Physics 2 · 14 min read · Updated 2026-05-11

Electric Systems — AP Physics 2

AP Physics 2 · AP Physics 2 CED Unit 3 · 14 min read

1. What Is an Electric System? ★★☆☆☆ ⏱ 3 min

An electric system is any defined collection of charged objects, conductors, and associated electric fields bounded by an explicit closed surface chosen for analysis. Unlike analyzing isolated charges, studying electric systems requires tracking what crosses the system boundary, applying conservation rules, and calculating net properties for the entire collection. This topic makes up ~3-5% of the total AP Physics 2 exam score, appearing in both multiple-choice and free-response sections.

2. Conservation of Charge in Electric Systems ★★☆☆☆ ⏱ 4 min

All analysis of electric systems starts with conservation of charge, the fundamental rule that charge cannot be created or destroyed, only transferred or rearranged. Systems are classified by their boundary:

\text{Closed System: } \sum Q_{\text{initial}} = \sum Q_{\text{final}}

\text{Open System: } \Delta Q_{\text{system}} = Q_{\text{in}} - Q_{\text{out}}

A common exam application is charge redistribution when two conducting spheres are brought into contact. Charge moves freely on conductors, so the system reaches electrostatic equilibrium with equal electric potential on both spheres. For identical conductors (same radius, same capacitance), charge splits equally between them.

📐 Worked Example

Three identical conducting spheres on insulating stands have initial charges of $+6 \mu\text{C}$, $-4 \mu\text{C}$, and $+2 \mu\text{C}$ respectively. Sphere A touches Sphere B, then they are separated. Then Sphere B touches Sphere C, then they are separated. What is the final charge on Sphere B?

This is a closed system (no charge enters or leaves the collection of spheres), so total charge is conserved at every step.
After A touches B: total charge for the pair is $+6 \mu\text{C} + (-4 \mu\text{C}) = +2 \mu\text{C}$. Since spheres are identical, charge splits equally:
$Q_{A2} = Q_{B2} = +1 \mu\text{C}$
After B touches C: total charge for the pair is $+1 \mu\text{C} + (+2 \mu\text{C}) = +3 \mu\text{C}$. Again, identical spheres split charge equally:
$Q_{B3} = Q_{C3} = +1.5 \mu\text{C}$
Final charge on Sphere B is $+1.5 \mu\text{C}$.

3. Electric Potential Energy of Multi-Charge Systems ★★★☆☆ ⏱ 4 min

The total electric potential energy of a system of point charges is equal to the total work required to assemble the system from infinite separation, where all charges are initially at rest infinitely far apart. To calculate this, add the potential energy for every unique pair of charges, because potential energy is a scalar quantity.

U_{\text{total}} = \frac{1}{4\pi\epsilon_0} \sum_{i<j} \frac{q_i q_j}{r_{ij}} = k \sum_{i<j} \frac{q_i q_j}{r_{ij}}

where $k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2$, $q_i$ and $q_j$ are the charges of the pair, $r_{ij}$ is the distance between them, and the $i<j$ convention ensures we count each pair only once, avoiding double-counting. A negative total potential energy means the system is bound: net work is done by the electric field during assembly, so you must add external energy to pull all charges apart to infinity. A positive total means the system is unbound, with net repulsive interactions.

4. Gauss's Law for Enclosed Charge in Electric Systems ★★★☆☆ ⏱ 3 min

Gauss's law connects the net electric flux through a closed Gaussian surface (our system boundary) to the net charge enclosed by that surface. This is the primary tool for finding induced charge on conducting surfaces in electrostatic systems.

\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}

A key property of this law is that only charge inside the Gaussian surface contributes to the net flux. Any charge outside the surface produces zero net flux, because every electric field line that enters the surface also exits it. For conductors in electrostatic equilibrium, the electric field inside the conducting material is always zero, which lets us solve for induced charge by placing a Gaussian surface inside the conductor material.

📐 Worked Example

A neutral hollow conducting spherical shell has a point charge of $+Q$ placed at the center of the inner cavity. What is the charge on the inner surface of the shell, and what is the charge on the outer surface?

Choose a Gaussian surface that lies entirely within the conducting material of the shell, between the inner cavity surface and the outer surface of the shell.
For a conductor in electrostatic equilibrium, the electric field everywhere inside the conductor material is zero, so the net flux through the Gaussian surface is zero.
By Gauss's law, $\Phi_E = 0 = \frac{Q_{\text{enclosed}}}{\epsilon_0}$, so total enclosed charge is zero. The point charge at the center is $+Q$, so the inner surface must carry $-Q$ to give a total enclosed charge of $+Q + (-Q) = 0$.
The shell is originally neutral, so total charge of the shell is zero. If inner surface has $-Q$, the outer surface must carry $+Q$ to give a total shell charge of zero.

Common Pitfalls

Why: Students memorize the identical sphere case and incorrectly generalize it to any two conductors

Why: Students count interactions for each charge individually, leading to two entries for every pair

Why: Students confuse total charge in the entire problem with charge inside the defined system boundary

Why: Students forget induction only separates charge, it does not create new charge

Why: Students generalize conductor behavior to insulators, where charge is fixed in place

Quick Reference Cheatsheet

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