Electrostatic Potential & Capacitance

The charge $+q$ sits a distance $a$ from O along $\vec{p}$, and $-q$ sits a distance $a$ from O the opposite way. Apply the cosine rule to the triangle each charge makes with O and P. The angle at O is $\theta$ for $+q$’s triangle and $180^\circ-\theta$ for $-q$’s, and $\cos(180^\circ-\theta) = -\cos\theta$, so:

A real dipole is tiny compared with the distances we care about, so take $r \gg a$. Then $a^2$ is negligible next to the other terms, and keeping only terms of first order in $a/r$:

We need $1/r_1$ and $1/r_2$, so raise each to the power $-\tfrac{1}{2}$ and use the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$, here with $n=-\tfrac{1}{2}$:

Subtract them. The leading $1/r$ terms cancel, and only the small correction terms survive:

Multiply by $kq$ and recognise $q\times2a = p$:

Two things set this apart from a single charge’s potential. It depends on direction, through $\cos\theta$, not just on distance. And it falls off as $1/r^2$, a full power of $r$ faster than $kQ/r$, for the same reason the dipole’s field falls off as $1/r^3$ instead of $1/r^2$: seen from far away, $+q$ and $-q$ are almost on top of each other, and their contributions very nearly cancel. The approximation gets better the farther away P is, and for an ideal point dipole ($2a$ shrinking to zero while $p$ stays fixed) the terms we dropped vanish completely and the formula is exact.

On the axis, beyond $+q$, $\theta=0$ and $\cos\theta=1$, so the general formula gives $V=kp/r^2$. On the axis, though, we don’t need the approximation at all, just as with the axial field in the previous chapter: P is exactly $r-a$ from $+q$ and $r+a$ from $-q$, so:

Writing $q\times2a = p$:

which agrees with the general formula at $\theta=0$. On the other side, beyond $-q$ ($\theta=180^\circ$, $\cos\theta=-1$), the nearer charge is now $-q$, so the potential has the same size but is negative: $-kp/(r^2-a^2)$. There’s also a neat consistency check waiting in §2.4, which shows that the field is minus the rate of change of potential, $E=-dV/dr$. Applying that to the exact axial potential:

exactly the axial field the previous chapter found by adding vectors, now recovered from a scalar in one line.

On the equatorial line (the perpendicular bisector of the dipole), $\theta=90^\circ$ and $\cos\theta=0$, so the general formula predicts zero. Once again this is exact, not an approximation: every point on the equatorial line is equally far from both charges, $r_1 = r_2 = \sqrt{r^2+a^2}$, so:

at every distance, near or far. This is a good place to see that zero potential does not mean zero field: the previous chapter found a field of $kp/(r^2+a^2)^{3/2}$ at these very points. There’s no contradiction. The whole equatorial plane is an equipotential surface ($V=0$ everywhere on it), so by §2.3 the field must cross it at right angles, and it does: the equatorial field points opposite to $\vec{p}$, straight through the plane. Bringing a test charge in from infinity along the equatorial line takes zero work, because the field is perpendicular to the path at every step.

The second consequence above, that a conductor’s net charge always ends up on its outer surface, is the whole working principle of the Van de Graaff generator, a machine that builds up potentials of around a million volts. What makes it work is a stronger version of that statement: charge placed inside a hollow conductor moves to the outer surface even when that outer surface is already at an enormous potential. To see why, we first need the potential of a charged spherical shell.

Take a thin spherical shell of radius $R$ carrying charge $Q$ spread uniformly over it, and let $x$ be the distance from its centre. By symmetry the field can only point radially, with the same magnitude at every point a given distance from the centre, so apply Gauss’s law (previous chapter) to an imaginary concentric sphere of radius $x$. Outside the shell ($x>R$) that sphere encloses all of $Q$; inside ($x<R$) it encloses nothing:

Outside, then, the field is exactly that of a point charge $Q$ at the centre, so bringing a test charge in from infinity to any $x \ge R$ takes the same work as for a point charge, and the potential there is $kQ/x$, reaching $kQ/R$ at the surface. Inside, the field is zero, so moving the test charge around costs no further work at all: the potential stays at its surface value everywhere in the interior.

Now place a small conducting sphere of radius $r$, carrying charge $q$, at the centre of a large hollow conducting shell of radius $R$, carrying charge $Q$. The small sphere’s charge also sits on its surface, so it is a charged shell too, and the formula above applies to both. Potential is a scalar, so each conductor’s potential is just the sum of what the two charges contribute there. The outer shell (at distance $R$ from the centre) gets $kQ/R$ from its own charge, plus $kq/R$ from the small sphere, since it lies outside the small sphere:

The small sphere (at distance $r$) gets $kq/r$ from its own charge, plus $kQ/R$ from the outer shell, since it sits inside the shell, where the shell’s potential is the same everywhere:

Subtract, inner minus outer:

Look at what’s missing: $Q$. The outer shell’s own charge raises both potentials by exactly the same amount, $kQ/R$, so it cancels out of the difference. And since $r<R$, the bracket is always positive. So as long as the inner sphere holds any positive charge at all, it sits at a higher potential than the outer shell, however much charge the shell already carries. Join the two with a wire and positive charge flows from the inner sphere to the shell, downhill in potential, and keeps flowing until the difference is zero, which by the formula happens only when $q=0$. Every bit of the inner charge ends up on the outer shell.

A Van de Graaff generator is built around exactly this arrangement. A large hollow metal sphere, the dome, sits on top of a tall insulating column. An insulating belt, driven by a motor, loops over two pulleys, one at the base and one inside the dome. Near the bottom pulley, a comb of sharp metal points connected to a high-voltage supply sprays positive charge onto the belt (sharp points concentrate the field enough to ionise the air right next to them, which is what lets charge leap across). The belt carries this charge up into the dome, where a second comb, connected to the dome, collects it. Because that hand-off happens inside the dome, the collected charge plays the role of $q$ above: it moves straight to the outer surface, however high the dome’s potential has already climbed. The belt keeps delivering charge, and the dome’s potential, $kQ/R$ by the shell formula, keeps rising as $Q$ grows.

The energy isn’t free, of course. Outside the dome, the positive charge on the belt is repelled by the dome’s positive charge, so the motor has to do work hauling each bit of charge upward against that repulsion, and that work is what ends up stored as electrostatic potential energy. Only inside the dome, where the dome’s own field vanishes, does the final hand-off cost nothing.

What finally stops the potential rising is the air around the dome. At the surface of a charged sphere the field is the point-charge field at $x=R$, $E = kQ/R^2$, while the potential is $V = kQ/R$, so the two are tied together:

Air stops insulating once the field reaches its dielectric strength, about $3\times10^6$ V/m, and the dome then leaks charge away through sparks. So the highest potential a dome can hold is $V_{\max} = E_{\max}R$, which grows with the radius: that is why Van de Graaff generators are built with large, smoothly rounded domes, with no sharp edges for breakdown to start from. The million-volt potentials they reach are used to accelerate charged particles, such as protons, to the high energies needed for experiments in nuclear physics.

A capacitor doesn’t strictly need two conductors side by side. A single isolated conductor can store charge too, and it still has a capacitance: treat the “second plate” as being at infinity, where the potential is zero, so the potential difference $V$ in $C=Q/V$ is just the conductor’s own potential.

Take a conducting sphere of radius $R$ carrying charge $Q$. All of that charge sits on its outer surface (§2.6), spread uniformly by symmetry, so it is exactly the charged shell of §2.6.1, and the whole sphere, surface and interior alike, sits at the shell’s surface potential. (A solid sphere and a hollow one behave identically for the same reason: the charge is on the surface either way.)

Substitute this into $C = Q/V$:

The charge $Q$ cancels, just as §2.7 promised: capacitance depends only on geometry, and here the geometry is a single length, the radius. A sphere twice as large holds twice the charge at the same potential, so it has twice the capacitance.

What happens when $d$ changes depends entirely on whether a battery is still attached:

• Battery connected: $V$ is held fixed by the battery. As $d$ grows, $C$ falls, so $Q=CV$ falls too (charge flows back through the battery), and with it $E=V/d$ and the stored energy.
• Battery disconnected: $Q$ is trapped, fixed. Since $E=Q/(\varepsilon_0 A)$ depends only on the charge density, the field cannot change as $d$ grows. But $V=Ed$ with $E$ fixed and $d$ growing means $V$ climbs in direct proportion. The concept page’s simulation shows exactly this, see the interactive capacitor for the live version.

Same charge $Q$ sits on every capacitor in the chain; voltages add:

always smaller than the smallest individual capacitance, the same combination rule as resistors in parallel.

Same voltage $V$ across every capacitor; charges add:

always larger than the largest individual capacitance, more plate area effectively available to store charge.

Where, physically, is that energy? A powerful way to answer is to say it is stored in the electric field itself, in the space between the plates where the field exists. To see why that picture works, take a parallel-plate capacitor (plate area $A$, gap $d$) and rewrite its stored energy entirely in terms of the field $E$ between the plates. Start from

and substitute the two parallel-plate results from §2.8, the capacitance $C = \varepsilon_0 A/d$ and the uniform-field relation $V = Ed$:

One factor of $d$ cancels, and what is left multiplying $\tfrac{1}{2}\varepsilon_0E^2$ is $Ad$, the volume of the region between the plates, which is precisely the region where the field lives (outside an ideal parallel-plate capacitor the field is zero). Divide the energy by that volume to get the energy per unit volume, the energy density $u$:

measured in J/m³ (a quick check: $\varepsilon_0E^2$ has units $\text{C}^2\text{N}^{-1}\text{m}^{-2}\times\text{N}^2\text{C}^{-2} = \text{N/m}^2 = \text{J/m}^3$). The plate area and the gap have cancelled out completely: the result depends only on the field strength at that spot. That is what makes it more than a fact about capacitors. Any electric field, however non-uniform, can be chopped into boxes small enough that the field across each one is effectively uniform, with two opposite faces lying along equipotential surfaces, and each such box behaves like a miniature parallel-plate capacitor holding $\tfrac{1}{2}\varepsilon_0E^2$ of energy per unit volume. So wherever there is an electric field, between capacitor plates, around a charged sphere, or in empty space, energy is stored at a density of $\tfrac{1}{2}\varepsilon_0E^2$.