02Electric field continued: field lines, continuous charge distributions, and dipoles

The claim links two curves: the field line and the path. Ask what the field line controls at each instant, and whether that is the same thing as the path.

The field fixes the force and therefore the acceleration, $\vec a = q\vec E/m$, not the velocity.

Here the charge starts from rest, so its first movement is indeed along the field line, but by the time it has moved it carries velocity in that original direction while the line has already curved away.

False: only on a straight field line do the path and the line coincide.

Do not attack the premise, which is correct: the wire really is made of point charges. Attack the step from the pieces to the whole, and ask what changes about the assembly as you move away.

For a long wire the assembled result is $E = 2k\lambda/x$, an inverse first power, even though every element obeys an inverse square.

Here, moving from one distance to twice that distance makes each element four times weaker, but the stretch of wire that contributes appreciably becomes twice as long.

False: the two effects combine to give one over the distance, not one over its square.

Sort the four claims into two kinds: those that would merely be an odd choice by the artist, and those that would make the field itself ill defined at some point.

At any point the field has one direction, so exactly one line passes through each point; and lines begin on positive charge and end on negative charge or at infinity.

Here you are told the charges are at rest and that the drawing is a plane section, so the usual rules apply without exception.

The impossible one is the diagram in which two lines cross.

A rigid body has two independent things it can do. Settle each one separately rather than asking a single question about what happens.

Net force $= q\vec E + (-q)\vec E = 0$, while the torque is $\tau = pE\sin\theta$ about the centre.

Here the field is the same at both charges, so the two forces are exact opposites, and the dipole starts at a nonzero angle so the sine is not zero.

It rotates towards alignment with the field, and its centre does not accelerate.

Three parts, one idea. In each case ask what the charge is spread over, and divide by the total amount of that thing.

$\lambda = Q/L$ for a length, $\sigma = Q/A$ for an area, and then the charge of a piece is the density times the size of the piece.

Here $Q = 12.0\times10^{-9}$ C throughout; the rod is $0.300$ m long, the ring has circumference $2\pi(0.0400) = 0.2513$ m, and the disc has area $\pi(0.0400)^{2} = 5.027\times10^{-3}\ \mathrm{m^{2}}$.

The three densities are $4.00\times10^{-8}\ \mathrm{C/m}$, $4.77\times10^{-8}\ \mathrm{C/m}$ and $2.39\times10^{-6}\ \mathrm{C/m^{2}}$.

The same formula serves both points, so the work is in the bookkeeping. Write down the near and far distances for each case before touching the calculator.

$E = kQ/[a(a+L)]$ for the rod, and $E_{\rm est} = kQ/d^{2}$ with $d = a + L/2$ for the estimate.

Here $kQ = (8.99\times10^{9})(7.20\times10^{-9}) = 64.73$, $L = 0.180$ m, and the two values of $a$ are $0.0400$ m and $0.400$ m.

The fields are $7.36\times10^{3}\ \mathrm{N/C}$ and $2.79\times10^{2}\ \mathrm{N/C}$, and the centred point charge estimate is low by a factor $1.92$ at the near point and only $1.035$ at the far one.

Nothing new is needed; the same relation is being read in the other direction. Decide which symbol is unknown before rearranging anything.

$E = kQ/[a(a+L)]$, so $Q = E\,a(a+L)/k$.

Here $E = 1.85\times10^{3}\ \mathrm{N/C}$, $a = 0.100$ m, $a+L = 0.350$ m, and $k = 8.99\times10^{9}$.

The rod carries $+7.20\ \mathrm{nC}$, so $\lambda = 2.88\times10^{-8}\ \mathrm{C/m}$.

Two questions about the same function. The first is a substitution; the second is a property of the shape of the function and does not depend on the charge at all.

$E = kQx/(x^{2}+R^{2})^{3/2}$, with the maximum at $x = R/\sqrt2$ where $E = 0.385\,kQ/R^{2}$.

Here $kQ = (8.99\times10^{9})(30.0\times10^{-9}) = 269.7$, $R = 0.120$ m and the first field point is at $x = 0.160$ m.

The field at 16.0 cm is $5.39\times10^{3}\ \mathrm{N/C}$, and the peak is $7.21\times10^{3}\ \mathrm{N/C}$ at $x = 8.49\ \mathrm{cm}$.

Neither part needs the length of the wire, and neither needs its total charge. Ask which single property of the wire the reading actually determines.

$E = 2k\lambda/x$, so $\lambda = Ex/(2k)$ and the distance for a given field is $x = 2k\lambda/E$.

Here $E = 2.40\times10^{3}\ \mathrm{N/C}$ at $x = 0.0500$ m, and $2k = 1.798\times10^{10}$ in SI units.

The density is $+6.67\ \mathrm{nC/m}$ and the second reading occurs at $15.0\ \mathrm{cm}$.

The two formulas differ by one factor. Identify that factor first, and you will be able to predict the size of the error before computing either field.

$E_{\infty} = 2k\lambda/x$ and the exact result is that multiplied by $L/\sqrt{x^{2}+L^{2}}$.

Here $2k\lambda = (1.798\times10^{10})(5.00\times10^{-9}) = 89.9$, $x = 0.0500$ m and $L = 0.120$ m, so $\sqrt{x^{2}+L^{2}} = 0.130$ m.

The shortcut gives $1.80\times10^{3}\ \mathrm{N/C}$ and the exact value is $1.66\times10^{3}\ \mathrm{N/C}$, so the shortcut is 8.3 per cent too high.

Both parts are built from the same product. Form it once and keep it on the page.

$p = q\ell$, $\tau = pE\sin\theta$, and $W_{\rm ext} = pE(\cos\theta_1-\cos\theta_2)$.

Here $q = 12.0\times10^{-9}$ C, $\ell = 0.0250$ m, $E = 4.80\times10^{4}$ N/C, $\theta_1 = 25.0^{\circ}$ and $\theta_2 = 90.0^{\circ}$.

The moment is $3.00\times10^{-10}\ \mathrm{C\,m}$, the torque is $6.09\times10^{-6}\ \mathrm{N\,m}$ and the work is $+1.31\times10^{-5}\ \mathrm{J}$.

Two formulas that differ only by a factor. Decide which one belongs to which direction before substituting anything.

$E_{\rm bis} = kp/r^{3}$ and $E_{\rm axis} = 2kp/r^{3}$, both valid when $r \gg \ell$.

Here $kp = (8.99\times10^{9})(6.00\times10^{-11}) = 0.5394$, $r = 0.0800$ m so $r^{3} = 5.12\times10^{-4}\ \mathrm{m^{3}}$, and $\ell = 1.00\times10^{-3}$ m.

The bisector value is $1.05\times10^{3}\ \mathrm{N/C}$ and the axial value is twice that, $2.11\times10^{3}\ \mathrm{N/C}$.

Test each candidate against two things you already know without integrating: what it gives at the centre, and what it gives very far away.

At the centre the field must vanish by symmetry, and far away the ring must look like a point charge, so any correct expression must reduce to $kQ/x^{2}$ for large $x$.

Here every element is $\sqrt{x^{2}+R^{2}}$ from the field point, and the surviving fraction of each contribution is $\cos\theta = x/\sqrt{x^{2}+R^{2}}$.

The correct expression is $\int k\,dq\,x/(x^{2}+R^{2})^{3/2}$, which is $kQx/(x^{2}+R^{2})^{3/2}$.

Assemble the geometry before touching the charge. The two lengths are equal here, which makes the slant distance easy but also makes wrong answers look plausible.

$E = kQx/(x^{2}+R^{2})^{3/2}$, and $(x^{2}+R^{2})^{3/2}$ means the square root of the bracket, cubed.

Here $x = R = 0.100$ m, so $x^{2}+R^{2} = 0.0200\ \mathrm{m^{2}}$, its square root is $0.1414$ m, and $kQ = 359.6$ in SI units.

The field is $1.27\times10^{4}\ \mathrm{N/C}$.

Before computing, decide whether the dipole is being moved towards alignment or away from it, because that fixes the sign of the answer.

$W_{\rm ext} = pE(\cos\theta_1 - \cos\theta_2)$, and $pE$ is the natural energy scale of the problem.

Here $pE = (2.00\times10^{-10})(5.00\times10^{4}) = 1.00\times10^{-5}$, with $\cos 60.0^{\circ} = 0.500$ and $\cos 120.0^{\circ} = -0.500$.

The work is $+1.00\times10^{-5}\ \mathrm{J}$.

Do not fit a formula. Form the ratio of the two readings and the ratio of the two distances, and ask what power connects them.

A point charge gives $E \propto 1/r^{2}$, a long wire gives $1/r$, and a dipole gives $1/r^{3}$.

Here the distance doubles from $4.00$ cm to $8.00$ cm while the field falls from $500$ to $62.5\ \mathrm{N/C}$.

The field falls by a factor of eight when the distance doubles, so the source is a small electric dipole.

Both sources are positive, so ask first in which region of the line their fields could possibly point in opposite directions.

$E_{\rm wire} = 2k\lambda/x$ pointing away from the wire, and $E_{\rm point} = kq/(d-x)^{2}$ pointing away from the charge, where $d$ is the separation.

Here $\lambda = 4.00\times10^{-9}\ \mathrm{C/m}$, $q = 3.00\times10^{-9}\ \mathrm{C}$ and $d = 0.100$ m, so the condition is $2(4.00)(d-x)^{2} = (3.00)x$.

The zero lies at $x = 1.79\ \mathrm{cm}$ from the wire.

Two steps that must not be merged. Ask what the rod does at that point first, and only afterwards what happens to a charge placed there.

$E = kQ/[a(a+L)]$ for the rod, and then $\vec F = q\vec E$ for the sphere, with the sign of $q$ deciding the direction.

Here $Q = 10.0\times10^{-9}$ C, $a = 0.0500$ m, $a+L = 0.250$ m, and the sphere carries $-3.00\times10^{-9}$ C.

The field is $7.19\times10^{3}\ \mathrm{N/C}$ away from the rod, and the force is $2.16\times10^{-5}\ \mathrm{N}$ towards the rod.

Two steps again, and they belong to two different parts of the course. Get the field from the ring, then hand it to Newton's second law.

$E = kQx/(x^{2}+R^{2})^{3/2}$, then $F = qE$ and $a = F/m$.

Here $R = 0.0600$ m, $Q = 45.0\times10^{-9}$ C, $x = 0.0800$ m, $q = 2.00\times10^{-9}$ C and $m = 5.00\times10^{-4}$ kg.

The field is $3.24\times10^{4}\ \mathrm{N/C}$ and the acceleration is $0.129\ \mathrm{m/s^{2}}$, away from the ring.

Two sources on one line, so the vector addition is one dimensional. Fix a positive direction on the line before computing anything.

$E_{\rm wire} = 2k\lambda/x$ pointing away from the wire, $E_{\rm point} = kq/s^{2}$ pointing away from the charge, and the two are added along the line.

Here $\lambda = 3.00\times10^{-9}$ C/m with $x = 0.0400$ m, and the point charge is $4.00\times10^{-9}$ C at a distance $s = 0.100-0.0400 = 0.0600$ m from the field point.

The wire gives $1.35\times10^{3}\ \mathrm{N/C}$ away from the wire and the charge gives $9.99\times10^{3}\ \mathrm{N/C}$ towards it, so the total is $8.64\times10^{3}\ \mathrm{N/C}$ towards the wire.

The dipole formula for torque was derived for a uniform field, and the field of a point charge is not uniform, so the first thing to settle is whether it is uniform enough over the molecule.

$E = kQ/r^{2}$ at the molecule, then $\tau = pE\sin\theta$ using the local field.

Here $Q = 80.0\times10^{-9}$ C at $r = 0.200$ m, and the molecule has $p = 5.00\times10^{-12}\ \mathrm{C\,m}$ at $35.0^{\circ}$.

The field is $1.80\times10^{4}\ \mathrm{N/C}$ and the torque is $5.16\times10^{-8}\ \mathrm{N\,m}$.