Unit 10 · Topic 10.2

10.2 Redistribution of Charge Between Conductors

When conductors touch or are joined by a wire, charge flows between them until they're all at the same potential. The total charge is conserved, but the bigger conductor ends up with more, so the split is equal only when the conductors are identical. Grounding uses the Earth as a huge conductor at zero potential.

Key terms

electrical contact
equal potential
charge sharing
ground
charging by induction

Charge flows until potentials match

Connect two conductors and they become one conductor. In equilibrium, one conductor has one potential (10.1). So charge flows from the higher-potential conductor to the lower one until their potentials are equal. Only then does the flow stop.

Two rules decide the final charges: the potentials end up equal, and the total charge stays the same.

Two spheres joined by a long wire

Take spheres of radius $R_1$ and $R_2$ , far enough apart that they don't affect each other, joined by a thin wire. Equal potentials mean $\dfrac{kq_1}{R_1} = \dfrac{kq_2}{R_2}$ , so $\dfrac{q_1}{q_2} = \dfrac{R_1}{R_2}$ . The bigger sphere ends up with more charge.

Surface charge density goes the other way. Since $\sigma = \dfrac{q}{4\pi R^2}$ and q is proportional to R, σ is proportional to 1/R. The smaller sphere has the larger σ, and so the stronger field at its surface ( $E = \sigma/\varepsilon_0$ ). This is the reason charge piles up at sharp points on a conductor: a sharp point acts like a tiny sphere.

Identical conductors

If two conductors are identical, symmetry says equal potentials need equal charges. They split the total evenly. That's the rule from 8.2, and it's a special case of the general one. It doesn't work for spheres of different sizes.

Grounding

The Earth acts like an enormous conductor whose potential barely changes however much charge it takes or gives. We call its potential zero. Ground a conductor and charge flows until the conductor is at zero potential too.

If there are no other charges nearby, grounding drains a conductor to zero net charge. If a charged object is nearby, the conductor ends up with whatever induced charge makes its potential zero, and that charge is opposite to the nearby object. Break the ground connection first and the conductor keeps that charge; this is charging by induction.

You can even find how much. Ground a sphere of radius R while a point charge q sits a distance D from its center. The potential at the center must be zero. Every bit of the sphere's induced charge $q'$ is the same distance R from the center, however unevenly it's spread, so $\dfrac{kq}{D} + \dfrac{kq'}{R} = 0$ and $q' = -q\dfrac{R}{D}$ . The closer the charge, the more charge the sphere pulls up from the ground.

A small conductor touching a big one

Because the bigger conductor takes the bigger share (for spheres, charge is proportional to radius), a small conductor touched to a big charged one takes only a small share. Touch a marble-sized metal ball to a large charged dome and the dome keeps almost all its charge. That's why a tiny test sphere can sample the charge on a big conductor without changing it much, and why grounding works: next to the Earth, any object is tiny.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Two different spheres joined by a wire
Two metal spheres, of radius 0.10 m and 0.30 m, are far apart and joined by a long thin wire. Together they carry +8.0 nC. Find the charge on each sphere, their common potential, and the ratio of their surface charge densities.
Show the solutionHide the solution
1. Step 1: Equal potentials: $\dfrac{q_1}{0.10} = \dfrac{q_2}{0.30}$ , so $q_2 = 3q_1$ .
2. Step 2: Conservation: q₁ + 3q₁ = 8.0 nC, so q₁ = 2.0 nC and q₂ = 6.0 nC.
3. Step 3: Potential: $V = \dfrac{(9.0\times10^{9})(2.0\times10^{-9})}{0.10} = 180$ V. Check with the big sphere: $\dfrac{(9.0\times10^{9})(6.0\times10^{-9})}{0.30} = 180$ V.
4. Step 4: Since σ is proportional to 1/R, $\dfrac{\sigma_{\text{small}}}{\sigma_{\text{large}}} = \dfrac{0.30}{0.10} = 3$ . The small sphere has 3 times the surface charge density.
Answer: 2.0 nC on the small sphere, 6.0 nC on the large one; both at 180 V; σ is 3 times larger on the small sphere
Example 2Calculator allowed
Identical spheres, before and after touching
Two identical small metal spheres carry +7.0 nC and −3.0 nC. They attract each other with force F when a distance r apart. They are touched together, then returned to distance r. Find the new force in terms of F.
Show the solutionHide the solution
1. Step 1: Before: $F = \dfrac{k(7.0)(3.0)}{r^2}$ (in nC²), attractive.
2. Step 2: Touching: identical spheres share the total equally. (7.0 − 3.0) ÷ 2 = +2.0 nC each.
3. Step 3: After: $F' = \dfrac{k(2.0)(2.0)}{r^2}$ , repulsive. So $F' = \dfrac{4}{21}F$ .
Answer: 4/21 of F (about 0.19F), and now repulsive
Example 3Calculator allowed
Induction with two touching spheres (classic trap)
Two neutral metal spheres on insulating stands touch each other. A positively charged rod is brought near sphere A, on the side away from B. With the rod still there, the spheres are pulled apart; then the rod is removed. What charge does each sphere have?
Show the solutionHide the solution
1. Step 1: While touching, the spheres act as one conductor. The rod attracts electrons toward A, so A becomes negative and B, on the far side, becomes positive.
2. Step 2: Separating the spheres while the rod is still there traps those charges. Conservation of charge says they're equal and opposite, since the pair started neutral.
3. Step 3: Removing the rod afterward doesn't change the net charge on either sphere; each one's charge just spreads evenly over its own surface.
4. Step 4: The trap is removing the rod before separating, which lets the electrons flow back and leaves both neutral.
Answer: A is negative and B is positive, with equal amounts of charge

Common mistakes

Splitting charge equally between conductors of different sizes. Equal potentials, not equal charges, decide the split.
Expecting the larger sphere to have the larger surface charge density. It has more charge but less charge per area.
Thinking a grounded conductor always ends up neutral. With a charged object nearby, it holds an induced charge.

On the exam

Questions often ask which sphere has more charge, more surface charge density or a stronger surface field after connection. Write the equal-potential condition first, then compare.
For induction procedures, describe electron movement step by step and say why the order of steps matters.

Connected topics

Videos

Topic 10.2 - Redistribution of Charge Between Conductors
Lessons With LondotWatch on YouTube (opens in a new tab)
AP Physics C E&M - Unit 10 - Lesson 1 - Charging Mechanisms
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Two Charged Spherical Conductors
Physics NinjaWatch on YouTube (opens in a new tab)
Conservation of Charge Example Problems
Flipping PhysicsWatch on YouTube (opens in a new tab)
Grounding
The Physics ClassroomWatch on YouTube (opens in a new tab)
Charging By Induction - Electrostatics
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 10.2 Redistribution of Charge Between Conductors. Pick an answer to see if you got it, and why.

A metal sphere of radius R and a metal sphere of radius 3R are far apart and joined by a long, thin wire. Together they hold total charge Q, and they are in equilibrium.

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

Described situation

Question 1 of 4Calculator allowed

How much charge is on the smaller sphere?

Question 2 of 4Calculator allowed

How does the field just outside the smaller sphere compare with the field just outside the larger sphere?

Question 3 of 4Calculator allowed

What is the potential of the two spheres?

Question 4 of 4Calculator allowed

A positively charged metal sphere, with no other charges nearby, is connected to ground by a wire. What happens?

0 of 4 answered