Potential of distributions & conductors

A whole metal sits at one voltage. Its surface is an equipotential — which is why field lines always meet a conductor at right angles.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §25.5–25.6.

Before you start

What you need first

Topic 10 — $V = kQ/r$, adding potentials, and equipotential surfaces.
Topic 8 — inside a conductor $E = 0$; the charge is on the surface.
Topic 7 — a charged ball looks like a point charge from outside.

What you'll be able to do

Explain why a conductor is all at one potential.
Give the potential of a charged sphere: constant inside, $1/r$ outside.
Explain why field lines meet a conductor at $90^\circ$.
Find the potential of spread-out charge by adding contributions.

The key fact

A conductor sits at one potential

Inside a conductor in equilibrium $E = 0$ (Topic 8). Moving a charge from one point to another inside takes work $W = q\,\Delta V$ — but with no field there is no work, so $\Delta V = 0$. Every point of the conductor, surface included, is at the same potential.

A conductor is an equipotential. Its surface is an equipotential surface — and from Topic 10, field lines cross an equipotential at right angles. That is why field lines always meet a conductor's surface perpendicular to it.

The potential of a charged sphere

Outside, a charged conducting sphere (radius $R$, charge $Q$) looks like a point charge. Inside, the potential cannot change (it is a conductor), so it stays at its surface value:

$$ V = \frac{kQ}{r}\;(r \ge R), \qquad V = \frac{kQ}{R}\;(r \le R) $$

So $V$ is flat inside and falls off as $1/r$ outside, while the field $E$ is zero inside and falls off as $1/r^2$ outside.

Charged sphere: $V$ is flat inside $R$ then $1/r$; $E$ is zero inside then $1/r^2$. $R$ marks the surface.

📐 Worked example 1

Potential of a charged metal sphere

A metal sphere of radius $R = 0.10$ m carries $Q = 4.0\ \mu\text{C}$. Find the potential (a) at its surface, (b) at $r = 0.30$ m, (c) at its centre.

1At the surface, $V = kQ/R$:

$$ V = \frac{(8.99\times10^{9})(4.0\times10^{-6})}{0.10} \approx 3.6\times10^{5}\ \text{V} $$

2Outside at $r = 0.30$ m, $V = kQ/r$:

$$ V = \frac{(8.99\times10^{9})(4.0\times10^{-6})}{0.30} \approx 1.2\times10^{5}\ \text{V} $$

3At the centre — anywhere inside — $V$ equals its surface value:

$$ V_{\text{centre}} \approx 3.6\times10^{5}\ \text{V} $$

Note $E = 0$ inside but $V \ne 0$: zero field just means $V$ does not change there.

Potential of spread-out charge

For continuous charge, add up the potential of every tiny piece — a scalar sum, so much easier than the field integral of Topic 4 (no components, no angles):

$$ V = k\!\int \frac{dq}{r} $$

Same pieces as the field integral, but you add plain numbers. Once you have $V$ everywhere, you can even get the field back from $E = -\Delta V/\Delta x$ (Topic 10).

✏️ Try it yourself

A metal sphere of radius $R = 0.20$ m is raised to a potential of $9000$ V.

(a) What charge is on it?
(b) What is the potential at its centre?

(a) From $V = kQ/R$: $Q = \dfrac{VR}{k} = \dfrac{(9000)(0.20)}{8.99\times10^{9}} \approx 2.0\times10^{-7}\ \text{C} = 0.20\ \mu\text{C}.$ (b) The centre is inside the conductor, so it is at the same potential: $9000$ V.

Common mistakes

Mistake	Fix
Thinking $V = 0$ inside a conductor because $E = 0$.	$E = 0$ means $V$ does not change — $V$ is constant (and usually non-zero) inside.
Using $V = kQ/r$ with $r$ inside the sphere.	Inside, $V$ is fixed at the surface value $kQ/R$.
Expecting field lines to graze a conductor.	The surface is an equipotential, so field lines meet it at $90^\circ$.
Adding potentials of a distribution as vectors.	$V = k\!\int dq/r$ is a scalar sum — just add the numbers.

Recap — the whole topic on one screen

Conductor	All one potential; surface is an equipotential.
Field lines	Meet a conductor's surface at $90^\circ$.
Charged sphere	$V = kQ/r$ outside; $V = kQ/R$ (constant) inside.
Distribution	$V = k\!\int dq/r$ — a scalar sum.

Next topic

Capacitance

Put two conductors at different potentials and they store charge. How much charge per volt a pair of conductors holds is its capacitance — the start of capacitors.

→ Topic 12 · Capacitance & calculating it

Mistake	Fix
Thinking \(V = 0\) inside a conductor because \(E = 0\).	\(E = 0\) means \(V\) does not change — \(V\) is constant (and usually non-zero) inside.
Using \(V = kQ/r\) with \(r\) inside the sphere.	Inside, \(V\) is fixed at the surface value \(kQ/R\).
Expecting field lines to graze a conductor.	The surface is an equipotential, so field lines meet it at \(90^\circ\).
Adding potentials of a distribution as vectors.	\(V = k\!\int dq/r\) is a scalar sum — just add the numbers.