to rearrange internally, which every real material is to some degree
in a conductor the rearrangement is whole electrons travelling across the object; in an insulator each molecule distorts slightly in place and nothing travels far
the effect needs no contact and does not change the neutral object's total charge, which stays exactly zero
Split the neutral object into the induced charge that ended up nearer the source and the equal induced charge that ended up further away. The near one is attracted and the far one is repelled by exactly the same amount of charge, so if the force did not depend on distance they would cancel perfectly and nothing would move. They do not cancel, because the near one sits where the force is stronger, and the attraction wins by the difference between two inverse squares.
The scrap carries no net charge, and it still moves. The minus that ended up on the near face sits where the force is stronger than the plus on the far face does, and the difference between the two inverse squares is the whole effect.
Looks like this, but is not
This is induction: hold a positive rod near an isolated metal sphere without touching it. Electrons in the sphere crowd onto the near face, the far face is left positive, the sphere is attracted, and the moment you take the rod away everything relaxes back and the sphere is neutral again, as it always was.
This looks like the same experiment and is a different one: touch the sphere with the rod. Now charge actually crosses the gap, the sphere keeps a net positive charge of its own after the rod leaves, and it will repel the next positive rod you bring near. Same equipment, same attraction at the start, permanently different sphere at the end, and the only difference was contact.
The pull on a 1 mm scrap of paper, and whether it can lift it
The rubbed ruler carries $Q = 5.0\times 10^{-8}\ \mathrm{C}$ and is held $1.00\ \mathrm{cm}$ above a square scrap of paper $1.0\ \mathrm{mm}$ on a side, cut from $80\ \mathrm{g/m^{2}}$ paper. Model the polarised scrap as two point charges of size $q = 2.0\times 10^{-11}\ \mathrm{C}$, the negative one at the near face and the positive one at the far face, $0.05\ \mathrm{cm}$ further away. Find the net pull and compare it with the weight of the scrap.
Given
$Q = 5.0\times 10^{-8}\ \mathrm{C}$ on the ruler
$q = 2.0\times 10^{-11}\ \mathrm{C}$ induced on each face
The net electric force on the scrap, and its ratio to the weight.
Solution
Keep the common factor $kQq$ outside the bracket and subtract the two inverse squares inside it. Computing the two forces separately and subtracting five-figure numbers at the end would throw away exactly the small difference the whole answer consists of.
Independent order of magnitude check on the mechanism rather than on the arithmetic: if the two induced charges sat at the same distance, the bracket would be exactly zero and the scrap would not move at all. Shrinking the offset from $0.05\ \mathrm{cm}$ to $0.01\ \mathrm{cm}$ should therefore cut the force by roughly a factor of five, and it does, giving $1.8\times 10^{-6}\ \mathrm{N}$, still more than twice the weight.
Two inverse squares, one subtraction, one weight. The subtraction is the physics; everything else is bookkeeping.
Ten times its own weight is why the punchings jump rather than merely lean. It also tells you why the demonstration fails on a wet day: a film of water on the paper is conducting enough to leak the induced charge away before the pull can act.
Charging a sphere by induction without ever touching it
A neutral metal sphere sits on an insulating stand. A positively charged rod is brought near it but never touches it. While the rod is held in place, the sphere is briefly connected to the ground by a wire, the wire is then removed, and only afterwards is the rod taken away. During the , $2.5\times 10^{11}$ electrons flow up the wire into the sphere. Find the sign and size of the sphere's final charge.
Given
Sphere initially neutral, on an insulating stand
Rod is positive and never touches the sphere
Order of operations: rod near, ground, remove ground, remove rod
$2.5\times 10^{11}$ electrons enter the sphere
Find
The final charge of the sphere.
Solution
Track the electrons rather than the attraction. The attraction is the same at every stage and therefore tells you nothing about which stage did the permanent work.
What the grounding wire is for
$$\text{rod near} \Rightarrow \text{near face } -,\ \text{far face } +,\ Q_{\text{sphere}} = 0$$
before grounding this is only a rearrangement, so the total is still zero and nothing permanent has happened
$$\text{ground} \Rightarrow \text{electrons climb to the far face and stay}$$
the far positive face is what the earth responds to; electrons come up to neutralise it and the near face is held in place by the rod
Count what stayed behind
$$Q = -N e = -(2.5\times 10^{11})(1.602\times 10^{-19}\ \mathrm{C})$$
the wire has been removed, so those electrons are now trapped on the sphere whatever the rod does next
$$Q = -4.0\times 10^{-8}\ \mathrm{C}$$
two significant figures from the given count
Answer $$\boxed{Q_{\text{sphere}} = -4.0\times 10^{-8}\ \mathrm{C},\ \text{opposite in sign to the rod}}$$
Check
Independent check through conservation applied to a bigger system: the sphere gained $4.0\times 10^{-8}\ \mathrm{C}$ of negative charge and the earth lost exactly that much, so the sphere and the earth together are unchanged, and the rod, which never touched anything, still carries whatever it started with.
No force calculation at all. The entire answer is the order in which the wire and the rod were removed.
Reverse the last two steps, taking the rod away before the wire, and the electrons simply run back down to the earth and the sphere ends up neutral. That is the whole trick, and it is why induction questions are always about the order of operations.
Checkpoint
§01.2 — what an attraction does and does not prove●●○○○
Half a minute on the point this concept exists for. A small ball hanging on a thread is attracted to a positively charged rod held nearby, and it swings towards it every time.
Given
The rod is positively charged
The ball is attracted, not repelled
No contact is made at any point
Find
(a) What can be concluded about the ball's net charge?
Hint 1/4
List every state the ball could be in, then ask which of them would produce an attraction. The question is what survives, not what is likeliest.
Hint 2/4
Repulsion happens only between two charges of the same sign; attraction happens both between opposite charges and between a charge and a neutral body.
Hint 3/4
Substitute the three cases with a positive rod: negative ball attracts, neutral ball attracts through polarisation, positive ball repels.
Hint 4/4
So the ball is negative or neutral, and only a repulsion would have settled it.
Show solution
Work through all three cases rather than reasoning from the observed attraction backwards, because the backwards route silently assumes the ball is charged.
like signs, which is the only case the observation excludes
Answer $$\boxed{q_{\text{ball}} \le 0}$$
Check
Independent check by running the experiment the other way: bring up a negatively charged rod instead. A negative ball now repels and a neutral ball is still attracted, so the pair of observations separates the two cases that this single one cannot.
Attraction is consistent with two situations, repulsion with only one. Whenever a question hands you an observed attraction, list the neutral case alongside the opposite sign case before concluding anything.
⚠ Treating attraction as proof of charge
the two body rule is learned first and gets applied to every attraction, including the ones where only one body is charged
wrong$$\text{attracted} \Rightarrow q \ne 0$$
right$$\text{repelled} \Rightarrow q \ne 0,\ \text{same sign as the source}$$
⚠ Getting the sign backwards after induction
the rod is positive and stays positive throughout, so it feels as though what it leaves behind should be positive too
The pull or push between two charges is proportional to each of them, so doubling either charge doubles the force, and it is inversely proportional to the square of the gap, so doubling the separation cuts the force to a quarter. The force acts along the line joining the two charges and nowhere else, and each charge feels the same size of force as the other, however unequal the two charges are.
The pair of forces Coulomb's law describes. They are equal in size and opposite in direction, they act on different bodies, and the fact that one charge is nearly twice the other changes neither of those things.
Looks like this, but is not
This is Coulomb's law working: two metal spheres a centimetre across, charged and held with their centres $20.0\ \mathrm{cm}$ apart. Their size is one twentieth of the gap, so treating each as a point at its centre costs almost nothing.
This looks like the same experiment and the formula no longer applies: the same two spheres brought until their centres are $1.5\ \mathrm{cm}$ apart, so they nearly touch. Each sphere's charge is free to move, and it crowds towards or away from the other sphere, so there is no longer a single distance $r$ to put in the formula. The number the formula returns is then not a small error, it is an answer to a different question.
separation
force
factor from the row above
$5.0\ \mathrm{cm}$
$53.9\ \mathrm{N}$
—
$10.0\ \mathrm{cm}$
$13.5\ \mathrm{N}$
divided by 4
$20.0\ \mathrm{cm}$
$3.37\ \mathrm{N}$
divided by 4
$40.0\ \mathrm{cm}$
$0.843\ \mathrm{N}$
divided by 4
Each row doubles the separation and each force is a quarter of the one above it, which is what an inverse square means in numbers rather than in symbols. Notice how fast this falls: moving the spheres from five centimetres to forty, a factor of eight, drops the force by a factor of sixty four.
The force between charges of +3.0 and −5.0 microcoulombs at 20.0 cm
Two small spheres carry $q_1 = +3.0\ \mathrm{\mu C}$ and $q_2 = -5.0\ \mathrm{\mu C}$ and their centres are $20.0\ \mathrm{cm}$ apart. Find the force each one feels, in size and direction.
The magnitude of the force and its direction on each sphere.
Solution
Convert the prefixes and the centimetres before touching the formula. Doing it inside the formula is where the factor of a million and the factor of a hundred get lost, and both mistakes leave an answer that still looks like a force.
the metres squared cancel and leave newtons, which is the unit check built into the working
Decide the direction in one sentence
$$q_1 q_2 < 0 \Rightarrow \text{attraction along the line joining them}$$
unlike signs, so each sphere is pulled towards the other, and this is the only place the signs are used
Answer $$\boxed{F = 3.37\ \mathrm{N}\ \text{on each sphere, each pulled towards the other}}$$
Check
Independent check by rescaling from a case that needs no calculator: two charges of $1.0\ \mathrm{\mu C}$ at $1.00\ \mathrm{m}$ give $F = 8.99\times 10^{-3}\ \mathrm{N}$. Multiplying the charges by 3 and 5 raises that by 15, and cutting the separation to a fifth of a metre raises it by another 25, giving $8.99\times 10^{-3}\times 375 = 3.37\ \mathrm{N}$.
Two prefix conversions, one squaring, one division, one sentence about signs. The sentence is worth as many marks as the arithmetic.
Three and a half newtons is roughly the weight of a bag of sugar, produced by charges you cannot see on two objects the size of a marble. That is the scale to remember: laboratory charges make forces you can feel, which is why the topic is worth a term.
Electric versus gravitational force between two protons
Two protons sit $1.0\times 10^{-15}\ \mathrm{m}$ apart, about the separation inside a nucleus. Find the electric repulsion, the gravitational attraction, and the ratio of the two.
Given
$q = +e = 1.602\times 10^{-19}\ \mathrm{C}$ each
$m = 1.67\times 10^{-27}\ \mathrm{kg}$ each
$r = 1.0\times 10^{-15}\ \mathrm{m}$
$k = 8.99\times 10^{9}$ and $G = 6.674\times 10^{-11}$ in SI units
Find
$F_E$, $F_G$, and $F_E/F_G$.
Solution
Compute the ratio symbolically before putting numbers in. The two laws have the same shape, so the separation cancels, and a ratio that does not depend on $r$ is a far stronger statement than two numbers at one particular distance.
Independent plausibility check on the first number: $230\ \mathrm{N}$ is the weight of a $23\ \mathrm{kg}$ mass, sitting on two particles with a mass of $10^{-27}\ \mathrm{kg}$ each. If that seems absurd, it is the right reaction, and it is exactly why something other than these two forces has to be holding a nucleus together.
Two evaluations of the same shape of formula, then one division. The division is the only part worth remembering.
The number $10^{36}$ is the reason gravity never appears again in this half of the course. Between two objects that both carry charge, the gravitational force is not small, it is negligible beyond any measurement you could make.
Checkpoint
§01.3 — scaling both charges and the separation at once●●○○○
Thirty seconds on the shape of the law, with no calculator. Two point charges attract each other with a force of magnitude $F$. Somebody then doubles both charges and, at the same time, doubles the distance between them.
Given
Original force magnitude $F$
Both charges doubled
Separation doubled
Find
(a) What is the new force magnitude?
Hint 1/4
Nothing has to be computed. Track what each change does to the force separately, then multiply the two factors together.
Hint 2/4
In $F = k|q_1||q_2|/r^{2}$, each charge enters to the first power and the separation to the second.
Hint 3/4
Substitute the changes from the question, both charges doubled and $r$ doubled: the numerator grows by $2\times 2 = 4$ and the denominator by $2^{2} = 4$.
Hint 4/4
So the two factors of four cancel and the force is unchanged at $F$.
Show solution
Form the ratio of new to old rather than computing either one, because none of $k$, $q_1$, $q_2$ or $r$ is given a value and all of them cancel.
each factor of two in the charges appears once, each factor in the separation appears twice, and here they happen to balance
Answer $$\boxed{F' = F}$$
Check
Independent check with numbers: $1.0\ \mathrm{\mu C}$ and $1.0\ \mathrm{\mu C}$ at $1.00\ \mathrm{m}$ give $8.99\ \mathrm{mN}$, and $2.0\ \mathrm{\mu C}$ with $2.0\ \mathrm{\mu C}$ at $2.00\ \mathrm{m}$ give $(8.99\times 10^{9})(4\times 10^{-12})/4 = 8.99\ \mathrm{mN}$, the same value.
Count the numerator factor and the denominator factor separately, then multiply them together at the end. Almost every wrong answer to a scaling question comes from changing one of the two and forgetting the other.
⚠ Forgetting to square the separation
the two charges are written next to each other on top and the single $r$ underneath looks symmetric with them, so the exponent gets dropped when the numbers go in
1.4Three charges at once: adding the pair forces as vectors
Extends the two charge formula to any number of charges without adding a single new idea to it.
Coulomb's law answers a question about two charges. Real problems hand you three or four, and nothing so far says what the extra ones do to the pair you already know how to handle.
RuleRule: the for forces
Conditions
each pair force is computed exactly as if the other charges were not there
no charge screens, weakens or blocks the force between two others, however it is placed between them
the sum is a vector sum: magnitudes may be added only when the forces happen to lie along the same line
Work out the force from the first charge on its own, then from the second on its own, and so on, each time pretending the rest of them do not exist. Then add those answers as arrows: all the sideways parts into one column, all the up and down parts into another. The formula for a pair never changes, no matter what else is in the room.
Two pair forces on the same charge, drawn to scale from the point where they act, and the arrow that replaces them. Neither pair force knows about the other charge, and that is exactly what makes the sum legitimate.
Looks like this, but is not
This is superposition done correctly: two forces on the same charge, one of $1.00\ \mathrm{N}$ to the right and one of $1.00\ \mathrm{N}$ to the left. They lie on one line, so the arithmetic is $1.00 - 1.00 = 0$, and adding the numbers really is adding the vectors.
This looks like the same addition and is not: the same two forces of $1.00\ \mathrm{N}$, but now at $90^{\circ}$ to each other. The sum is $1.41\ \mathrm{N}$, not $2.00\ \mathrm{N}$. Turn them to $120^{\circ}$ apart and the sum is $1.00\ \mathrm{N}$, the same as one of them alone. Three different answers from the same two magnitudes, so the magnitudes were never the whole story.
force on $q_3$
$x$ component (N)
$y$ component (N)
from $q_1$, straight down
$0$
$-1.798$
from $q_2$, along the diagonal
$+0.690$
$-0.518$
net
$+0.690$
$-2.316$
Two columns and three rows, and the last row is the answer. The reason to write it out like this rather than in a line of prose is that the final magnitude comes from the bottom row only, and a table makes it impossible to reach for the wrong two numbers.
Net force on the right hand charge of three charges on a line
Three point charges lie on the $x$ axis: $q_1 = -8.0\ \mathrm{\mu C}$ at $x = 0$, $q_2 = +3.0\ \mathrm{\mu C}$ at $x = 0.30\ \mathrm{m}$ and $q_3 = +4.0\ \mathrm{\mu C}$ at $x = 0.50\ \mathrm{m}$. Find the net force on $q_3$.
Given
$q_1 = -8.0\ \mathrm{\mu C}$ at $x=0$
$q_2 = +3.0\ \mathrm{\mu C}$ at $x=0.30\ \mathrm{m}$
$q_3 = +4.0\ \mathrm{\mu C}$ at $x=0.50\ \mathrm{m}$
The magnitude and direction of $\vec{F}$ on $q_3$.
Solution
Everything lies on one line, so the vector sum collapses to a signed sum along $x$. Setting up components in two dimensions here would be correct and would take three times as long.
Independent check on which force should win, done without the numbers: $q_2$ is smaller than $q_1$ by a factor $3/8$, but it is closer by a factor $0.20/0.50$, which raises its force by $(0.50/0.20)^{2} = 6.25$. The product $0.375\times 6.25 = 2.34$ says the near charge should win by a factor of about two and a third, and $2.70/1.15 = 2.35$.
Two applications of one formula and one subtraction. The direction sentence after each application is what makes the subtraction rather than an addition correct.
The lesson to carry forward is that distance beats charge, because charge enters once and distance enters twice. When a problem asks which of several charges dominates, look at the separations first.
Net force on a charge at the corner of a 3-4-5 triangle
Three point charges sit in a plane: $q_1 = +6.0\ \mathrm{\mu C}$ at the origin, $q_2 = +8.0\ \mathrm{\mu C}$ at $(0.40\ \mathrm{m},\,0)$ and $q_3 = -3.0\ \mathrm{\mu C}$ at $(0,\,0.30\ \mathrm{m})$. Find the net force on $q_3$, as a magnitude and a direction.
Given
$q_1 = +6.0\ \mathrm{\mu C}$ at $(0,0)$
$q_2 = +8.0\ \mathrm{\mu C}$ at $(0.40,\,0)\ \mathrm{m}$
$q_3 = -3.0\ \mathrm{\mu C}$ at $(0,\,0.30)\ \mathrm{m}$
$\vec{F}$ on $q_3$: magnitude, and angle from the $+x$ direction.
Solution
Resolve into components rather than trying to add the two arrows head to tail with the cosine rule. The cosine rule works and needs the angle between the arrows, which is itself a calculation, whereas the components are read straight off the geometry.
Independent check by bounding rather than recomputing: the largest the sum of a $1.798\ \mathrm{N}$ and a $0.863\ \mathrm{N}$ force can ever be is $2.66\ \mathrm{N}$, when they are parallel, and the smallest is $0.94\ \mathrm{N}$, when they are opposed. The answer $2.42\ \mathrm{N}$ sits inside that range and near the top of it, which is right, because the two arrows in the figure are only about $53^{\circ}$ apart.
Coulomb's law twice, one hypotenuse, one unit vector, two column sums, one arctangent. The hypotenuse and the unit vector are the two lines nobody writes down and everybody needs.
Notice that the answer was never assembled from the two magnitudes $1.798$ and $0.863$. They were converted into components immediately and never used again, and that habit is what keeps two dimensional problems from turning into guesswork.
Checkpoint
§01.4 — reading a symmetric arrangement without computing●●○○○
Half a minute, and the answer is available without a calculator if the symmetry is spotted. Two identical charges of $+q$ sit at $(-a,0)$ and $(+a,0)$. A third charge $+Q$ is placed on the $y$ axis at $(0,b)$ with $b>0$.
Given
Equal charges $+q$ at $(-a,\,0)$ and $(+a,\,0)$
A charge $+Q$ at $(0,\,b)$ with $b>0$
Find
(a) In which direction does the net force on $+Q$ point?
Hint 1/4
Do not compute either force. Draw the two arrows on the third charge and ask what the mirror symmetry of the picture forces to happen.
Hint 2/4
Superposition adds components, so a component that appears twice with opposite signs vanishes and one that appears twice with the same sign doubles.
Hint 3/4
Substitute the arrangement given, $+q$ at $(-a,0)$ and $(+a,0)$ with $+Q$ at $(0,b)$: the two $x$ components are equal and opposite, and both $y$ components point away from the axis, upwards.
Hint 4/4
So the net force points straight up, along $+y$.
Show solution
Use the mirror symmetry instead of computing, because the two magnitudes are equal by construction and only their directions differ.
the surviving column, with the factor $b/\sqrt{a^{2}+b^{2}}$ being the sine that projects each force onto the axis
Answer $$\boxed{\vec{F}_{\text{net}}\ \text{points along } +y}$$
Check
Independent check at a limit: let $b$ become very large. The expression falls off as $2kqQ/b^{2}$, which is what a single charge of size $2q$ at the origin would give, and from far away two charges $2a$ apart should indeed look like one charge of twice the size.
A mirror symmetry kills the component across the mirror and says nothing at all about the component along it. Use it to delete work, never to conclude that the whole force vanishes.
⚠ Adding the magnitudes of two forces that are not parallel
both numbers came out of the same formula and both are called forces, so they look like two terms in an ordinary sum
a charge sitting between two others looks like an obstacle, and the everyday picture of blocking something gets carried over
wrong$$F_{13} = 0\ \text{because } q_2 \text{ lies between them}$$
right$$F_{13} = \frac{k|q_1||q_3|}{r_{13}^{2}},\ \text{unchanged by } q_2$$
1.5The electric field: what the empty point already knows
Splits every force problem into a property of the source and a property of whatever you later put there.
Coulomb's law needs two charges before it says anything. That is awkward, because the ruler was doing something to the space above the desk before any paper arrived.
DefinitionDefinition: the electric field at a point
Conditions
the test charge $q_0$ is positive and small enough that it does not move the charges producing the field
the field is a property of the point, so it has a value whether or not anything is sitting there
the second line is the field of one point source only, and several sources are handled in the next concept
$$\boxed{\begin{aligned}\vec{E} &= \frac{\vec{F}}{q_0}, \qquad [\,E\,] = \mathrm{N/C}\\[3pt] E &= k\,\frac{|Q|}{r^{2}}, \qquad \text{away from } +Q,\ \text{towards } -Q\end{aligned}}$$
Put a small positive charge at the point, measure the force on it, and divide by how much charge you used. The number you get does not depend on the probe, only on everything else, so it is fair to call it a property of the point itself. Once you have it, the force on any charge placed there is that field times the charge, and nothing about the sources has to be looked at again.
The field of a single charge of $+4.0\ \mathrm{\mu C}$ sampled at three distances, with the arrows drawn to scale: $3.60\times 10^{6}$, $8.99\times 10^{5}$ and $4.00\times 10^{5}\ \mathrm{N/C}$. The third arrow is nine times shorter than the first because the distance is three times larger.
Looks like this, but is not
This is the definition working: a probe of $+3.0\ \mathrm{nC}$ at a point feels $1.2\times 10^{-4}\ \mathrm{N}$ to the right, so the field there is $4.0\times 10^{4}\ \mathrm{N/C}$ to the right. Swap the probe for one of $+6.0\ \mathrm{nC}$ and the force doubles, the division halves it back, and the field comes out the same.
This looks like the same statement and is a different quantity: saying that the field at the point is $1.2\times 10^{-4}$. That is the force on one particular probe, in newtons, and it changes the moment you change the probe. A field of $4.0\times 10^{4}\ \mathrm{N/C}$ and a force of $1.2\times 10^{-4}\ \mathrm{N}$ differ by a factor of the probe charge and by their units, and no amount of context makes them interchangeable.
Field 30.0 cm from a 4.0 microcoulomb charge, and what it does to an electron
A point charge $Q = +4.0\ \mathrm{\mu C}$ sits alone. Find the electric field at a point $30.0\ \mathrm{cm}$ from it, then find the force and the acceleration of an electron released at that point.
$E$ with its direction, then $F$ and $a$ for the electron.
Solution
Find the field first and the force second, even though Coulomb's law could give the force in one step. The field is the part that stays true when the electron is replaced by anything else, and this is the habit the rest of the course is built on.
the same second law as in mechanics, with the electron mass, and it is the mass rather than the force that makes this number extreme
Answer $$\boxed{E = 4.00\times 10^{5}\ \mathrm{N/C}\ \text{outwards},\quad F = 6.40\times 10^{-14}\ \mathrm{N}\ \text{inwards},\quad a = 7.03\times 10^{16}\ \mathrm{m/s^{2}}}$$
Check
Independent check on the field by a different route: the force between $Q$ and a $1.0\ \mathrm{C}$ charge at that distance would be $(8.99\times 10^{9})(4.0\times 10^{-6})(1.0)/(0.300)^{2} = 4.00\times 10^{5}\ \mathrm{N}$, and dividing by the one coulomb returns the same field, which is the definition read backwards.
One field, one multiplication, one division by a mass. Note that the source charge was used once and then never again.
Ten to the sixteen metres per second squared is not a misprint; it is what a laboratory field does to something as light as an electron. Whenever a number like this appears, check the mass before suspecting the field: the field here is ordinary, and the electron is not.
Getting the field from a measurement made with a negative probe
At a point P, a probe charge of $q_0 = -2.0\ \mathrm{nC}$ is observed to feel a force of $8.0\times 10^{-5}\ \mathrm{N}$ in the $+x$ direction. Find the electric field at P. Then find the force on a charge of $+5.0\ \mathrm{nC}$ placed at the same point.
$\vec{F} = 8.0\times 10^{-5}\ \mathrm{N}$ in the $+x$ direction
Second charge: $+5.0\ \mathrm{nC}$ at the same point
Find
$\vec{E}$ at P, and the force on the second charge.
Solution
Divide the magnitudes and settle the direction in a separate sentence. Dividing the vector by a negative number does work, but it hides the reversal inside a sign that is easy to lose, and the whole point of this concept is that the direction is physics rather than bookkeeping.
Independent consistency check between the two probes: the second charge is $2.5$ times the size of the first and of the opposite sign, so its force should be $2.5$ times larger and reversed. It is, $2.0\times 10^{-4}$ against $8.0\times 10^{-5}$, and along $-x$ against $+x$.
One division, one sentence about the sign, one multiplication. The sentence is where the exam mark is.
Note what was never needed: the charges producing this field, where they are, or how many of them there are. That is exactly what defining a field buys, and it is why the next concept can add fields without ever mentioning forces.
Checkpoint
§01.5 — swapping the probe at a fixed point●●●○○
Thirty seconds on what the field does and does not depend on. A charge of $+2.0\ \mathrm{nC}$ placed at a point P feels a force of $6.0\times 10^{-5}\ \mathrm{N}$. The probe is then removed and replaced by a charge of $-4.0\ \mathrm{nC}$ at the very same point, with the sources untouched.
Given
First probe $+2.0\ \mathrm{nC}$, force $6.0\times 10^{-5}\ \mathrm{N}$
Second probe $-4.0\ \mathrm{nC}$ at the same point
The charges producing the field are not moved
Find
(a) What happens to the field at P, and what force does the second probe feel?
Hint 1/4
Two separate questions are being asked. One is about the point, the other about the object placed at it, and only one of them can have changed.
Hint 2/4
The field is fixed by the sources alone, and the force on a charge placed in it is $F = |q|E$, along the field for a positive charge and against it for a negative one.
Hint 3/4
Substitute the data given: $E = (6.0\times 10^{-5})/(2.0\times 10^{-9}) = 3.0\times 10^{4}\ \mathrm{N/C}$, then $F = (4.0\times 10^{-9})(3.0\times 10^{4})$.
Hint 4/4
So the field is unchanged and the new force is $1.2\times 10^{-4}\ \mathrm{N}$, pointing the opposite way to the first one.
Show solution
Get the field out of the first measurement before touching the second probe, so that the second answer is one multiplication rather than a fresh problem.
Independent check by ratio rather than by recomputation: the second probe is twice the size of the first, so its force must be exactly twice as large whatever the field happens to be, and $1.2\times 10^{-4}$ is twice $6.0\times 10^{-5}$.
The field belongs to the sources; the probe only reads it. Any question that swaps the object at the point while leaving the sources alone is one multiplication, not a fresh problem.
⚠ Dividing by the signed probe charge and then also reversing the direction
both moves are individually correct, and doing them together feels safer than doing either alone
right$$E = \frac{F}{|q_0|},\ \text{then reverse once because } q_0<0 \Rightarrow \vec{E}\ \text{along } -x$$
⚠ Quoting a field in newtons
the field was computed by dividing a force, and the newtons in the numerator are the unit that stays in mind
wrong$$E = 4.0\times 10^{4}\ \mathrm{N}$$
right$$E = 4.0\times 10^{4}\ \mathrm{N/C}$$
⚠ Thinking the field disappears when no charge is there
the definition is written with a test charge in it, so the test charge looks like part of the situation rather than part of the measurement
wrong$$\text{no charge at P} \Rightarrow E_P = 0$$
right$$E_P = \frac{k|Q|}{r^{2}}\ \text{regardless of what is at P}$$
1.6The field of several point charges, and the place where it vanishes
Adds the fields of several sources at one point, and finds the point where they cancel exactly.
The field was defined from one source charge. Real arrangements have several, and the rule for combining them is the one already used for forces, because a field is a force divided by a number.
RuleRule: superposition of electric fields
Conditions
each source contributes $k|Q_i|/r_i^{2}$ with $r_i$ measured from that source to the field point, never between the sources
the direction of each contribution is away from a positive source and towards a negative one, decided one source at a time
the sizes may be added or subtracted directly only when the contributions lie along one line
Compute the field each charge would produce at the point if it were alone, complete with direction, then add those arrows. Nothing in the recipe changes when the charges are unlike or unequal; what changes is whether the arrows reinforce or fight each other, and where along the line they manage to cancel completely.
The field at P from a positive and a negative source, added as arrows. The two contributions are nearly the same size and point almost opposite ways, which is why the sum is smaller than either of them.
Looks like this, but is not
Two like charges have a null point between them: with $+5.0\ \mathrm{\mu C}$ at the origin and $+2.0\ \mathrm{\mu C}$ at $x = 0.60\ \mathrm{m}$, the two fields point in opposite directions everywhere between them, so somewhere in the gap they must be equal in size and cancel.
Change one sign and the search has to move outside the gap: with $+5.0\ \mathrm{\mu C}$ and $-2.0\ \mathrm{\mu C}$ at the same two places, both fields point the same way at every point between the charges, so they can never cancel there. The null point is out beyond the smaller charge, at $x = 1.63\ \mathrm{m}$, and looking for it between them wastes the whole question.
$x$ from the larger charge
$E_1$, pointing right
$E_2$, pointing left
$0.30\ \mathrm{m}$
$4.99$
$2.00$
$0.368\ \mathrm{m}$
$3.32$
$3.34$
$0.40\ \mathrm{m}$
$2.81$
$4.50$
The first row has the stronger charge winning and the last row has the weaker one winning, so the crossing has to be between them, and the middle row is sitting on it. Bracketing like this is worth doing even when you intend to solve the equation, because it tells you immediately whether the algebraic answer is in the right place.
Where the field vanishes between charges of +5.0 and +2.0 microcoulombs
A charge $q_1 = +5.0\ \mathrm{\mu C}$ sits at $x=0$ and $q_2 = +2.0\ \mathrm{\mu C}$ at $x = 0.60\ \mathrm{m}$. Find the point on the $x$ axis where the net electric field is zero.
Given
$q_1 = +5.0\ \mathrm{\mu C}$ at $x = 0$
$q_2 = +2.0\ \mathrm{\mu C}$ at $x = 0.60\ \mathrm{m}$
The value of $x$ where $\vec{E}_{\text{net}} = 0$.
Solution
Decide the region before doing any algebra. Outside the pair the two fields point the same way and can never cancel, so only the gap between them is worth searching, and knowing that turns a quadratic with two roots into a single sensible one.
Locate the region where cancellation is possible
$$0 < x < 0.60\ \mathrm{m}$$
between two positive charges the two fields point away from their own source, hence in opposite directions, which is the only way to cancel
Independent check by evaluating both fields there rather than by re-solving: $E_1 = (8.99\times 10^{9})(5.0\times 10^{-6})/(0.368)^{2} = 3.32\times 10^{5}\ \mathrm{N/C}$ and $E_2 = (8.99\times 10^{9})(2.0\times 10^{-6})/(0.232)^{2} = 3.34\times 10^{5}\ \mathrm{N/C}$. They agree to the rounding of the position, and they point opposite ways.
One region argument, one square root, one linear equation. The region argument is what stops the quadratic producing a second, meaningless answer outside the gap.
The null point is always nearer the smaller charge, and here it sits at $0.368$ against $0.232$, which is a ratio of $1.58$, and $\sqrt{5/2} = 1.58$. That relation is worth remembering as a check: the distances are in the ratio of the square roots of the charges.
Net field at a point off the axis of two unequal charges
A charge $q_1 = +4.0\ \mathrm{\mu C}$ sits at the origin and $q_2 = -6.0\ \mathrm{\mu C}$ at $(0.30\ \mathrm{m},\,0)$. Find the electric field at the point $P = (0,\,0.40\ \mathrm{m})$, and then the force on a charge of $-2.0\ \mathrm{nC}$ placed at P.
Given
$q_1 = +4.0\ \mathrm{\mu C}$ at $(0,0)$
$q_2 = -6.0\ \mathrm{\mu C}$ at $(0.30,\,0)\ \mathrm{m}$
$P = (0,\,0.40)\ \mathrm{m}$
Test charge at P: $-2.0\ \mathrm{nC}$
Find
$\vec{E}$ at P as a magnitude and angle, then $\vec{F}$ on the test charge.
Solution
Get the field first and attach the test charge at the very end. Computing two Coulomb forces on the test charge directly would give the same answer with two extra multiplications by $2.0\times 10^{-9}$, and would hide the fact that the field at P is the same whatever is put there.
the field is already known, so the sources play no further part
$$\text{direction: } 21.9^{\circ}\ \text{below the } -x \text{ direction}$$
the charge is negative, so the force is exactly opposite to the field
Answer $$\boxed{E = 1.40\times 10^{5}\ \mathrm{N/C}\ \text{at } 21.9^{\circ}\ \text{above } +x,\qquad F = 2.79\times 10^{-4}\ \mathrm{N}\ \text{opposite to it}}$$
Check
Independent route by the cosine rule instead of components: the two contributions are $2.25$ and $2.16$ in units of $10^{5}$, and the angle between them is $143^{\circ}$, since one points along $+y$ and the other along $(0.60,-0.80)$. Then $E^{2} = 2.25^{2}+2.16^{2}+2(2.25)(2.16)\cos 143^{\circ}$ gives $E = 1.40\times 10^{5}\ \mathrm{N/C}$, the same number from a calculation that shares no step with the first.
Two field magnitudes, one hypotenuse, one unit vector, two column sums, one arctangent, one multiplication. Six of those eight lines are geometry.
Two contributions of almost the same size gave a resultant well below either of them. That is the general lesson for fields near unlike charges, and it is why an answer bigger than the largest single contribution should always be treated as an arithmetic error until proved otherwise.
Checkpoint
§01.6 — which side of the gap the null point sits on●●●○○
Half a minute, with no arithmetic needed. Two positive charges are held apart on a line: $+9.0\ \mathrm{\mu C}$ on the left and $+1.0\ \mathrm{\mu C}$ on the right. There is exactly one point between them where the net field vanishes.
Given
$+9.0\ \mathrm{\mu C}$ at the left end
$+1.0\ \mathrm{\mu C}$ at the right end
The null point lies somewhere between them
Find
(a) Whereabouts between the charges does the null point lie?
Hint 1/4
The point has to be further from the strong charge than from the weak one. The only question is by how much.
Hint 2/4
Setting $kQ_1/r_1^{2} = kQ_2/r_2^{2}$ gives $r_1/r_2 = \sqrt{Q_1/Q_2}$, so distances go as the square roots of the charges.
Hint 3/4
Substitute the data given, $Q_1 = 9.0\ \mathrm{\mu C}$ and $Q_2 = 1.0\ \mathrm{\mu C}$: $r_1/r_2 = \sqrt{9} = 3$, so the gap splits three to one.
Hint 4/4
So the null point is three quarters of the way from the large charge towards the small one.
Show solution
Work with the ratio of the distances rather than with a value for $d$, because $d$ is not given and cancels out of the condition anyway.
the two distances have to fill the gap, which converts the ratio into two positions
Answer $$\boxed{r_1 = \tfrac{3}{4}d\ \text{from the large charge}}$$
Check
Independent check by evaluating the two fields at the quarter point instead: there $r_1 = d/4$ and $r_2 = 3d/4$, giving $E_1 \propto 9/(1/16) = 144$ against $E_2 \propto 1/(9/16) = 1.78$, nowhere near equal, which confirms that the null point is not on the strong side.
The null point always lies on the side of the smaller charge, and the two distances go as the square roots of the charges rather than as the charges. That single ratio settles every "whereabouts" version of this question without a calculator.
⚠ Hunting for a null point between two unlike charges
cancellation feels like something that should happen in the middle, and the words between the charges are in every worked example of the like charge case
the charges are the numbers in the question and the square root only appears after the equation is written down, so it gets skipped when the answer is guessed
Any time exactly two charges are involved, and as the inner loop of every problem with more than two.
Convert before you compute
Turn every charge into coulombs and every length into metres, on their own line, before the formula is written. Micro is $10^{-6}$, nano is $10^{-9}$, and a centimetre is $10^{-2}\ \mathrm{m}$ which becomes $10^{-4}$ once it is squared.
Find the distance between the two charges
Not from the origin, and in two dimensions not from a coordinate difference. If the charges are at $(x_1,y_1)$ and $(x_2,y_2)$, the distance is the hypotenuse, and it is worth writing down as its own line.
Put magnitudes into the formula
Evaluate $F = k|q_1||q_2|/r^{2}$ with the sizes of the charges only. The number that comes out is positive, and if it is not, a sign got in where it does not belong.
Decide the direction in one sentence
Like signs push apart, unlike signs pull together, always along the line joining the two charges. Write that sentence down; it is worth marks and it is the only place the signs are used.
Check the size against something you know
Two microcoulombs at a metre give about $18\ \mathrm{mN}$. Scale from that: if your answer is a million newtons or a nanonewton, a prefix or a square is missing.
Where it goes wrong
The separation is squared in the formula but the number typed in was not squared, which shows up as an answer too large by the value of $r$.
A charge went in as 3.0 rather than $3.0\times 10^{-6}$, giving an answer larger by $10^{6}$ or $10^{12}$ that still has newtons on it.
Signed charges were used, so the answer carries a minus sign that gets read as a direction and then contradicts the direction sentence.
Net force or field from several charges, by components
Three or more charges, or any arrangement that is not on a single straight line.
Draw the axes and mark every charge
Put the coordinates on the diagram, including the point you are asked about. Half the errors in this topic are geometry errors, and they are all visible on a sketch.
Take the sources one at a time
For each source, compute its own contribution as if it were the only charge present. Nothing screens anything, and a charge sitting between two others changes neither of their contributions.
Attach a direction to each contribution
For a force, like repels and unlike attracts. For a field, away from a positive source and towards a negative one. Write the unit vector as the displacement divided by its own length.
Resolve into two columns
Multiply each magnitude by its unit vector to get an $x$ part and a $y$ part. A table with one row per source and two columns is the cheapest insurance available in this topic.
Add each column separately
Sum the $x$ entries, sum the $y$ entries. This is the only step where the different sources are allowed to interact, and magnitudes never meet each other here.
Rebuild the arrow and sanity check it
Magnitude from the summed components, angle from their ratio, quadrant from their signs. The result must lie between the difference and the sum of the contributions; outside that range something is wrong.
Where it goes wrong
Magnitudes are added directly, which is right only when every contribution lies on one line and is otherwise always too large.
A coordinate difference is used as the distance, so the diagonal source gets a distance that is too small and a contribution that is too big.
The arctangent is taken without looking at the signs, putting the answer in the wrong quadrant, typically off by $180^{\circ}$.
Finding the point where the net field is zero
Two point charges on a line, and a question asking where a third charge would feel no force, or where the field vanishes.
Decide the region first
For two like charges the null point is between them. For two unlike charges it is outside the pair, on the side of the smaller magnitude. Getting this wrong wastes the whole question, because the algebra will still produce a number.
Name one unknown distance
Call the distance from one charge $x$ and write the other distance in terms of it, as $d-x$ inside the gap or $x-d$ outside it. Two unknowns where one will do is the commonest way to get stuck.
Set the two magnitudes equal
$k|Q_1|/r_1^{2} = k|Q_2|/r_2^{2}$. The directions were settled in step one, so only sizes are compared here and $k$ cancels immediately.
Take the square root before cross multiplying
Rearranging to $r_1/r_2 = \sqrt{|Q_1|/|Q_2|}$ keeps the equation linear and avoids the second, meaningless root that squaring introduces.
Check by evaluating both fields there
Put the answer back and compute both contributions. They should agree to the rounding you kept, and the point should be nearer the smaller charge.
Where it goes wrong
The gap is split in the ratio of the charges rather than of their square roots, which is a large error whenever the charges differ by more than a little.
A null point is looked for between two unlike charges, where the two fields point the same way and can never cancel.
Squaring both sides produces a quadratic and the wrong root is chosen, giving a point outside the region argued for in step one.
Field at the midpoint between +2.0 and +8.0 microcoulombs
Charges $q_1 = +2.0\ \mathrm{\mu C}$ at $x=0$ and $q_2 = +8.0\ \mathrm{\mu C}$ at $x=0.40\ \mathrm{m}$. Find the electric field at the midpoint $x = 0.20\ \mathrm{m}$.
Given
$q_1 = +2.0\ \mathrm{\mu C}$ at $x=0$
$q_2 = +8.0\ \mathrm{\mu C}$ at $x=0.40\ \mathrm{m}$
Field point at $x = 0.20\ \mathrm{m}$
Find
$\vec{E}$ at the midpoint.
Solution
Both distances are the same here, so the two contributions differ only through the charges, and the arithmetic collapses to one subtraction.
The distances are equal, so the two fields must be in the ratio of the charges, $8.0$ to $2.0$, and $1.798/0.450 = 4.0$ exactly. The answer must then be three quarters of the larger contribution, and $0.75\times 1.798 = 1.35$.
A field question never names an object sitting at the point of interest, so no test charge may enter the working and the answer comes out in newtons per coulomb. Adding fields is the same component sum as adding forces, with the charge at the point taken out of it.
Force on a −5.0 nC charge at that same midpoint
Same two charges: $q_1 = +2.0\ \mathrm{\mu C}$ at $x=0$ and $q_2 = +8.0\ \mathrm{\mu C}$ at $x=0.40\ \mathrm{m}$. A charge of $-5.0\ \mathrm{nC}$ is now placed at $x = 0.20\ \mathrm{m}$. Find the force on it.
Given
$q_1 = +2.0\ \mathrm{\mu C}$ at $x=0$
$q_2 = +8.0\ \mathrm{\mu C}$ at $x=0.40\ \mathrm{m}$
$q_3 = -5.0\ \mathrm{nC}$ at $x = 0.20\ \mathrm{m}$
Find
$\vec{F}$ on the placed charge.
Solution
Reuse the field from the previous problem rather than computing two Coulomb forces. The geometry has not changed, so the only new information is the charge that was put there.
Independent route, computing the two Coulomb forces directly: $F_1 = k(2.0\times 10^{-6})(5.0\times 10^{-9})/(0.20)^{2} = 2.25\times 10^{-3}\ \mathrm{N}$ towards $-x$ and $F_2 = 8.99\times 10^{-3}\ \mathrm{N}$ towards $+x$, whose difference is $6.74\times 10^{-3}\ \mathrm{N}$ towards $+x$.
Once the field at a point is known, the force on anything placed there is a single multiplication, and the sign of that charge decides the direction. Compute the field first even when the question asks for a force: it is the half of the work that can be reused.
Identical geometry, identical sources, and two answers that differ by a factor of $5.0\times 10^{-9}$ and by a reversal: the field is a property of the empty point and comes out in newtons per coulomb, the force belongs to the object you put there and comes out in newtons.
How to tell them apart
Count what the question mentions. If it names a charge sitting at the point of interest, the answer is a force in newtons and the sign of that charge decides the direction. If it names only a location, the answer is a field in newtons per coulomb, and no test charge may appear anywhere in the working.
Scaffolding comes off
The common skeleton
Sketch the charges with their coordinates, and mark the one the question asks about.
Convert every charge to coulombs and every distance to metres before writing any formula.
For each source separately, find the distance to the target and evaluate $k|q||Q|/r^{2}$ with magnitudes only.
Give each contribution a direction: like repels, unlike attracts, always along the line joining the pair.
Resolve into components if the contributions are not collinear, then add each column on its own.
Rebuild the magnitude and angle, and check the result lies between the difference and the sum of the contributions.
1 · fully worked
Net force on the far charge of three charges on a line
Three point charges lie on the $x$ axis: $q_1 = +3.0\ \mathrm{\mu C}$ at $x = 0$, $q_2 = -2.0\ \mathrm{\mu C}$ at $x = 0.40\ \mathrm{m}$, and $q_3 = +5.0\ \mathrm{\mu C}$ at $x = 0.70\ \mathrm{m}$. Find the net force on $q_3$.
Given
$q_1 = +3.0\ \mathrm{\mu C}$ at $x = 0$
$q_2 = -2.0\ \mathrm{\mu C}$ at $x = 0.40\ \mathrm{m}$
$q_3 = +5.0\ \mathrm{\mu C}$ at $x = 0.70\ \mathrm{m}$
Find
The magnitude and direction of the net force on $q_3$.
Solution
Everything is collinear, so the components reduce to signs along $x$, and the whole vector step is one subtraction. Setting up a full two dimensional table here would be correct and pointless.
Independent check on which contribution should dominate: $q_2$ is smaller than $q_1$ by $2/3$ but nearer by $0.30$ against $0.70$, which multiplies its force by $(0.70/0.30)^{2} = 5.44$. The product $0.667\times 5.44 = 3.63$, and the computed ratio $0.999/0.275 = 3.63$ agrees.
Coulomb's law twice, two direction sentences, one subtraction.
The charge that wins is the near one even though it is the smaller one, because charge enters the formula once and distance enters it twice.
2 · you write the reasoning
Easier than the one above: the target charge sits between the other two, and both forces on it point the same way, so nothing cancels. Charges are $q_1 = +2.0\ \mathrm{\mu C}$ at $x=0$, $q_2 = -4.0\ \mathrm{\mu C}$ at $x = 0.50\ \mathrm{m}$, and $q_3 = +1.0\ \mathrm{\mu C}$ at $x = 0.25\ \mathrm{m}$. The steps are done for you. Write, in the empty column, why each one is allowed.
$$r_{13} = r_{23} = 0.25\ \mathrm{m}$$
reasoning
The target sits exactly halfway between the two sources, so both separations are $0.50/2 = 0.25\ \mathrm{m}$. Writing this once saves computing it twice and makes the two forces directly comparable, since only the charges will differ.
Coulomb's law with magnitudes, so the number is a size. The direction is decided separately: $q_1$ and $q_3$ are both positive, so $q_3$ is pushed away from $q_1$, and $q_1$ is on its left, so the push is towards $+x$.
The same law with the same separation, so the only change is the charge, which is twice as large and gives twice the force. The direction here comes from unlike signs: $q_3$ is pulled towards $q_2$, which sits on its right, so this force also points towards $+x$.
$$F = 0.288 + 0.575 = 0.863\ \mathrm{N}$$ towards $+x$
reasoning
Both forces lie on one line and point the same way, so the vector sum really is the arithmetic sum. This is the one situation where adding magnitudes is legitimate, and it is legitimate because the directions were checked first and found to agree.
3 · find the buried error
Harder than the rung above, because the charges are no longer on one line, and the work has been done for you badly. Charges are $q_1 = +6.0\ \mathrm{\mu C}$ at $(0,\,0)$, $q_2 = -4.0\ \mathrm{\mu C}$ at $(0.40,\,0)\ \mathrm{m}$, and the target is $q_3 = +2.0\ \mathrm{\mu C}$ at $(0.40,\,0.30)\ \mathrm{m}$. Exactly two of the four steps below are wrong. Find both.
Step 1. $q_2$ is directly below $q_3$, so $r_{23} = 0.30\ \mathrm{m}$ and $F_{23} = 0.799\ \mathrm{N}$, pulling $q_3$ downwards.
Step 2. $q_1$ is $0.40\ \mathrm{m}$ away, so $F_{13} = 0.674\ \mathrm{N}$.
Step 3. $q_1$ and $q_3$ are both positive, so $\vec{F}_{13}$ points away from the origin, along $(0.80,\ 0.60)$.
Step 4. The net force is $0.674 + 0.799 = 1.47\ \mathrm{N}$.
the two buried errors (2)
⚠ step 2
The separation between $q_1$ and $q_3$ was taken as $0.40\ \mathrm{m}$, which is only the horizontal gap. The charges are at $(0,0)$ and $(0.40,\,0.30)$, so the separation is the hypotenuse, $r_{13} = 0.50\ \mathrm{m}$.
The number $0.40$ is written on the diagram and the number $0.50$ is not; it has to be computed, and a step that has to be computed is a step that can be skipped without noticing.
right
$r_{13} = \sqrt{(0.40)^{2}+(0.30)^{2}} = 0.50\ \mathrm{m}$, so $F_{13} = 0.432\ \mathrm{N}$, not $0.674\ \mathrm{N}$.
⚠ step 4
The two magnitudes were added arithmetically. They are not parallel: one points along $(0.80,\ 0.60)$ and the other straight down, so only their components may be added.
Both numbers came out of the same formula and both are labelled as forces, so they look like two terms of an ordinary sum, and the sum is even bigger than the correct answer, which feels safe.
right
$F_x = 0.345$, $F_y = 0.259 - 0.799 = -0.540$, so $F = \sqrt{0.345^{2}+0.540^{2}} = 0.641\ \mathrm{N}$ at $57.4^{\circ}$ below the $+x$ direction.
4 · the bare problem
§01.4 — net force at the apex of a symmetric arrangement●●●○○
The same skeleton, with none of it written out. Two equal positive charges are held at the ends of a base line and a negative charge sits above the midpoint, at the apex of an isosceles triangle. Nothing here needs a new idea, only the six steps used above.
Given
$q_1 = +3.0\ \mathrm{\mu C}$ at $(0,\,0)$
$q_2 = +3.0\ \mathrm{\mu C}$ at $(0.60,\,0)\ \mathrm{m}$
$q_3 = -5.0\ \mathrm{\mu C}$ at $(0.30,\,0.40)\ \mathrm{m}$
(b) State its direction, and say which step of the working made the sideways parts disappear.
Hint 1/4
Before computing anything, look at the arrangement in a mirror placed through the apex. Which component of the answer can survive that reflection?
Hint 2/4
Each pair force is $k|q||Q|/r^{2}$ along the line joining the pair, attractive here since the signs are unlike, and the two contributions are added by components.
Hint 3/4
Substitute the data given, $q_1 = q_2 = +3.0\ \mathrm{\mu C}$ at $(0,0)$ and $(0.60,0)$ with $q_3 = -5.0\ \mathrm{\mu C}$ at $(0.30,\,0.40)$: each distance is $\sqrt{0.30^{2}+0.40^{2}} = 0.50\ \mathrm{m}$, so each force is $0.539\ \mathrm{N}$ with a downward part of $0.539\times 0.80$.
Hint 4/4
So the net force is $2\times 0.539\times 0.80 = 0.863\ \mathrm{N}$, straight down.
Show solution
Exploit the symmetry before computing, so that only one distance and one magnitude have to be worked out and the second contribution comes for free.
Independent check on the size by bounding: each contribution is $0.539\ \mathrm{N}$ and the angle between them is $2\arctan(0.60/0.80) = 73.7^{\circ}$, so the sum must be less than $1.08\ \mathrm{N}$ and more than zero, and closer to the upper bound than the lower since the arrows are less than a right angle apart. The value $0.863\ \mathrm{N}$ is $80\%$ of the bound, which fits.
In a symmetric arrangement, work out one distance and one magnitude and let the mirror supply the second contribution. Knowing which step performs the cancellation, the column addition, is what lets you skip that half of the arithmetic on purpose rather than by luck.
Full exam-style question
Two positive charges on a line: field, force and the null pointexam format
Point charges $Q_1 = +9.0\ \mathrm{\mu C}$ and $Q_2 = +4.0\ \mathrm{\mu C}$ are fixed on the $x$ axis at $x = 0$ and $x = 0.50\ \mathrm{m}$. (a) Find the electric field at the midpoint of the line joining them. (b) Find the force on a charge of $-4.0\ \mathrm{nC}$ placed at that midpoint. (c) Find the point on the axis where the net field is zero.
Given
$Q_1 = +9.0\ \mathrm{\mu C}$ at $x=0$
$Q_2 = +4.0\ \mathrm{\mu C}$ at $x=0.50\ \mathrm{m}$
$\vec{E}$ at the midpoint, the force on the placed charge, and the position where $\vec{E}=0$.
Solution
Part (b) is answered from part (a) rather than from two fresh Coulomb forces, and part (c) is answered by a ratio rather than by a quadratic. Both choices exist because a three part question has to be finished under time pressure, and both shortcuts are safer as well as faster.
(a) The two contributions at the midpoint
$$r_1 = r_2 = 0.25\ \mathrm{m}$$
the midpoint is equidistant from both, so a single distance serves both contributions
Independent check on part (c) by evaluating both fields at $x = 0.30\ \mathrm{m}$: $E_1 = (8.99\times 10^{9})(9.0\times 10^{-6})/(0.30)^{2} = 8.99\times 10^{5}\ \mathrm{N/C}$ and $E_2 = (8.99\times 10^{9})(4.0\times 10^{-6})/(0.20)^{2} = 8.99\times 10^{5}\ \mathrm{N/C}$. They are equal to three figures and point opposite ways, so the net field there really is zero.
Three parts, and only part (a) contains a Coulomb calculation. Parts (b) and (c) are one multiplication and one ratio, which is what the structure of the question is testing.
The shape to carry into the exam is that field questions come in this order: field first, then anything you place in it, then the special points. Answering part (b) from scratch is not wrong, it just costs the minutes that part (c) needs.
Practice
A · concept 4 questions
1§01.1 — does rubbing make new charge●○○○○
A classmate explains the ruler demonstration by saying that friction generates electric charge, in the way that friction generates heat, and that the harder you rub the more charge you make.
Given
Claim: rubbing a plastic rod with a cloth creates electric charge on the rod.
Find
(a) True or false, with one sentence of justification.
Hint 1/4
Do not look only at the rod. Ask what state the cloth is in afterwards, and whether the pair together is different from before.
Hint 2/4
The total charge of an isolated system is constant; charging is a transfer between its parts and never a creation.
Hint 3/4
Substitute the case from earlier in this section: the rod ends at $-1.6\times 10^{-7}\ \mathrm{C}$ and the cloth at $+1.6\times 10^{-7}\ \mathrm{C}$, and the sum is zero, as it was at the start.
Hint 4/4
So the claim is false: rubbing separates charge that was already there.
Show solution
Argue from the pair rather than from the rod, because the whole illusion comes from testing only one of the two objects.
the total is unchanged, which is what creating something would have to violate
Answer $$\boxed{\text{False: charge is transferred, not created}}$$
Check
Independent check that does not use conservation as a premise: hang the cloth up as well, and it attracts the same paper punchings the rod does. If charge had been created on the rod alone, the cloth would have stayed inert.
Charge is moved, never made, so any account of a charging demonstration has to name both objects. When a claim about charge sounds like a claim about heat, look for the partner nobody tested.
2§01.2 — where the charge goes on a metal bar●●○○○
A long metal bar rests on two insulating supports. Somebody touches one end of it briefly with a heavily charged negative rod, then takes the rod away without touching anything else.
Given
The bar is metal, on insulating supports
The rod is negative and touches one end only
The rod is then removed
Find
(a) Where is the charge on the bar a moment later?
Hint 1/4
Two separate questions: how much charge is on the bar, and where on the bar it sits. The material decides only the second one.
Hint 2/4
In a conductor the excess charge is free to move, and since like charges repel it settles as far apart as the object allows.
Hint 3/4
Substitute the case given, a metal bar on insulating supports touched by a negative rod: the extra electrons can travel along the bar, and the supports give them no way off it.
Hint 4/4
So the whole bar ends up negatively charged, with the surplus spread along it.
Show solution
Separate the amount from the arrangement. The amount is fixed by the contact, and only the arrangement depends on the material.
How much, then where
$$Q_{\text{bar}} < 0\ \text{after contact}$$
contact lets electrons cross, so the bar keeps a genuine surplus rather than a rearrangement
$$\text{conductor} \Rightarrow \text{charge free to move}$$
the defining property of a metal, and the reason the answer differs from the plastic case
$$\text{like charges repel} \Rightarrow \text{spread over the bar}$$
the surplus arranges itself to be as far apart as it can, which for an isolated bar means along its whole length
Answer $$\boxed{\text{negative, spread over the whole bar}}$$
Check
Independent check with a second measurement: bring a small charged tester near the far end. It responds, which it would not do if the charge had stayed at the touched end, and the response is a repulsion for a negative tester rather than the attraction a neutral bar would give.
Keep the amount of charge apart from its arrangement: the contact fixes the amount, the material fixes the arrangement. That split answers every "where does the charge sit" question, in this section and later ones.
3§01.3 — scaling one charge and the separation together●●○○○
Two point charges sit a fixed distance apart and repel each other with a force of $12\ \mathrm{N}$. Somebody proposes to triple one of the charges and, at the same time, triple the separation, expecting the two changes to cancel.
Given
Original force $12\ \mathrm{N}$
One charge tripled
Separation tripled
Find
(a) True or false: the force is unchanged at $12\ \mathrm{N}$.
Hint 1/4
The intuition being tested is that a factor of three up and a factor of three down must cancel. Check whether both threes enter the formula in the same way.
Hint 2/4
In Coulomb's law each charge appears to the first power and the separation to the second.
Hint 3/4
Substitute the changes from the question, one charge times 3 and $r$ times 3: the force is multiplied by $3/3^{2} = 1/3$, so $12\ \mathrm{N}$ becomes $4.0\ \mathrm{N}$.
Hint 4/4
So the statement is false; the force drops to $4.0\ \mathrm{N}$.
Show solution
Use the ratio of new to old so that $k$, both charges and $r$ all cancel; none of them is given a value and none is needed.
applying the ratio to the one force that was measured
Answer $$\boxed{F' = 4.0\ \mathrm{N}}$$
Check
Independent check with concrete numbers: $1.0\ \mathrm{\mu C}$ and $1.0\ \mathrm{\mu C}$ at $1.00\ \mathrm{m}$ give $8.99\ \mathrm{mN}$; tripling one charge and the distance gives $(8.99\times 10^{9})(3\times 10^{-12})/9 = 3.00\ \mathrm{mN}$, which is a third of the original.
Charge enters the force once and distance enters it twice, so equal factors never cancel. Scale a charge and a separation by the same number and the force changes by that number, which is the general form of what happened here.
4§01.3 — the pair of forces between very unequal charges●●○○○
A charge of $+1.0\ \mathrm{\mu C}$ and a charge of $+9.0\ \mathrm{\mu C}$ are held $10\ \mathrm{cm}$ apart. A classmate says the big one must obviously push harder, since it is nine times the charge.
Given
$q_1 = +1.0\ \mathrm{\mu C}$
$q_2 = +9.0\ \mathrm{\mu C}$
Separation $10\ \mathrm{cm}$
Find
(a) How do the two forces compare?
Hint 1/4
Write down the force on the first charge and the force on the second as two separate expressions, then compare them symbol by symbol.
Hint 2/4
Coulomb's law contains the product $|q_1||q_2|$, which is the same number whichever charge you regard as the one being pushed.
Hint 3/4
Substitute the data from the question: both forces are $k(1.0\times 10^{-6})(9.0\times 10^{-6})/(0.10)^{2} = 8.1\ \mathrm{N}$.
Hint 4/4
So the two forces are both $8.1\ \mathrm{N}$, pointing in opposite directions.
Show solution
Compute both explicitly rather than appealing to the third law, because the point of the question is that the formula already contains it.
Independent check on the consequences rather than the algebra: if the forces were unequal, the pair on its own would have a net force and would accelerate away with nothing pushing it, which nobody has ever seen a pair of charges do.
The law contains the product of the two charges, and a product does not care which side you read it from. Whenever a question invites you to say the bigger charge pushes harder, the forces are equal and it is the accelerations that differ.
B · computation 7 questions
1§01.1 — counting the electrons behind a measured charge●○○○○
A small metal pellet is charged in a laboratory experiment and an electrometer reports its charge as $-3.2\ \mathrm{\mu C}$. The demonstrator asks how many electrons that surplus corresponds to.
Given
$Q = -3.2\ \mathrm{\mu C}$
$e = 1.602\times 10^{-19}\ \mathrm{C}$
Find
(a) Find the number of excess electrons on the pellet.
Hint 1/4
You are being asked how many grains make up a given pile, so the answer is a division and not a multiplication.
Hint 2/4
Charge is quantised, $Q = ne$, so the count is $n = |Q|/e$.
Hint 3/4
Substitute the data from the question, $|Q| = 3.2\times 10^{-6}\ \mathrm{C}$ and $e = 1.602\times 10^{-19}\ \mathrm{C}$: $n = (3.2\times 10^{-6})/(1.602\times 10^{-19})$.
Hint 4/4
So $n = 2.0\times 10^{13}$ electrons.
Show solution
Divide magnitudes and report the sign in words, because a negative count of particles would be meaningless.
Convert, then divide
$$|Q| = 3.2\times 10^{-6}\ \mathrm{C}$$
the elementary charge is in coulombs, so the prefix has to go before anything else happens
Independent check by scaling from the worked example: one microcoulomb was $6.2\times 10^{12}$ electrons, and $3.2$ microcoulombs should be $3.2$ times that, giving $2.0\times 10^{13}$.
Divide magnitudes and carry the sign in words. A count of particles is a positive whole number, and the sign of the charge lives in the phrase "excess electrons" or "missing electrons", never in the count.
2§01.3 — the force between two charged spheres●●○○○
Two small spheres hang from insulating threads with their centres $15.0\ \mathrm{cm}$ apart. One carries $+6.0\ \mathrm{\mu C}$ and the other $-2.0\ \mathrm{\mu C}$, and each is small enough to be treated as a point.
(a) Find the magnitude of the force on each sphere, and say whether the threads hang towards or away from each other.
Hint 1/4
One number and one word are wanted. The number comes from the formula with magnitudes, the word from the two signs.
Hint 2/4
$F = k|q_1||q_2|/r^{2}$, with unlike signs meaning attraction along the line joining the centres.
Hint 3/4
Substitute the data from the question, $|q_1| = 6.0\times 10^{-6}$, $|q_2| = 2.0\times 10^{-6}$ and $r = 0.150\ \mathrm{m}$: $F = (8.99\times 10^{9})(1.2\times 10^{-11})/(0.0225)$.
Hint 4/4
So $F = 4.79\ \mathrm{N}$ on each sphere, and the threads hang towards each other.
Show solution
Convert the centimetres before squaring rather than after; squaring first and converting later is where the factor of ten thousand goes missing.
Independent check by rescaling a known case: $1.0\ \mathrm{\mu C}$ pairs at $1.00\ \mathrm{m}$ give $8.99\ \mathrm{mN}$. Multiplying the charges by $6\times 2 = 12$ and shrinking the distance to $0.150\ \mathrm{m}$, which multiplies by $1/0.0225 = 44.4$, gives $8.99\times 10^{-3}\times 12\times 44.4 = 4.79\ \mathrm{N}$.
Convert every length to metres before it goes anywhere near the square. A factor of a hundred lost in the distance becomes a factor of ten thousand lost in the answer, which is the most expensive slip available in this section.
3§01.5 — the field close to a small charge●●○○○
A charge of $+2.0\ \mathrm{nC}$ sits alone on an insulating stand, and a field probe is placed $5.0\ \mathrm{cm}$ from it with nothing else nearby.
(a) Find the magnitude and direction of the electric field at the probe.
Hint 1/4
Nothing is placed at the point, so no test charge appears anywhere in the working. Only the source and the distance matter.
Hint 2/4
$E = k|Q|/r^{2}$, pointing away from a positive source.
Hint 3/4
Substitute the data given, $|Q| = 2.0\times 10^{-9}\ \mathrm{C}$ and $r = 0.050\ \mathrm{m}$: $E = (8.99\times 10^{9})(2.0\times 10^{-9})/(0.0025)$.
Hint 4/4
So $E = 7.2\times 10^{3}\ \mathrm{N/C}$, pointing away from the charge.
Show solution
Use the point charge formula directly rather than imagining a test charge and dividing, since the two routes agree and the direct one has fewer places to lose a factor.
Independent check through the definition: a $1.0\ \mathrm{nC}$ probe placed there would feel $k(2.0\times 10^{-9})(1.0\times 10^{-9})/(0.0025) = 7.2\times 10^{-6}\ \mathrm{N}$, and dividing that by the probe charge returns the same field.
Take a field straight from its source rather than imagining a test charge and dividing it back out. The two routes agree, and the direct one has one place to drop a factor instead of three.
4§01.5 — an electron released in a known field●●○○○
An electron is released from rest at a point where the electric field has magnitude $2.5\times 10^{4}\ \mathrm{N/C}$ and points in the $+x$ direction. Gravity is negligible here, and that claim is worth testing at the end.
Given
$E = 2.5\times 10^{4}\ \mathrm{N/C}$ along $+x$
$e = 1.602\times 10^{-19}\ \mathrm{C}$
$m_e = 9.11\times 10^{-31}\ \mathrm{kg}$
Find
(a) Find the magnitude and direction of the force on the electron.
(b) Find its acceleration.
Hint 1/4
Two steps in a fixed order: the field gives the force, the force and the mass give the acceleration. The second step is ordinary mechanics.
Hint 2/4
$F = |q|E$, opposite to the field for a negative charge, and then $a = F/m$.
Hint 3/4
Substitute the data given, $E = 2.5\times 10^{4}\ \mathrm{N/C}$, $e = 1.602\times 10^{-19}\ \mathrm{C}$ and $m_e = 9.11\times 10^{-31}\ \mathrm{kg}$: $F = (1.602\times 10^{-19})(2.5\times 10^{4})$ and then $a = F/m_e$.
Hint 4/4
So $F = 4.0\times 10^{-15}\ \mathrm{N}$ along $-x$ and $a = 4.4\times 10^{15}\ \mathrm{m/s^{2}}$ in the same direction.
Show solution
Take the magnitude first and settle the reversal in words, because carrying a signed charge through two steps gives two chances to flip the direction and only one of them is correct.
Independent test of the claim that gravity is negligible: the electron's weight is $m_eg = (9.11\times 10^{-31})(9.80) = 8.9\times 10^{-30}\ \mathrm{N}$, which is smaller than the electric force by a factor of about $4\times 10^{14}$, so ignoring it was safe.
Get the size from the magnitude of the charge times the field, and settle the direction in a separate sentence from the sign. Carrying a signed charge through the arithmetic gives two chances to flip the arrow, and only one of them is correct.
5§01.4 — net force from two sources on a line●●●○○
Three point charges are fixed on the $x$ axis in a laboratory rig: $+4.0\ \mathrm{\mu C}$ at the origin, $+4.0\ \mathrm{\mu C}$ at $x = 0.20\ \mathrm{m}$, and $-2.0\ \mathrm{\mu C}$ at $x = 0.50\ \mathrm{m}$.
Given
$q_1 = +4.0\ \mathrm{\mu C}$ at $x = 0$
$q_2 = +4.0\ \mathrm{\mu C}$ at $x = 0.20\ \mathrm{m}$
$q_3 = -2.0\ \mathrm{\mu C}$ at $x = 0.50\ \mathrm{m}$
Find
(a) Find the net force on $q_3$, in magnitude and direction.
Hint 1/4
Two separate pair problems that are then combined. Notice before computing that both sources are positive and both lie on the same side of the target.
Hint 2/4
Each pair gives $k|q||Q|/r^{2}$ with its own separation, and on a line the vector sum is a signed sum.
Hint 3/4
Substitute the data given: $r_{13} = 0.50\ \mathrm{m}$ gives $0.288\ \mathrm{N}$ and $r_{23} = 0.30\ \mathrm{m}$ gives $0.799\ \mathrm{N}$, both attractions pulling $q_3$ towards $-x$.
Hint 4/4
So the net force is $0.288 + 0.799 = 1.09\ \mathrm{N}$ in the $-x$ direction.
Show solution
Check the two directions before computing either magnitude. They turn out to agree, which is what makes the final step an addition rather than the subtraction most three charge problems need.
Independent check on the ratio of the two contributions: the charges are identical, so the forces should go purely as the inverse squares of $0.30$ and $0.50$, a factor $(0.50/0.30)^{2} = 2.78$. The computed ratio $0.799/0.288 = 2.77$ matches.
Fix the direction of every contribution before computing a single magnitude. Once you know whether the arrows agree or oppose, you know whether the last step is an addition or a subtraction, and that decision is where problems with three charges are usually lost.
6§01.6 — locating the null point between two like charges●●●○○
Two positive charges are clamped on an optical rail: $+12\ \mathrm{\mu C}$ at $x = 0$ and $+3.0\ \mathrm{\mu C}$ at $x = 0.90\ \mathrm{m}$. A student wants the place where a small test charge would feel nothing at all.
Given
$Q_1 = +12\ \mathrm{\mu C}$ at $x = 0$
$Q_2 = +3.0\ \mathrm{\mu C}$ at $x = 0.90\ \mathrm{m}$
Find
(a) Find the position on the axis where the net electric field is zero.
Hint 1/4
Decide the region before any algebra. Both charges are positive, so where do their two fields point in opposite directions?
Hint 2/4
Setting the magnitudes equal gives $r_1/r_2 = \sqrt{Q_1/Q_2}$, and the two distances must add up to the separation.
Hint 3/4
Substitute the data given, $Q_1 = 12\ \mathrm{\mu C}$, $Q_2 = 3.0\ \mathrm{\mu C}$ and a gap of $0.90\ \mathrm{m}$: $r_1/r_2 = \sqrt{4} = 2$ with $r_1 + r_2 = 0.90\ \mathrm{m}$.
Hint 4/4
So $r_1 = 0.60\ \mathrm{m}$, that is at $x = 0.60\ \mathrm{m}$.
Show solution
Take the square root of the ratio before cross multiplying, so the equation stays linear and the meaningless second root never appears.
Region and ratio
$$0 < x < 0.90\ \mathrm{m}$$
between two like charges the fields point in opposite directions, which is the only region where cancellation is possible
$$\frac{r_1}{r_2} = \sqrt{\frac{12}{3.0}} = 2$$
equal magnitudes and equal $k$, so only the ratio survives, and it is the root of the charge ratio
the null point is nearer the smaller charge, as it must be
Answer $$\boxed{x = 0.60\ \mathrm{m}}$$
Check
Independent check by evaluating both fields there: $E_1 = (8.99\times 10^{9})(12\times 10^{-6})/(0.60)^{2} = 3.00\times 10^{5}\ \mathrm{N/C}$ and $E_2 = (8.99\times 10^{9})(3.0\times 10^{-6})/(0.30)^{2} = 3.00\times 10^{5}\ \mathrm{N/C}$, equal and opposite.
The null point sits on the side of the smaller charge, and the distances are in the ratio of the square roots of the charges, not of the charges. Take that square root before cross multiplying: the equation stays linear and the meaningless root outside the gap never appears.
7§01.6 — field above the midpoint of two equal charges●●●○○
Two equal charges of $+5.0\ \mathrm{nC}$ are fixed at $(0,\,0)$ and $(0.40\ \mathrm{m},\,0)$. The field is wanted at the point $P = (0.20\ \mathrm{m},\,0.15\ \mathrm{m})$, directly above the midpoint of the pair.
Given
$Q_1 = Q_2 = +5.0\ \mathrm{nC}$
Positions $(0,0)$ and $(0.40,\,0)\ \mathrm{m}$
$P = (0.20,\,0.15)\ \mathrm{m}$
Find
(a) Find the magnitude and direction of the net field at P.
Hint 1/4
Look at the symmetry before computing. P sits on the perpendicular bisector of the pair, so one of the two components of the answer is already known.
Hint 2/4
Each source gives $k|Q|/r^{2}$ away from itself, and the components are found by multiplying by the unit vector from source to field point.
Hint 3/4
Substitute the data given: each distance is $\sqrt{(0.20)^{2}+(0.15)^{2}} = 0.25\ \mathrm{m}$, each field is $719\ \mathrm{N/C}$, and the vertical fraction of each is $0.15/0.25 = 0.60$.
Use the symmetry to kill the horizontal component instead of computing both contributions in full; that halves the work and removes the step where a sign is most often dropped.
Independent bound rather than a repeat: the answer must be less than the sum of the two contributions, $1438\ \mathrm{N/C}$, and more than their difference, zero. It comes out at $60\%$ of the upper bound, which is exactly the vertical fraction $0.15/0.25$, as it has to be when the horizontal parts cancel.
On a perpendicular bisector one component is dead before the arithmetic begins. Compute one contribution, keep only the part along the symmetry axis, and double it.
C · exam level 4 questions
1§01.3 — rescaling a measured force●●●○○
In a laboratory measurement two small charged spheres repel each other with a force of $12\ \mathrm{N}$. For the next run one of the spheres is recharged to three times its previous charge, and the spheres are moved to twice their previous separation.
Given
Original force $12\ \mathrm{N}$
One charge multiplied by 3
Separation multiplied by 2
Find
(a) What force is measured in the second run?
Hint 1/4
Handle the two changes separately and multiply the factors at the end. Nothing about the actual charges or the actual distance is needed.
Hint 2/4
A charge enters Coulomb's law once, the separation twice, so a factor $a$ on one charge and $b$ on the distance give a factor $a/b^{2}$ on the force.
Hint 3/4
Substitute the changes from the question, $a = 3$ and $b = 2$: the factor is $3/2^{2} = 0.75$, applied to $12\ \mathrm{N}$.
Hint 4/4
So the new force is $0.75\times 12 = 9.0\ \mathrm{N}$.
Show solution
Take the ratio of the two force expressions, because every unknown in them is common to both runs and cancels.
Independent check with invented values that reproduce the original: $q_1 = q_2 = 1.0\ \mathrm{\mu C}$ at $r = 0.0274\ \mathrm{m}$ gives about $12\ \mathrm{N}$; tripling one charge and doubling $r$ gives $(8.99\times 10^{9})(3\times 10^{-12})/(0.0548)^{2} = 8.98\ \mathrm{N}$, which is $9.0\ \mathrm{N}$ to two figures.
When two runs share all their unknowns, the ratio is the whole solution and nothing has to be evaluated. A measured value plus a described change is nearly always an instruction to take a ratio rather than to look up constants.
2§01.4 — net force on a charge off the axis●●●●○
An examination style arrangement: $q_1 = +5.0\ \mathrm{\mu C}$ sits at the origin, $q_2 = -8.0\ \mathrm{\mu C}$ at $(0.60\ \mathrm{m},\,0)$, and the charge of interest is $q_3 = +2.0\ \mathrm{\mu C}$ at $(0,\,0.80\ \mathrm{m})$. Work in components throughout.
Given
$q_1 = +5.0\ \mathrm{\mu C}$ at $(0,0)$
$q_2 = -8.0\ \mathrm{\mu C}$ at $(0.60,\,0)\ \mathrm{m}$
$q_3 = +2.0\ \mathrm{\mu C}$ at $(0,\,0.80)\ \mathrm{m}$
(a) Find the two pair forces on $q_3$, each with its components.
(b) Find the magnitude and direction of the net force on $q_3$.
Hint 1/4
Two pair problems and one addition. Before starting, note which of the two forces is vertical, because that one needs no resolving.
Hint 2/4
Each force is $k|q||Q|/r^{2}$ along the line joining the pair, repulsive for like signs and attractive for unlike, and components come from the unit vector between the two positions.
Hint 3/4
Substitute the data given: $r_{13} = 0.80\ \mathrm{m}$ and $r_{23} = \sqrt{(0.60)^{2}+(0.80)^{2}} = 1.00\ \mathrm{m}$, so the two magnitudes are $0.140\ \mathrm{N}$ and $0.144\ \mathrm{N}$.
Hint 4/4
So the net force is $9.00\times 10^{-2}\ \mathrm{N}$ at $16.4^{\circ}$ above the $+x$ direction.
Show solution
Resolve each force as magnitude times unit vector rather than juggling angles, because the triangle here has whole number sides and the unit vector comes out exactly without a single trigonometric function.
Independent check by the cosine rule: the two contributions are $0.140$ and $0.144\ \mathrm{N}$ and the angle between them is $143^{\circ}$, since one points along $+y$ and the other along $(0.60,-0.80)$. Then $F^{2} = 0.140^{2}+0.144^{2}+2(0.140)(0.144)\cos 143^{\circ}$ gives $F = 9.0\times 10^{-2}\ \mathrm{N}$, from a route that shares no step with the component sum.
When the geometry has whole number sides, resolve each force as a magnitude times a unit vector instead of reaching for angles: the components come out exactly and no trigonometric function is needed until the final direction.
3§01.2 — the order of operations in ●●●○○
A neutral metal sphere stands on an insulating pillar. A positively charged rod is brought close to it without touching. While the rod is held there, a wire is connected briefly from the sphere to the ground. The wire is then removed, and only after that is the rod carried away.
Given
Sphere initially neutral, on an insulating pillar
Rod positive, never touching
Order: rod near, ground, remove ground, remove rod
Find
(a) What is the final charge of the sphere?
Hint 1/4
Follow the electrons through the four steps in order. Only one of the four does anything that survives to the end.
Hint 2/4
Grounding lets charge flow between the object and the earth; whatever is left behind is trapped the moment the connection is broken.
Hint 3/4
Substitute the situation described: the positive rod draws electrons to the near face, the earth supplies more of them through the wire, the wire is cut with those electrons still on the sphere, and only then does the rod leave.
Hint 4/4
So the sphere is left negatively charged, and once the rod is gone that surplus spreads over the whole sphere.
Show solution
Work step by step through the sequence rather than reasoning from the final picture, because every wrong answer to this question comes from swapping two of the steps.
Four steps, one at a time
$$\text{rod near}: \ \text{near face } -,\ \text{far face } +,\ Q = 0$$
a rearrangement only, since no charge has entered or left the sphere
the earth responds to the far positive face and neutralises it, while the near face is held in place by the rod
$$\text{ground off}: \ Q < 0\ \text{trapped}$$
the path to the earth is broken, so whatever arrived can no longer leave
$$\text{rod off}: \ Q < 0\ \text{spreads out}$$
nothing holds the surplus at one face any more, and like charges repel, so it distributes itself over the sphere
Answer $$\boxed{Q_{\text{sphere}} < 0,\ \text{spread over the surface}}$$
Check
Independent check by swapping the last two steps: remove the rod first and the electrons that came up the wire are no longer held, so they run back to the earth and the sphere ends neutral. The two orders give different answers, which is the sign that the sequence, and not the rod, is doing the work.
In an induction question the order of the steps is the answer. Walk the sequence forwards, then test it by swapping the last two steps; if the result changes, the sequence was what the question was really about.
4§01.6 — a null point and a field value on the same axis●●●●○
Two positive charges are fixed on the $x$ axis: $+8.0\ \mathrm{\mu C}$ at the origin and $+2.0\ \mathrm{\mu C}$ at $x = 0.30\ \mathrm{m}$. An examination question asks for two different things about the same axis.
Given
$Q_1 = +8.0\ \mathrm{\mu C}$ at $x = 0$
$Q_2 = +2.0\ \mathrm{\mu C}$ at $x = 0.30\ \mathrm{m}$
(a) Find the point on the axis where the net field is zero.
(b) Find the net field at $x = 0.10\ \mathrm{m}$, with its direction.
Hint 1/4
Part (a) is a ratio question with no field values in it. Part (b) is an evaluation at a point that is not the answer to part (a), so the two fields there will not cancel.
Hint 2/4
For (a), $r_1/r_2 = \sqrt{Q_1/Q_2}$ with $r_1+r_2$ equal to the separation. For (b), evaluate $k|Q|/r^{2}$ for each source and combine along the line.
Hint 3/4
Substitute the data given: $\sqrt{8.0/2.0} = 2$ with $r_1+r_2 = 0.30\ \mathrm{m}$ for part (a); for part (b) the distances are $0.10\ \mathrm{m}$ and $0.20\ \mathrm{m}$.
Hint 4/4
So the null point is at $x = 0.20\ \mathrm{m}$, and the field at $x = 0.10\ \mathrm{m}$ is $6.74\times 10^{6}\ \mathrm{N/C}$ towards $+x$.
Show solution
Do part (a) by ratio and part (b) by direct evaluation. Trying to answer (b) by scaling from (a) is tempting and does not work, because the two fields there are not equal.
Independent consistency check between the two parts: at $x = 0.10\ \mathrm{m}$ we are on the large charge side of the null point, so the field must point away from the large charge, that is towards $+x$, and it must not be zero. Both hold, and moving out to $x = 0.20\ \mathrm{m}$ would make the two contributions equal at $1.80\times 10^{6}\ \mathrm{N/C}$ each, which is the null point found in part (a).
A null point says where two fields are equal and nothing about their size anywhere else. Answer a location by ratio and a value by direct evaluation, and never try to scale the second out of the first.
D · interleaved 3 questions
1§01.1 — a very small charge, counted and then used●●●○○
Mixed practice, and the type is not announced. A tiny metal sphere in a vacuum chamber is measured to carry a net charge of $-4.8\times 10^{-17}\ \mathrm{C}$. A second sphere carrying $+1.6\times 10^{-16}\ \mathrm{C}$ is brought to a centre to centre distance of $2.0\ \mathrm{cm}$.
(a) How many excess electrons does the first sphere carry?
(b) Find the force between the two spheres, with its sense.
Hint 1/4
Two unrelated questions about the same pair of objects. Decide for each one whether it is about counting grains or about a force.
Hint 2/4
For the count, $n = |Q|/e$. For the force, $F = k|Q_1||Q_2|/r^{2}$, attractive because the signs are unlike.
Hint 3/4
Substitute the data given: $n = (4.8\times 10^{-17})/(1.602\times 10^{-19})$, and $F = (8.99\times 10^{9})(4.8\times 10^{-17})(1.6\times 10^{-16})/(0.020)^{2}$.
Hint 4/4
So there are about $3.0\times 10^{2}$ excess electrons, and the force is $1.7\times 10^{-19}\ \mathrm{N}$, attractive.
Show solution
Answer the count from the charge as given, without converting it into a number of electrons first and multiplying back. Every extra conversion is a chance to drop a power of ten in numbers this small.
Independent plausibility check on the size of the force: dividing it by the elementary charge gives $1.1\ \mathrm{N/C}$, so this is the force a single electron would feel in a field of about one newton per coulomb. A few hundred electrons at two centimetres produce nothing a laboratory could weigh, which is the right conclusion and explains why real electrostatics experiments use microcoulombs.
Use the charge as it was given rather than converting it into a count and multiplying back; each extra trip through $e$ is another chance to lose a power of ten. Then ask whether the size of the answer is one an instrument could ever register.
2§01.5 — working backwards from a measured force●●●●○
Mixed practice, type not announced. At a point P inside an apparatus, a probe carrying $+2.0\ \mathrm{nC}$ is measured to feel a force of $3.6\times 10^{-5}\ \mathrm{N}$ in the $-y$ direction. The charges producing this are hidden inside the housing and are not moved during the experiment.
Given
Probe $+2.0\ \mathrm{nC}$, force $3.6\times 10^{-5}\ \mathrm{N}$ along $-y$
(b) The probe is replaced by a charge of $-5.0\ \mathrm{nC}$. Find the force on it.
(c) Suppose the field is produced by a single point charge sitting $0.15\ \mathrm{m}$ directly above P. Find its sign and size.
Hint 1/4
Three questions that go in one direction: from a measurement to the field, from the field to a different object, and from the field back to a source.
Hint 2/4
$E = F/|q_0|$, with the direction reversed for a negative probe; then $F = |q|E$; then $|Q| = Er^{2}/k$ for a single point source.
Hint 3/4
Substitute the data given: $E = (3.6\times 10^{-5})/(2.0\times 10^{-9})$, then $F = (5.0\times 10^{-9})E$, then $|Q| = E(0.15)^{2}/(8.99\times 10^{9})$.
Hint 4/4
So $E = 1.8\times 10^{4}\ \mathrm{N/C}$ along $-y$, the new force is $9.0\times 10^{-5}\ \mathrm{N}$ along $+y$, and the source would be $+4.5\times 10^{-8}\ \mathrm{C}$.
Show solution
Extract the field once and reuse it for both of the later parts. Answering part (c) from the original force directly would work too, but it would carry the probe charge into a calculation that has nothing to do with it.
Independent check on part (c) by going forwards instead of backwards: a charge of $4.5\times 10^{-8}\ \mathrm{C}$ at $0.15\ \mathrm{m}$ gives $E = (8.99\times 10^{9})(4.5\times 10^{-8})/(0.0225) = 1.8\times 10^{4}\ \mathrm{N/C}$, which is the field the measurement reported.
Extract the field once and reuse it. With $E$ in hand, a different probe or the source that produced the field is one multiplication or one division away, and the original probe charge never appears again.
3§01.2 — putting a number on the comb and paper trick●●●●○
Mixed practice, type not announced. A charged comb carrying $8.0\times 10^{-8}\ \mathrm{C}$ is held $1.50\ \mathrm{cm}$ above a small scrap of paper of mass $2.0\times 10^{-7}\ \mathrm{kg}$. Model the polarised scrap as two charges of size $1.2\times 10^{-11}\ \mathrm{C}$, opposite in sign, with the near one at $1.50\ \mathrm{cm}$ and the far one at $1.56\ \mathrm{cm}$ from the comb.
Given
$Q = 8.0\times 10^{-8}\ \mathrm{C}$ on the comb
$q = 1.2\times 10^{-11}\ \mathrm{C}$ induced on each face
Paper mass $2.0\times 10^{-7}\ \mathrm{kg}$, $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the net upward force on the scrap and say whether it lifts.
(b) The comb is raised to $3.00\ \mathrm{cm}$, with the induced charges and their $0.06\ \mathrm{cm}$ offset unchanged. Does it still lift?
Hint 1/4
The scrap carries no net charge, so the answer cannot come from one application of Coulomb's law. It comes from the difference between two.
Hint 2/4
$F_{\text{net}} = kQq\,(1/r_1^{2} - 1/r_2^{2})$, and it lifts when that exceeds $mg$.
Hint 3/4
Substitute the data given, $kQq = 8.63\times 10^{-9}$ with $r_1 = 0.0150\ \mathrm{m}$ and $r_2 = 0.0156\ \mathrm{m}$: the bracket is $4444 - 4109 = 335\ \mathrm{m^{-2}}$, and the weight is $1.96\times 10^{-6}\ \mathrm{N}$.
Hint 4/4
So the pull is $2.9\times 10^{-6}\ \mathrm{N}$, about $1.5$ times the weight, and it lifts; at $3.00\ \mathrm{cm}$ the pull falls to $3.7\times 10^{-7}\ \mathrm{N}$ and it does not.
Show solution
Subtract the two inverse squares inside the bracket rather than computing two forces and subtracting them. The two forces agree to three figures, so subtracting them at the end would destroy the very quantity being asked for.
Independent check on how the two answers relate: doubling the height cut the force by $2.9\times 10^{-6}/3.7\times 10^{-7} = 7.8$, close to $2^{3} = 8$. A force on a neutral object falls off one power faster than the force between two charges, and that is why the trick is so fussy about how close you hold the comb.
When two nearly equal inverse squares are subtracted, do the subtraction inside the bracket before multiplying anything out; subtracting two rounded forces at the end destroys the very quantity being asked for. The pull on a neutral object falls off one power of distance faster than the pull between two charges, which is why such tricks are so fussy about height.
Mistake ledger (18 entries)
⚠ Saying that rubbing creates charge
the rod was inert before and attracts things afterwards, so something obviously appeared, and only the rod ever gets tested
$q_0$ small enough not to disturb the sources; direction reversed if the probe is negative
Field of a point charge
$$E = k\,\frac{|Q|}{r^{2}}$$
one source, $r$ from the source to the field point, away from $+Q$ and towards $-Q$
Superposition of fields
$$\vec{E}_{\text{net}} = \sum_i \vec{E}_i$$
each contribution with its own distance and its own direction
Where the field of two charges vanishes
$$\frac{r_1}{r_2} = \sqrt{\frac{|Q_1|}{|Q_2|}}$$
between the charges if they have like signs, outside and beyond the smaller one if unlike
Check yourself
Close the page and write down, from memory: the three facts about charge and what each one forbids; what a conductor does that an insulator does not; why a neutral scrap of paper moves at all; Coulomb's law with its constant and the two things the vertical bars are there to prevent; how the force changes when a charge or a separation is scaled; the rule for combining two forces or two fields that are not parallel; the definition of the field and its unit; the field of a point charge with its direction; and the condition that locates a null point. Then reopen and compare. Whatever is missing is your reread list, and none of it is scored.
Turn a charge in microcoulombs into a number of electrons, and say in one line why $-1.0\times 10^{-19}\ \mathrm{C}$ cannot be anybody's charge?
c-charge
Explain why a neutral scrap of paper is attracted, and say what a sphere is left carrying after a positive rod, a grounding wire and the removal of both in that order?
c-conductors-induction
Compute the force between two given charges at a given separation, state its direction in one sentence, and say what happens to it when one charge is tripled and the distance doubled?
c-coulomb-law
Take three charges with coordinates, find the two pair forces on one of them with their components, and report the sum as a magnitude and an angle?
c-superposition-forces
Go from a measured force on a negative probe to the field, and from a field to the force on any other charge, without reversing the direction twice?
c-electric-field
Find the point where two like charges give zero field, and evaluate the net field at an off axis point from two sources?
c-field-superposition
Glossary (21 terms)
electric chargeelektrik yükü
The property of matter that makes objects attract or repel electrically, coming in two signs and in whole multiples of one fixed amount.
coulombkulon
The SI unit of charge, written C. One coulomb is an enormous charge by laboratory standards; charges you can produce by rubbing are of order a tenth of a microcoulomb.
elementary chargetemel yük
The size of the charge on a proton or an electron, $1.602\times 10^{-19}\ \mathrm{C}$, written $e$ and always taken as a positive number.
quantisation of chargeyükün kuantumlanması
The fact that every measured charge is a whole number of elementary charges, so no object can carry a fraction of one.
conservation of chargeyükün korunumu
The rule that the total charge of an isolated system never changes, so charging one object always leaves an equal and opposite charge somewhere else.
conductoriletken
A material in which charge can travel freely, so a surplus placed anywhere on it spreads over the whole object.
insulatoryalıtkan
A material in which charge cannot travel, so a surplus placed on it stays where it was put.
serbest elektron
An electron in a metal that is not bound to any one atom and is able to move through the material, which is what makes a metal a conductor.
charging by contactdokunmayla yükleme
Transferring charge by touching a charged object to another, which leaves both with the same sign of charge.
charging by inductionetkiyle yükleme
Charging an object without touching it, by holding a charged object nearby, grounding briefly, then removing the ground before the charged object; the result has the opposite sign to what was held nearby.
groundingtopraklama
Connecting an object to the earth so that charge can flow to or from a reservoir large enough for the exchange not to change it.
polarisationkutuplanma
The internal rearrangement of a neutral object near a charge, which leaves one face with a surplus of one sign and the opposite face with a surplus of the other while the total stays zero.
electrostaticselektrostatik
The study of charges that are at rest, or moving slowly enough that their motion can be ignored.
point chargenoktasal yük
A charged object whose size is negligible compared with the distances involved, so that all its charge can be treated as sitting at one point.
Coulomb's lawCoulomb yasası
The rule that the force between two point charges is proportional to each charge and inversely proportional to the square of the distance between them, acting along the line joining them.
Coulomb constantCoulomb sabiti
The constant $k = 8.99\times 10^{9}\ \mathrm{N\cdot m^{2}/C^{2}}$ appearing in Coulomb's law, equal to one over four pi times the .
permittivity of free spaceboşluğun elektriksel geçirgenliği
The constant $\varepsilon_0 = 8.85\times 10^{-12}\ \mathrm{C^{2}/(N\cdot m^{2})}$, an alternative way of writing the same constant that appears in Coulomb's law.
principle of superpositionüst üste binme ilkesi
The rule that the total force or field from several charges is the vector sum of what each one would produce on its own, with no charge affecting what any other contributes.
electric fieldelektrik alan
The force per unit charge at a point, in newtons per coulomb, defined by the sources alone and existing whether or not anything is placed there.
test chargedeneme yükü
A small positive charge imagined at a point in order to define the field there, chosen small enough not to disturb the charges producing it.
null point
A place where the fields of several charges cancel exactly, so a charge placed there would feel no electric force at all.
What comes next
§02 · Electric field continued: field lines, continuous charge distributions, and dipoles
Everything here was done with a handful of separate charges, each one contributing its own arrow to a sum with a small number of terms. Next week the number of terms becomes infinite, because the charge is smeared along a rod or around a ring rather than sitting at points, and the sum becomes an integral. The definition of the field does not change at all; only the way of adding up does.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition, the chapter carrying the same title as this week's syllabus line The required book for the course. The treatment of charge, conductors and insulators, induction, Coulomb's law and the electric field of point charges follows this text. No section numbers are quoted anywhere in these notes, because the syllabus week line gives none and inventing them would be worse than leaving them out.
PHYS 102 syllabus: the week 1 line and the published assessment weights The week line reads Electric Charge and Electric Field. The weights quoted in the sixty second card are the published ones, and nothing beyond them is claimed about how this material will be examined.
The International System of Units, as revised in 2019 Used only for the value of the elementary charge, which the 2019 revision fixes exactly, and for the definition of the coulomb that follows from it.