Electrostatics

A point charge $+q$ is held at distance $d$ from a large grounded metal plate. The plate is neutral and connected to earth. What force acts on the charge?

Not zero. The charge is pulled towards the plate with

$F=\frac{q^2}{16\pi {\epsilon }_0{d}^2}$

which is exactly the force a charge $-q$ would exert from a distance $d$ behind the plate.

The reasoning is a uniqueness argument. The grounded plate is an equipotential at $V=0$, and the field in the region containing the charge is determined completely by the charge distribution there plus that boundary condition. Any arrangement reproducing the same boundary gives the same field — and a mirror charge $-q$ makes the plane midway between them a $V=0$ surface automatically.

Main gives you three symmetric charge distributions and asks you to substitute. Advanced gives you a distribution that is not symmetric, a boundary that is not obvious, or an energy question where the answer depends on which energy you mean. This chapter is the machinery for those.

1. Fields by integration

Symmetry is a luxury. When it is absent, integrate — and integrate the potential first whenever possible, because $V$ is a scalar.

A finite line charge at perpendicular distance $r$, with its ends subtending angles ${\theta }_1$ and ${\theta }_2$ from the foot of the perpendicular:

$E_{\perp }=\frac{\lambda }{4\pi {\epsilon }_{0}r}\left(\sin {\theta }_1+\sin {\theta }_2\right),E_{\parallel }=\frac{\lambda }{4\pi {\epsilon }_{0}r}\left(\cos {\theta }_2-\cos {\theta }_1\right)$

Setting both angles to $90^{\circ }$ recovers the infinite-wire result $\lambda /2\pi {\epsilon }_{0}r$, and the parallel component vanishes by symmetry only when the point faces the middle of the rod.

A ring on its axis gives $E=\frac{kQx}{{\left(x^2+R^2\right)}^{3/2}}$, zero at the centre and zero at infinity, so it peaks in between — at $x=R/\sqrt{2}$, where $E_{max}=\frac{2kQ}{3\sqrt{3}{R}^2}$.

A disc, built from rings, gives $E=\frac{\sigma }{2{\epsilon }_0}\left(1-\frac{x}{\sqrt{x^2+R^2}}\right)$, tending to $\sigma /2{\epsilon }_0$ as $R\to \infty$ — the infinite sheet, independent of distance.

Illustration 1

A rod of length $L$ carries uniform charge $\lambda$. Find the field at a point on its perpendicular bisector at distance $r$.

By symmetry ${\theta }_1={\theta }_2=\theta$ with $\sin \theta =\frac{L/2}{\sqrt{r^2+L^2/4}}$, and $E_{\parallel }=0$.

$E=\frac{2k\lambda }{r}\sin \theta =\frac{k\lambda L}{r\sqrt{r^2+L^2/4}}$

Two limits check it. For $r\gg L$ this becomes $kQ/r^2$ with $Q=\lambda L$, a point charge. For $r\ll L$ it becomes $2k\lambda /r$, the infinite wire.

Illustration 2

A uniformly charged semicircular arc of radius $R$ carries total charge $Q$. Find the field at its centre.

Take an element at angle $\theta$ carrying $dq=\frac{Q}{\pi }d\theta$. The components along the diameter cancel in pairs; the perpendicular components add:

$E={\int }_0^{\pi }\frac{kQd\theta }{\pi R^2}\sin \theta =\frac{kQ}{\pi R^2}{\left[-\cos \theta \right]}_0^{\pi }=\frac{2kQ}{\pi R^2}$

Note that the answer is $2/\pi$ times what a point charge $Q$ at distance $R$ would give. Spreading the charge around the arc costs the field a factor of about $0.64$, which is the average of $\sin \theta$.

2. The potential is usually the easier object

Because $V$ is a scalar, its integral over a charge distribution needs no components resolved. Recover the field afterwards:

$\vec{E}=-\nabla V=-\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right)$

Equipotential surfaces are everywhere perpendicular to field lines, and no work is done moving a charge along one. Where equipotentials crowd together, the field is strong.

The ring is the clean demonstration. Every element of the ring is the same distance from an axial point, so

$V=\frac{kQ}{\sqrt{x^2+R^2}}\Rightarrow E=-\frac{dV}{dx}=\frac{kQx}{{\left(x^2+R^2\right)}^{3/2}}$

which reproduces the earlier result in two lines instead of an integral over components.

Inside a uniformly charged insulating sphere the potential is not constant, unlike a conductor:

$V_{in}=\frac{kQ\left(3R^2-r^2\right)}{2R^3},V_{centre}=\frac{3kQ}{2R}$

so the centre sits at one and a half times the surface potential. A conducting sphere of the same charge is at $kQ/R$ throughout.

Illustration 3

In a region the potential is $V=4x^2$ volts, with $x$ in metres. Find the field at $x=2$ m, and describe the equipotential surfaces.

$E_x=-\frac{dV}{dx}=-8x$

At $x=2$: $E_x=-16$ V m${}^{-1}$, pointing in the negative $x$ direction.

The equipotentials are the surfaces $x=$ constant, that is, planes perpendicular to the $x$ axis, and they crowd closer as $x$ grows because the field is getting stronger.

The field is not uniform even though the equipotentials are planes. Flatness of an equipotential says nothing about the spacing between successive ones.

Illustration 4

Find the potential at the centre of a uniformly charged disc of radius $R$ and surface density $\sigma$.

Build the disc from rings of radius $s$ and width $ds$, each carrying $dq=\sigma 2\pi sds$ at distance $s$ from the centre:

$V={\int }_0^R\frac{k\sigma 2\pi sds}{s}=2\pi k\sigma R=\frac{\sigma R}{2{\epsilon }_0}$

The integrand's $s$ cancels completely, which is why the potential integral is finite even though each ring sits at a different distance. Attempting the same by components would have been far messier.

3. Gauss's law past the three standard cases

Gauss's law is exact for any closed surface; what fails without symmetry is the ability to pull $E$ out of the integral. Two Advanced settings restore that ability.

A non-uniformly charged sphere, where $\rho$ depends only on $r$, is still spherically symmetric, so $E$ is constant on a concentric sphere. Only the enclosed charge changes:

$q_{enc}={\int }_0^r\rho \left(r^{\prime }\right)4\pi r^{\prime 2}d{r}^{\prime },E=\frac{q_{enc}}{4\pi {\epsilon }_0{r}^2}$

A spherical cavity in a uniformly charged sphere is handled by superposition with a negative density, exactly as in gravitation. The field inside the cavity is uniform:

${\vec{E}}_{cavity}=\frac{\rho \vec{d}}{3{\epsilon }_0}$

where $\vec{d}$ joins the two centres.

Illustration 5

A sphere of radius $R$ carries charge density $\rho ={\rho }_0\frac{r}{R}$. Find the field at $r<R$ and at $r>R$.

$q_{enc}={\int }_0^r\frac{{\rho }_0{r}^{\prime }}{R}4\pi r^{\prime 2}d{r}^{\prime }=\frac{\pi {\rho }_0{r}^4}{R}$

$E_{in}=\frac{{\rho }_0{r}^2}{4{\epsilon }_{0}R}$, growing as $r^2$ rather than linearly.

Total charge is $\pi {\rho }_0{R}^3$, so $E_{out}=\frac{{\rho }_0{R}^3}{4{\epsilon }_0{r}^2}$.

Continuity at $r=R$ is the check: both expressions give ${\rho }_{0}R/4{\epsilon }_0$, as they must, since there is no surface charge.

4. Energy: self, interaction, and density

Three quantities are called electrostatic energy, and Advanced questions turn on telling them apart.

Interaction energy counts pairs: $U=\sum \frac{k{q}_i{q}_j}{r_{ij}}$ over $i<j$, which is what "energy of a system of point charges" means.

Self-energy is the work to assemble a body from infinitely dispersed charge:

$U_{sphere}=\frac{3k{Q}^2}{5R},U_{shell}=\frac{k{Q}^2}{2R}$

Energy density distributes the same total through space:

$u=\frac{1}{2}{\epsilon }_0{E}^2$

and integrating $u$ over all space reproduces the self-energy — which is a useful way to obtain it without assembling anything.

Illustration 6

Find the self-energy of a uniformly charged spherical shell by integrating the energy density.

Outside, $E=\frac{kQ}{r^2}$; inside, $E=0$.

$U={\int }_R^{\infty }\frac{1}{2}{\epsilon }_0{\left(\frac{kQ}{r^2}\right)}^{2}4\pi r^{2}dr=\frac{{\epsilon }_0{k}^2{Q}^2}{2}\left(4\pi \right){\int }_R^{\infty }\frac{dr}{r^2}$

$=\frac{k{Q}^2}{2R}$

The whole energy lives in the field outside, since the interior contributes nothing. This is the cleanest demonstration that field energy is not a bookkeeping fiction.

Illustration 7

Two identical conducting spheres of radius $r$, each carrying charge $q$, are held with centres $d$ apart, where $d\gg r$. Find the total electrostatic energy.

Self-energies: $2\times \frac{k{q}^2}{2r}=\frac{k{q}^2}{r}$

Interaction energy: $\frac{k{q}^2}{d}$

$U_{total}=k{q}^2\left(\frac{1}{r}+\frac{1}{d}\right)$

Only the interaction term changes when the spheres are moved, which is why forces can be found by differentiating it alone. Self-energies are constants of the geometry and drop out.

5. Conductors, cavities and earthed shells

Three properties do all the work. The field inside conducting material is zero; the surface is an equipotential; and any charge placed in a cavity induces an equal and opposite charge on the cavity wall, with the balancing charge appearing on the outer surface, distributed as though the cavity did not exist.

Earthing fixes a conductor's potential at zero and lets charge flow. Which surface loses charge, and how much, follows from writing $V=0$ for that conductor.

Illustration 8

Concentric conducting shells of radii $a$ and $b$ carry charges $q$ and $Q$. The outer shell is now earthed. Find the charge remaining on it and the potential of the inner shell.

Earthing sets $V_b=0$. With inner charge $q$ and outer charge $Q^{\prime }$:

$V_b=\frac{k\left(q+Q^{\prime }\right)}{b}=0\Rightarrow Q^{\prime }=-q$

The outer shell's original charge $Q$ has drained away entirely, and

$V_a=\frac{kq}{a}-\frac{kq}{b}=kq\left(\frac{1}{a}-\frac{1}{b}\right)$

The result is independent of $Q$, which is the whole point of a grounded screen: whatever it started with, it ends up carrying exactly what is needed to cancel the field outside.

Illustration 9

A point charge $q$ sits off-centre inside a cavity in an uncharged conducting block. Describe the charge distribution and the field outside the block.

The cavity wall carries $-q$, distributed non-uniformly, crowding towards the nearer wall.

The outer surface carries $+q$, distributed as if the cavity and the charge did not exist — for a spherical block, uniformly.

The external field is therefore that of a point charge $q$ at the block's centre, regardless of where inside the cavity the charge actually sits.

The conductor erases the internal geometry. That is electrostatic shielding stated as a positive result rather than as an absence of field.

6. The method of images

For a grounded plane, replace the conductor by a mirror charge $-q$ at the mirror position, and use the mirror only for the region containing the real charge.

$F=\frac{k{q}^2}{{\left(2d\right)}^2}=\frac{q^2}{16\pi {\epsilon }_0{d}^2},\sigma (s)=-\frac{qd}{2\pi {\left(s^2+d^2\right)}^{3/2}}$

The induced charge integrates to exactly $-q$. The work to remove the charge to infinity is not the interaction energy of the pair, because the field exists only on one side:

$W={\int }_d^{\infty }\frac{k{q}^2}{4x^2}dx=\frac{k{q}^2}{4d}$

which is half of what a genuine pair of charges would require.

Illustration 10

A charge $q=2\mu$C is held $0.10$ m above a large grounded plate. Find the force on it and the work needed to pull it to a height of $0.20$ m.

$F=\frac{9\times 10^9{\left(2\times 10^{-6}\right)}^2}{4{\left(0.10\right)}^2}=\frac{3.6\times 10^{-2}}{0.04}=0.9$ N, attractive.

$W=\frac{k{q}^2}{4}\left(\frac{1}{0.10}-\frac{1}{0.20}\right)=9\times 10^{-3}\left(10-5\right)=0.045$ J

Using the pair formula would double both answers. Only half of space contains field, so only half the energy is there to be paid for.

7. Dipoles, in full

A dipole's field at distance $r$ and polar angle $\theta$ from its axis is

$E=\frac{kp}{r^3}\sqrt{1+3{\cos }^2\theta },\tan \alpha =\frac{1}{2}\tan \theta$

where $\alpha$ is the angle between $\vec{E}$ and the radius vector. The axial and equatorial results, $2kp/r^3$ and $kp/r^3$, are the cases $\theta =0$ and $\theta =\pi /2$.

In a uniform field a dipole feels a torque but no net force:

$\vec{\tau }=\vec{p}\times \vec{E},U=-\vec{p}\cdot \vec{E}$

In a non-uniform field there is also a translational force $F=p\frac{dE}{dx}$ along the dipole. This is why any dipole, permanent or induced, is drawn towards regions of stronger field — and why an uncharged scrap of paper jumps to a charged comb.

Illustration 11

Two identical dipoles of moment $p$ lie on the same axis, separated by $r$, pointing the same way. Find the force between them.

The second sits in the axial field of the first, $E=\frac{2kp}{r^3}$, so

$F=p\frac{dE}{dr}=p\frac{d}{dr}\left(\frac{2kp}{r^3}\right)=-\frac{6k{p}^2}{r^4}$

The magnitude is $\frac{6k{p}^2}{r^4}$, and the sign shows it is attractive.

Note the fourth power. Dipole-dipole forces fall off far faster than Coulomb forces, which is exactly why neutral matter holds together only at short range.

Illustration 12

A dipole of moment $p$ is placed at angle $60^{\circ }$ to a uniform field $E$. Find the torque and the work needed to rotate it to $180^{\circ }$.

$\tau =pE\sin 60^{\circ }=0.866pE$

$W=U_{180}-U_{60}=\left(-pE\cos 180^{\circ }\right)-\left(-pE\cos 60^{\circ }\right)$

$=pE+0.5pE=1.5pE$

The maximum possible work is $2pE$, taken from full alignment to full anti-alignment. Starting part-way round always costs less.

8. Capacitors: forces, dielectrics and lost energy

Each plate sits in the field of the other, which is $\sigma /2{\epsilon }_0$, not $\sigma /{\epsilon }_0$. So the attraction is

$F=\frac{Q^2}{2{\epsilon }_{0}A}=\frac{1}{2}QE$

A dielectric slab part-way in is pulled further in, whether the charge or the voltage is held fixed, because in both cases the system lowers its energy by increasing the capacitance. At constant $V$, for a slab of width $b$ entering a gap $d$,

$F=\frac{{\epsilon }_{0}b{V}^2\left(K-1\right)}{2d}$

Connecting two charged capacitors always dissipates energy, no matter how good the wires:

$V_{common}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2},\Delta U=\frac{C_1{C}_2{\left(V_1-V_2\right)}^2}{2\left(C_1+C_2\right)}$

Illustration 13

A $4\mu$F capacitor charged to $200$ V is connected across an uncharged $6\mu$F capacitor. Find the common voltage and the energy lost.

$V=\frac{4\times 200+0}{10}=80$ V

$\Delta U=\frac{\left(4\right)\left(6\right){\left(200\right)}^2}{2\left(10\right)}\times 10^{-6}=\frac{24\times 4\times 10^4}{20}\times 10^{-6}=4.8\times 10^{-2}$ J

Half the original energy is gone, and it goes as heat in the wires and as radiation. The loss is independent of the wire resistance, which only changes how long the dissipation takes.

Illustration 14

A parallel-plate capacitor of area $A$ and separation $d$ carries charge $Q$. Find the work needed to double the separation, with the plates isolated.

$F=\frac{Q^2}{2{\epsilon }_{0}A}$, which is independent of separation, so

$W=Fd=\frac{Q^{2}d}{2{\epsilon }_{0}A}$

Check against energy: $U=\frac{Q^{2}d}{2{\epsilon }_{0}A}$ initially and doubles when $d$ does, so $\Delta U$ equals $W$ exactly. The field is unchanged; only the volume containing it has doubled.

Illustration 15

A slab of dielectric constant $K=3$ fills half the gap of a parallel-plate capacitor, parallel to the plates. Find the capacitance relative to the empty value.

The two halves are capacitors in series, each of separation $d/2$:

$\frac{1}{C}=\frac{d/2}{{\epsilon }_{0}A}+\frac{d/2}{3{\epsilon }_{0}A}=\frac{d}{2{\epsilon }_{0}A}\left(1+\frac{1}{3}\right)$

$C=\frac{3}{2}\frac{{\epsilon }_{0}A}{d}=1.5C_0$

Filling half the gap edge-to-edge instead would give parallel halves and $2C_0$, so the orientation of the slab matters as much as its constant.

Summary

• Integrate the potential first when you can; $V$ is a scalar and $\vec{E}=-\nabla V$ recovers the field.
• Finite line: $E_{\perp }=\frac{k\lambda }{r}\left(\sin {\theta }_1+\sin {\theta }_2\right)$, $E_{\parallel }=\frac{k\lambda }{r}\left(\cos {\theta }_2-\cos {\theta }_1\right)$.
• Ring peaks at $x=R/\sqrt{2}$ with $E_{max}=\frac{2kQ}{3\sqrt{3}{R}^2}$; a disc tends to $\sigma /2{\epsilon }_0$, independent of distance.
• $\vec{E}=-\nabla V$; equipotentials are perpendicular to field lines, and crowding means a strong field.
• Ring potential $\frac{kQ}{\sqrt{x^2+R^2}}$ differentiates straight to the field; inside an insulating sphere $V_{centre}=\frac{3kQ}{2R}$.
• Non-uniform $\rho (r)$ keeps spherical symmetry, so only $q_{enc}$ changes; check continuity at the surface.
• Cavity in a charged sphere: the internal field is uniform, $\rho \vec{d}/3{\epsilon }_0$.
• Three energies differ: interaction (pairs), self-energy ($3k{Q}^2/5R$ for a sphere, $k{Q}^2/2R$ for a shell), and density $\frac{1}{2}{\epsilon }_0{E}^2$.
• Charge in a cavity induces $-q$ on the wall and $+q$ on the outer surface, spread as if the cavity did not exist.
• Earthing sets $V=0$ for that conductor; solve for the charge that makes it so, and the original charge is irrelevant.
• Image charge: $F=\frac{k{q}^2}{4d^2}$, $\sigma (s)=-\frac{qd}{2\pi {\left(s^2+d^2\right)}^{3/2}}$, and $W_{\infty }=\frac{k{q}^2}{4d}$ — half the pair value.
• Dipole: $E=\frac{kp}{r^3}\sqrt{1+3{\cos }^2\theta }$, $\vec{\tau }=\vec{p}\times \vec{E}$, $U=-\vec{p}\cdot \vec{E}$, and $F=pdE/dx$ in a non-uniform field.
• Two axial dipoles attract with $\frac{6k{p}^2}{r^4}$ — a much faster fall-off than Coulomb.
• Plate attraction $F=\frac{Q^2}{2{\epsilon }_{0}A}$, since each plate sits in the other's field, $\sigma /2{\epsilon }_0$.
• Connecting two charged capacitors always loses $\frac{C_1{C}_2{\left(V_1-V_2\right)}^2}{2\left(C_1+C_2\right)}$, whatever the wire resistance.