Potential of continuous distributions, charged conductors and the Millikan experiment

So far you have added up potentials for point charges. In real objects the charge is spread over a rod, a ring or a disk. The method is the same: divide the object into small pieces $dq$ and add up (that is, integrate) the contribution $k\,dq/r$ of each. Compared with the field calculation there is a big advantage: because the potential is a scalar, there is no splitting into components and no looking for cancellation by symmetry. The second half of the week is conductors: all points of a charged conductor are at the same potential, and this simple fact explains the sharp-tip effect, the lightning rod and the Faraday cage. Finally, the Millikan experiment shows that charge is quantized. On the exam, the ring and disk formulas, spheres joined by a wire, and the Millikan balance question come up very often.

Golden rules:

1. $V = k\int dq/r$. First write $dq$ ($\lambda\,dx$, $\sigma\,dA$), then write $r$ in terms of the integration variable.
2. On the axis of a ring and a disk everything is a ready-made formula. Memorize them, and see the derivation once.
3. Conductor = equipotential. Inside, $E = 0$ and $V$ equals the surface value (not zero).
4. Conductors joined by a wire come to the same potential. Charge is shared in proportion to the radius.
5. Millikan: $qE = mg$. The charge always turns out to be an integer multiple of $e$.

Potential of continuous distributions

Each small piece acts like a point charge:

$dV = k\frac{dq}{r} \quad\Rightarrow\quad V = k\int\frac{dq}{r}$

Charge densities: linear charge density $\lambda = Q/L$, $dq = \lambda\,dx$; surface charge density $\sigma = Q/A$, $dq = \sigma\,dA$. The reference is again $V = 0$ at infinity.

There is a second way too: if the field is already known (like the sphere and shell you found with Gauss’s law), you integrate with $\Delta V = -\int\vec E\cdot d\vec s$. For symmetric objects the second way is faster.

Ring, on the axis

A charge $Q$ is distributed uniformly on a ring of radius $a$. For a point on the axis at distance $x$ from the center, every piece of the ring is at the same distance: $r = \sqrt{x^2 + a^2}$. This constant comes out of the integral:

$V = \frac{k}{\sqrt{x^2 + a^2}}\int dq = \frac{kQ}{\sqrt{x^2 + a^2}}$

At the center ($x = 0$) $V = kQ/a$; far away ($x \gg a$) $V \approx kQ/x$, a point charge. The field along the axis comes from the derivative:

$E_x = -\frac{dV}{dx} = \frac{kQx}{(x^2 + a^2)^{3/2}}$

This is the same result you found in the field week by long component work. Through the potential it came in two lines.

Disk, on the axis

Divide a disk of radius $R$ into rings of radius $r$ and thickness $dr$. The charge of each ring is $dq = \sigma\,2\pi r\,dr$, and its distance to the point on the axis is $\sqrt{x^2 + r^2}$:

$V = \int_0^R \frac{k\,\sigma\,2\pi r\,dr}{\sqrt{x^2 + r^2}} = 2\pi k\sigma\left(\sqrt{x^2 + R^2} - x\right)$

At the center $V = 2\pi k\sigma R$.

Rod, on its extension

A rod of length $L$ lies between $x' = 0$ and $x' = L$, and the point is a distance $d$ beyond the end of the rod ($x = L + d$). The distance from a piece to the point is $L + d - x'$:

$V = k\lambda\int_0^L \frac{dx'}{L + d - x'} = k\lambda\ln\frac{L + d}{d}$

PITFALL · Collecting the charge at one point. A rod or disk is not a point charge when seen from close up. Putting all the charge at the center and writing $kQ/r$ gives the wrong result; the integral is required. This approximation is valid only when the distance is much larger than the size of the object.

Solid sphere and shell

You know the fields from Gauss’s law. If you integrate inward from infinity:

At the center of the insulating sphere $V = 1.5\,kQ/R$. Because the field inside is not zero, the potential keeps increasing from the surface toward the center.

Potential of a charged conductor

Inside a conductor in electrostatic equilibrium $\vec E = 0$, and at the surface the field is perpendicular to the surface. Walking along the surface, $\vec E\cdot d\vec s = 0$, and inside $\vec E = 0$. The result: the whole conductor (inside and surface) is at a single potential.

PITFALL · If the field is zero inside, the potential is zero. No. $E = 0$ means the potential does not change. The value inside equals the value at the surface: $kQ/R$ for a charged sphere.

The sharp-tip effect

Connect two conducting spheres of radii $r_1$ and $r_2$, far from each other, with a thin wire. They come to the same potential:

$\frac{kq_1}{r_1} = \frac{kq_2}{r_2} \;\Rightarrow\; \frac{q_1}{q_2} = \frac{r_1}{r_2}, \qquad \frac{E_1}{E_2} = \frac{q_1/r_1^2}{q_2/r_2^2} = \frac{r_2}{r_1}$

The smaller sphere takes less charge, but the field at its surface is larger. The sharp tip of a conductor acts like a small-radius sphere: the charge density and field are largest there. If the field exceeds the limit air can withstand (about $3\times10^{6}$ V/m), the air ionizes and corona discharge occurs. This is why lightning rods are made pointed.

A cavity inside a conductor

If a charged conductor has a cavity with no charge in it, $E = 0$ in the cavity as well. The Faraday cage and protection from lightning inside a car rely on this.

The Millikan oil-drop experiment

Millikan studied small charged oil drops between two horizontal plates. When the potential difference between the plates is adjusted, a drop hangs in the air. At that moment:

$qE = mg, \qquad E = \frac{\Delta V}{d} \quad\Rightarrow\quad q = \frac{mgd}{\Delta V}$

In thousands of measurements $q$ always came out as an integer multiple of $1.60\times10^{-19}$ C: $q = ne$. This is the experimental proof that charge is quantized. In the real experiment the mass of the drop is found from the terminal speed with the field off; on the exam it is usually given.

Applications

• Van de Graaff generator: A belt carries charge into a conducting dome. Because charge delivered inside immediately moves to the outer surface, the potential of the dome can rise very high (millions of volts). The limit is set by corona discharge, which is why the dome is made large and smooth.
• Electrostatic precipitator: Dust particles in flue gas are charged in the corona region around a high-voltage wire, then attracted to grounded walls and collected.

Worked examples

Example 1. Ring. $a = 0.30$ m, $Q = +12$ nC. What is the potential on the axis at $x = 0.40$ m and at the center?

r = sqrt(0.40² + 0.30²) = 0.50 m
V(0.40) = kQ/r = (8.99e9)(12e-9)/0.50 = 216 V
V(0)    = kQ/a = (8.99e9)(12e-9)/0.30 = 360 V

Example 2. Disk. $R = 0.10$ m, $\sigma = 2.0\ \mu\text{C/m}^2$. What is the potential on the axis at $x = 0.050$ m?

sqrt(x² + R²) = sqrt(0.0025 + 0.0100) = 0.1118 m
V = 2π(8.99e9)(2.0e-6)(0.1118 - 0.050) = 6.98e3 V
Center: V = 2πkσR = 1.13e4 V

Example 3. Rod. $L = 1.0$ m, $Q = +6.0$ nC, the point is on the extension, $d = 0.50$ m beyond the end.

λ = 6.0e-9 C/m
V = kλ ln((1.0 + 0.50)/0.50) = (8.99e9)(6.0e-9)(ln 3) = 59.3 V

Example 4. Millikan. A drop of mass $3.3\times10^{-15}$ kg and charge $3e$ hangs between plates 1.2 cm apart. What voltage is needed?

ΔV = mgd/q = (3.3e-15)(9.80)(0.012)/(3 × 1.60e-19)
ΔV = 809 V

Does the field point up or down? If the drop is negatively charged (excess electrons), for the force to be upward the field must point down; that is, the upper plate is positive.

Example 5. Spheres joined by a wire. Radii 2.0 cm and 8.0 cm, total charge 50 nC.

q1/q2 = r1/r2 = 1/4  ->  q1 = 10 nC,  q2 = 40 nC
Common potential: V = kq1/r1 = (8.99e9)(10e-9)/0.020 = 4.50e3 V
Surface fields: E1 = kq1/r1² = 2.25e5 V/m,  E2 = kq2/r2² = 5.62e4 V/m
E1/E2 = 4 = r2/r1

Common mistakes