United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How does charge redistribute when conductors are connected, and what equalises between them?

Topic 10.2 Redistribution of Charge between Conductors: predict how charge redistributes when conductors are connected, using the equalisation of potential.

A calculus-based answer to AP Physics C E&M Topic 10.2, covering charge sharing between connected conductors, equalisation of potential, the role of size and curvature, and grounding.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

When conductors are connected (touched or joined by a wire), charge flows between them until they reach the same potential, because any potential difference would drive a current. Total charge is conserved during the redistribution. For two connected spheres, $V_1 = V_2$ means $\dfrac{kq_1}{r_1} = \dfrac{kq_2}{r_2}$ , so the charge splits in proportion to radius: the larger sphere holds more charge ( $q \propto r$ ). But the surface charge density $\sigma = \dfrac{q}{4\pi r^2} \propto \dfrac{1}{r}$ is higher on the smaller sphere, which is why charge concentrates at points and sharp edges. Grounding a conductor connects it to an effectively infinite reservoir at $V = 0$ , draining or supplying charge until the conductor sits at zero potential.

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What this topic is asking
Connecting conductors equalises potential
Sharing depends on size, density depends on curvature
Grounding
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What this topic is asking

The College Board (Topic 10.2) wants you to predict how charge redistributes when conductors are connected: charge flows until the connected conductors reach a common potential, with the larger conductor taking more charge and the smaller, more sharply curved one carrying the higher surface charge density. Charge is conserved throughout.

Connecting conductors equalises potential

Any potential difference between connected conductors would drive a current through the connection, so equilibrium is reached only when the potentials match. The redistribution is set by two equations: $V_1 = V_2$ , and $q_1 + q_2 = Q_{total}$ (conservation).

For two connected spheres of radii $r_1$ and $r_2$ , equal potential gives

\frac{kq_1}{r_1} = \frac{kq_2}{r_2} \quad\Longrightarrow\quad \frac{q_1}{q_2} = \frac{r_1}{r_2}

so charge is shared in proportion to radius: the bigger sphere takes more. The surface charge density, however, behaves oppositely:

\sigma = \frac{q}{4\pi r^2} \propto \frac{q}{r^2} \propto \frac{r}{r^2} = \frac{1}{r}

so the smaller sphere has the higher density. This is the physics of why charge piles up at points: a sharp tip is like a tiny-radius sphere, giving a large $\sigma$ and a strong local field (the basis of lightning rods and corona discharge).

Grounding

Grounding connects a conductor to the Earth, an essentially infinite reservoir of charge held at potential $V = 0$ . Charge flows on or off the conductor until it too sits at $V = 0$ . This is how charging by induction is completed and how unwanted charge is safely drained away.

Two connected spheres reaching equilibrium

Sphere A (radius $0.05$ m) holds $+12$ nC; sphere B (radius $0.15$ m) is neutral. A thin wire connects them. Find the final charge on each.

step 1 Write the two conditions

Equal potential: $\dfrac{kq_A}{0.05} = \dfrac{kq_B}{0.15}$ . Conservation: $q_A + q_B = 12$ nC.

step 2 Use the potential condition

$\dfrac{q_A}{0.05} = \dfrac{q_B}{0.15}$ gives $q_B = 3q_A$ (B's radius is three times A's, so it holds three times the charge).

step 3 Substitute into conservation

$q_A + 3q_A = 12$ nC, so $4q_A = 12$ nC, $q_A = 3.0$ nC and $q_B = 9.0$ nC.

step 4 Check the densities

$\sigma_A = \dfrac{3.0\,\text{nC}}{4\pi(0.05)^2}$ exceeds $\sigma_B = \dfrac{9.0\,\text{nC}}{4\pi(0.15)^2}$ ; the smaller sphere A carries the higher density, as expected.

Try this

Q1. Two identical conducting spheres carry $+8$ nC and $+2$ nC. They are connected by a wire. Find the final charge on each. [2 points]

Cue. Equal radii share equally: each ends with $\dfrac{8 + 2}{2} = +5$ nC.

Q2. State which has the larger surface charge density after connection: a small sphere or a large one. [1 point]

Cue. The small sphere; $\sigma \propto 1/r$ , so charge concentrates on smaller curvature.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I (multiple choice). Two isolated conducting spheres, radii

R

and

2R

, are connected by a long thin wire and share charge. The ratio of their surface charge densities

\sigma_{small}/\sigma_{large}

is (A)

1

(B)

2

(C)

1/2

(D)

4

. Justify your reasoning.

Show worked answer →

A 1-point MCQ on charge sharing between connected spheres. The answer is (B).

Connected conductors reach the same potential: $\dfrac{kq_1}{R} = \dfrac{kq_2}{2R}$ , so $q_2 = 2q_1$ (the larger sphere holds more charge). Surface density $\sigma = \dfrac{q}{4\pi r^2}$ : $\sigma_{small} = \dfrac{q_1}{4\pi R^2}$ , $\sigma_{large} = \dfrac{2q_1}{4\pi(2R)^2} = \dfrac{q_1}{8\pi R^2}$ . The ratio is $\dfrac{\sigma_{small}}{\sigma_{large}} = 2$ . The smaller sphere has the higher density (charge concentrates on sharp curvature).

AP 2024 (style)4 marksSection II (FRQ, quantitative). Sphere A (radius

0.10

m) carries

+9.0

nC; sphere B (radius

0.20

m) is neutral. They are connected by a long thin wire. (a) Explain what quantity equalises. (b) Calculate the final charge on each sphere. (c) State which sphere has the larger surface charge density.

Show worked answer →

A 4-point FRQ on charge redistribution by potential equalisation.

(a) Equalisation (1 point): connected conductors reach a common potential $V$ ; charge flows until $V_A = V_B$ .
(b) Charges (2 points): $\dfrac{kq_A}{0.10} = \dfrac{kq_B}{0.20}$ gives $q_B = 2q_A$ . With $q_A + q_B = 9.0$ nC: $q_A + 2q_A = 9.0$ , so $q_A = 3.0$ nC and $q_B = 6.0$ nC.
(c) Density (1 point): the smaller sphere A has the larger surface charge density (charge per area is higher on the smaller radius).

Markers reward equal potential, the proportional split, and identifying the higher-density sphere.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)