Unit 10 · Topic 10.1

10.1 Electrostatics with Conductors

A conductor's free charges keep moving until the forces on them vanish. Once they settle, the field inside the conductor is zero, any extra charge sits on the surface, the field just outside points straight out from the surface, and the whole conductor is at one potential.

Key terms

conductor
electrostatic equilibrium
surface charge
equipotential surface
electrostatic shielding

Why the field inside is zero

A conductor has electrons that are free to move. If there were any field inside the material, it would push those electrons, and charge would keep flowing. Electrostatic equilibrium is the state where charge has stopped moving, so it must be the state where the field inside the material is zero. In a metal, this happens in a tiny fraction of a second.

The electrons don't cancel the field by magic. They shift until the field made by the rearranged charges exactly cancels any outside field everywhere inside the material.

Where the extra charge goes

Draw a Gaussian surface just inside the conductor's surface. The field is zero everywhere on it, so the enclosed charge is zero (8.6). So any extra charge must sit on the surface, not inside the material.

Hollow conductors follow the same logic. If a charge +q sits inside a cavity, a Gaussian surface inside the metal around the cavity must enclose zero net charge, so the cavity's inner wall carries −q. Whatever is left of the conductor's own charge, plus +q, appears on the outer surface.

The field at the surface

Just outside a conductor, the field is perpendicular to the surface. If it had a component along the surface, surface charges would slide, and the conductor wouldn't be in equilibrium.

A small pillbox Gaussian surface straddling the surface gives the field's size: $E = \dfrac{\sigma}{\varepsilon_0}$ , where σ is the local surface charge density. Compare an isolated sheet of charge, $\dfrac{\sigma}{2\varepsilon_0}$ : for the conductor, all the flux goes out one side because the field inside is zero.

One potential throughout

Since E = 0 inside, $\Delta V = -\int\vec{E}\cdot d\vec{r} = 0$ between any two points in the conductor. The whole conductor, surface and inside, is one equipotential. For a sphere of radius R with charge Q, V = kQ/R everywhere inside and on it, and V = kQ/r outside. A graph of V against r is flat out to R, then curves down as 1/r. Zero field inside does not mean zero potential.

Sharp points and shielding

On an irregular conductor, charge crowds where the surface curves most sharply, so σ and the field just outside are largest at points and edges. Very strong fields there can ionize the air and let charge leak away, which is why high-voltage equipment is built with smooth, rounded surfaces. (10.2 explains why charge crowds there, using two connected spheres.)
A closed conducting shell with no charge in its cavity has zero field in the cavity, no matter what charges are outside. This is electrostatic shielding, and it's why a car or a metal-mesh cage protects what's inside from outside fields.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
A charge inside a conducting shell
A +3.0 nC point charge sits at the center of a thick conducting spherical shell with inner radius 0.10 m and outer radius 0.15 m. The shell carries a net charge of +5.0 nC. Find the charge on each surface of the shell and the field at r = 0.050 m, 0.12 m and 0.30 m.
Show the solutionHide the solution
1. Step 1: Inner surface: a Gaussian sphere inside the metal must enclose zero charge, so the inner surface has −3.0 nC.
2. Step 2: Outer surface: the shell's total is +5.0 nC, so the outer surface has 5.0 − (−3.0) = +8.0 nC.
3. Step 3: r = 0.050 m (in the cavity): only the point charge is enclosed. $E = \dfrac{(9.0\times10^{9})(3.0\times10^{-9})}{(0.050)^2} \approx 1.1\times10^{4}$ N/C, outward.
4. Step 4: r = 0.12 m (in the metal): E = 0.
5. Step 5: r = 0.30 m (outside): enclosed charge is 3.0 − 3.0 + 8.0 = +8.0 nC, so $E = \dfrac{(9.0\times10^{9})(8.0\times10^{-9})}{(0.30)^2} = 800$ N/C, outward.
Answer: Inner surface −3.0 nC, outer +8.0 nC; E ≈ 1.1 × 10⁴ N/C, 0 and 800 N/C (all outward where nonzero)
Example 2Calculator allowed
Field just outside a conductor
A region of a conductor's surface carries σ = 2.0 × 10⁻⁶ C/m². Find the field just outside that region.
Show the solutionHide the solution
1. Step 1: $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{2.0\times10^{-6}}{8.85\times10^{-12}} \approx 2.3\times10^{5}$ N/C.
2. Step 2: It points straight out from the surface, since σ is positive.
Answer: About 2.3 × 10⁵ N/C, perpendicular to the surface, pointing outward
Example 3Calculator allowed
Zero field, but not zero potential (classic trap)
A solid conducting sphere of radius 0.12 m carries +4.0 nC. Find the field and the potential at its center.
Show the solutionHide the solution
1. Step 1: The center is inside the conductor, so E = 0.
2. Step 2: The whole conductor is at the surface potential: $V = \dfrac{kQ}{R} = \dfrac{(9.0\times10^{9})(4.0\times10^{-9})}{0.12} = 300$ V.
3. Step 3: The trap is reasoning that zero field means zero potential. Zero field means the potential isn't changing, so it keeps its surface value.
Answer: E = 0; V = 300 V

Common mistakes

Saying the potential inside a charged conductor is zero. It's constant and equal to the surface value.
Putting extra charge throughout a conductor's volume. In equilibrium it's all on the surface (an insulator can hold charge inside).
Forgetting the induced charge on the inner wall of a shell that surrounds a charge.
Using σ/(2ε₀) for the field just outside a conductor. That's for a lone sheet; a conductor gives σ/ε₀.

On the exam

Expect graphs of E and V against r for conducting spheres and shells. E is zero inside the metal and V is flat there; check that V never jumps.
Free-response explanations should name the reason, such as "charges in a conductor move until the field inside is zero", rather than just stating the result.

Connected topics

Videos

Topic 10.1 - Electrostatics with Conductors
Lessons With LondotWatch on YouTube (opens in a new tab)
3 Properties of Conductors in Electrostatic Equilibrium
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics C - Conductors
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
Electrostatic shielding & Faraday cage | Electrostatic potential & capacitance | Khan Academy
Khan Academy India - EnglishWatch on YouTube (opens in a new tab)
Electric Fields in Metals | Physics with Professor Matt Anderson | M17-05
Physics with Professor Matt AndersonWatch on YouTube (opens in a new tab)
Irregularly Shaped Conductors in Electrostatic Equilibrium
Flipping PhysicsWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 10.1 Electrostatics with Conductors. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A solid metal sphere of radius R carries charge Q and is in electrostatic equilibrium. What are the field and the potential at its center? ( $k = \frac{1}{4\pi\varepsilon_0}$ )

A point charge +q is placed at the center of a thick, neutral, spherical metal shell. The shell's inner radius is a and its outer radius is b. Everything is in electrostatic equilibrium.

Use $k = \frac{1}{4\pi\varepsilon_0}$ .

Described situation

Question 2 of 4Calculator allowed

What charge is on the inner surface and on the outer surface of the shell?

Question 3 of 4Calculator allowed

What is the field magnitude inside the metal (a < r < b) and outside the shell (r > b)?

Question 4 of 4Calculator allowed

The point charge is moved off center but stays inside the hollow. What happens to the field outside the shell?

0 of 4 answered