Electric Systems — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: System boundary definition for electric systems, conservation of charge, charge redistribution on conductors, multi-charge electric potential energy, Coulomb’s law for interacting charge systems, and Gauss’s law applied to enclosed charge in systems.

You should already know: Conservation of charge as a fundamental physical law, Coulomb’s law for isolated point charges, electric potential energy for a single charge in an external field.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Electric Systems?

An electric system is any defined collection of charged objects, conductors, and associated electric fields bounded by an explicit closed surface we choose for analysis. Unlike analyzing isolated charges, studying electric systems requires us to track what crosses the system boundary, apply conservation rules, and calculate net properties for the entire collection. This subtopic is part of AP Physics 2 Unit 3, accounting for approximately 3-5% of the total exam score, and appears regularly across both multiple-choice (MCQ) and free-response (FRQ) sections. Common contexts include charged conductor systems, isolated multi-charge arrangements, and open systems connected to external voltage sources. The core goal of analyzing an electric system is almost always to find net charge distribution, total potential energy, or net electric field at a point inside or outside the boundary, using fundamental conservation and field rules that we break down below.

2. Conservation of Charge in Electric Systems

All analysis of electric systems starts with conservation of charge, the fundamental rule that charge cannot be created or destroyed, only transferred or rearranged. We first classify systems by their boundary: a closed electric system has a boundary that does not allow charge to enter or leave, so the total net charge of the system is always constant. An open electric system allows charge to cross the boundary, so the change in the system’s total charge equals the net charge that crosses into the boundary. The formal rules are: $$\text{Closed System: }\sum Q_{\text{initial}}=\sum Q_{\text{final}}$$ $$\text{Open System: }\Delta Q_{\text{system}}=Q_{\text{in}}-Q_{\text{out}}$$

A common textbook and exam application is charge redistribution when two conducting spheres are brought into contact. Charge can move freely on conductors, so the system will reach electrostatic equilibrium with equal electric potential on both spheres. If the spheres are identical (same radius, same capacitance), charge splits equally between them, since equal potential implies equal charge for identical conductors.

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?

1. This is a closed system (no charge enters or leaves the collection of spheres), so total charge is conserved at every step.
2. After A touches B: total charge for the pair is $Q_{A1}+Q_{B1}=+6\mu \text{C}+(-4\mu \text{C})=+2\mu \text{C}$. Since spheres are identical, charge splits equally: $Q_{A2}=Q_{B2}=+1\mu \text{C}$.
3. After B touches C: total charge for the pair is $Q_{B2}+Q_{C1}=+1\mu \text{C}+(+2\mu \text{C})=+3\mu \text{C}$. Again, identical spheres split charge equally: $Q_{B3}=Q_{C3}=+1.5\mu \text{C}$.
4. Final charge on Sphere B is $+1.5\mu \text{C}$.

Exam tip: If the problem does not explicitly state conductors are identical, you cannot split charge equally. Use the equal potential rule $V_1=V_2$ to find the charge ratio instead.

3. Electric Potential Energy of Multi-Charge Systems

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, we add the potential energy for every unique pair of charges, because each pair interacts electrostatically, and potential energy is a scalar quantity. The formula is: $$U_{\text{total}}=\frac{1}{4\pi {ϵ}_0}\sum _{i<j}\frac{q_i{q}_j}{r_{ij}}=k\sum _{i<j}\frac{q_i{q}_j}{r_{ij}}$$ where $k=8.99\times 10^9{\text{ Nm}}^2/{\text{C}}^2$, $q_i$ and $q_j$ are the charges of the two charges in 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.

Worked Example

Three point charges $+q$, $+q$, and $-q$ are placed at the vertices of an equilateral triangle of side length $s$. What is the total electric potential energy of the system?

1. For 3 charges, there are $\frac{3(3-1)}{2}=3$ unique pairs, so we calculate the potential energy for each.
2. Pair 1 ($+q,+q$, separation $s$): $U_1=k\frac{(+q)(+q)}{s}=\frac{k{q}^2}{s}$. Pair 2 ($+q,-q$, separation $s$): $U_2=k\frac{(+q)(-q)}{s}=-\frac{k{q}^2}{s}$. Pair 3 ($+q,-q$, separation $s$): $U_3=k\frac{(+q)(-q)}{s}=-\frac{k{q}^2}{s}$.
3. Sum the three potential energies: $U_{\text{total}}=\frac{k{q}^2}{s}-\frac{k{q}^2}{s}-\frac{k{q}^2}{s}=-\frac{k{q}^2}{s}$.
4. The negative sign confirms this is a bound system, as expected with two attractive interactions and one repulsive interaction.

Exam tip: If you count interactions for each charge individually (e.g., each charge interacts with every other charge), you will get twice the correct total. Remember to divide your result by 2 if you do not use the $i<j$ counting convention.

4. Gauss's Law for Enclosed Charge in Electric Systems

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. The formal law is: $${\Phi }_E=\oint \vec{E}\cdot d\vec{A}=\frac{Q_{\text{enclosed}}}{{ϵ}_0}$$ 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?

1. 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.
2. 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.
3. By Gauss's law, ${\Phi }_E=0=\frac{Q_{\text{enclosed}}}{{ϵ}_0}$, 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 $+Q+(-Q)=0$.
4. 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 $-Q++Q=0$.

Exam tip: Always place your Gaussian surface inside the conductor material when solving for induced charge. Never place it inside the cavity or outside the shell, as this will not give you the zero electric field condition you need to solve for enclosed charge.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Splitting charge equally between two non-identical conductors after contact. Why: Students memorize the identical sphere case and incorrectly generalize it to any two conductors. Correct move: Always confirm the problem states conductors are identical before splitting charge equally; for non-identical conductors, use $V_1=V_2$ to find the charge ratio.
• Wrong move: Double-counting pairs when calculating total potential energy of a 3+ charge system. Why: Students count interactions for each charge individually, leading to two entries for every pair. Correct move: For $n$ charges, count exactly $\frac{n(n-1)}{2}$ unique pairs before summing potential energy.
• Wrong move: Including charge outside the Gaussian surface when calculating $Q_{\text{enclosed}}$ for Gauss's law. Why: Students confuse total charge in the entire problem with charge inside the defined system boundary. Correct move: Only add up charges that lie strictly inside your Gaussian surface; ignore all charges outside entirely.
• Wrong move: Assigning a non-zero net charge to a neutral conductor after induced charge separation. Why: Students forget induction only separates charge, it does not create new charge. Correct move: For any originally neutral conductor, the sum of charge on all its surfaces must equal zero after induction.
• Wrong move: Assuming charge redistributes when two charged insulating spheres are brought into contact. Why: Students generalize conductor behavior to insulators, where charge is fixed in place. Correct move: Charge does not move on insulators, so the charge of each sphere remains unchanged after contact.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

Two uncharged identical conducting spheres mounted on insulating stands are initially touching each other. A glass rod with net charge $+Q$ is brought near (but does not touch) sphere 1, which is adjacent to the rod. After the rod is brought near, the rod is removed completely, then the two spheres are separated. What is the net charge on sphere 1 after separation? A) $+Q/2$ B) $0$ C) $-Q/2$ D) $+Q$

Worked Solution: The two-sphere system is closed, with no charge transferred from the rod to the spheres (since the rod never touches them). When the rod is brought near, negative charge is induced on sphere 1 and positive charge on sphere 2, but total net charge of the system remains zero. When the rod is removed before separating the spheres, the separated charge redistributes back to a uniform neutral distribution across the two touching spheres. After separation, each sphere has net charge zero. The correct answer is B.

Question 2 (Free Response)

A system of four point charges is arranged at the corners of a square of side length $a$, as follows: corner 1: $+q$, corner 2 (adjacent to 1): $+q$, corner 3 (opposite 1): $-q$, corner 4 (adjacent to 3): $-q$. (a) How many unique pairs of charges are in this system? (b) Calculate the total electric potential energy of the system in terms of $k$, $q$, and $a$. (c) Explain why the sign of your answer makes physical sense.

Worked Solution: (a) For $n=4$ charges, the number of unique pairs is $\frac{n(n-1)}{2}=\frac{4(3)}{2}=6$ unique pairs. (b) List all 6 pairs:

1. 1-2 (separation $a$): $U_1=\frac{k(+q)(+q)}{a}=\frac{k{q}^2}{a}$
2. 1-3 (separation $a\sqrt{2}$): $U_2=\frac{k(+q)(-q)}{a\sqrt{2}}=-\frac{k{q}^2}{a\sqrt{2}}$
3. 1-4 (separation $a$): $U_3=\frac{k(+q)(-q)}{a}=-\frac{k{q}^2}{a}$
4. 2-3 (separation $a$): $U_4=\frac{k(+q)(-q)}{a}=-\frac{k{q}^2}{a}$
5. 2-4 (separation $a\sqrt{2}$): $U_5=\frac{k(+q)(-q)}{a\sqrt{2}}=-\frac{k{q}^2}{a\sqrt{2}}$
6. 3-4 (separation $a$): $U_6=\frac{k(-q)(-q)}{a}=\frac{k{q}^2}{a}$

Summing all terms: $$U_{\text{total}}=2\left(\frac{k{q}^2}{a}\right)-3\left(\frac{k{q}^2}{a}\right)-2\left(\frac{k{q}^2}{a\sqrt{2}}\right)=-(1+\sqrt{2})\frac{k{q}^2}{a}\approx -2.41\frac{k{q}^2}{a}$$ (c) The negative sign means the system is bound: there are more attractive (opposite charge) interactions than repulsive (same charge) interactions, so energy must be added to the system to separate all charges to infinite distance. This matches the expected behavior for this arrangement.

Question 3 (Application / Real-World Style)

A small, isolated Van de Graaff generator has a hollow conducting spherical dome of radius 0.15 m with a total initial charge of $+4.0\mu \text{C}$ distributed uniformly on its outer surface. A small point charge of $-1.0\mu \text{C}$ is placed at rest at the center of the hollow cavity inside the dome. No charge can enter or leave the dome after the point charge is added. Calculate the charge on the inner surface, the charge on the outer surface, and the magnitude of the electric field just outside the outer surface of the dome. Interpret your result for the electric field in context.

Worked Solution: Draw a Gaussian surface within the conducting material of the dome, between the inner and outer surfaces. $E=0$ inside the conductor, so net flux is zero, meaning total enclosed charge is zero: $Q_{\text{point}}+Q_{\text{inner}}=0⟹Q_{\text{inner}}=-(-1.0\mu \text{C})=+1.0\mu \text{C}$. Total charge of the dome is $+4.0\mu \text{C}$, so $Q_{\text{inner}}+Q_{\text{outer}}=+4.0\mu \text{C}⟹Q_{\text{outer}}=+3.0\mu \text{C}$. The electric field just outside a spherical shell is $E=\frac{k{Q}_{\text{outer}}}{r^2}=\frac{(9\times 10^9)(3.0\times 10^{-6})}{(0.15{)}^2}=1.2\times 10^6\text{ N/C}$. In context, this field is just below the dielectric breakdown strength of air ($\approx 3\times 10^6\text{ N/C}$), so this small Van de Graaff will produce small visible sparks near its surface when the air becomes slightly ionized.

7. Quick Reference Cheatsheet

8. What's Next

Mastering electric systems is the critical foundation for the next topics in Unit 3, including electric potential of charged conductors, Gauss's law applications to symmetric charge distributions, and capacitance of multi-conductor systems. Without being able to correctly apply conservation of charge and account for induced charge on conductor surfaces, you will struggle to correctly calculate capacitance or potential difference between conductors, a heavily tested topic on the AP Physics 2 exam. This topic also feeds into later units, including DC circuits, where conservation of charge is the basis for Kirchhoff's junction rule, and electromagnetism, where Gauss's law for charge is extended to other electromagnetic quantities. Next steps for your review:

• Gauss's Law Applications
• Electric Potential of Conductors
• Capacitance and Dielectrics
• Kirchhoff's Laws for Circuits