Electric Systems

AP Physics 2· AP Physics 2 CED — Electric Force, Field, and Potential· 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 $+ 6 μ C$ , $- 4 μ C$ , and $+ 2 μ C$ 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 $+ 6 μ C + (- 4 μ C) = + 2 μ C$ . Since spheres are identical, charge splits equally:
3
$Q_{A2} = Q_{B2} = +1 \mu\text{C}$
4
After B touches C: total charge for the pair is $+ 1 μ C + (+ 2 μ C) = + 3 μ C$ . Again, identical spheres split charge equally:
5
$Q_{B3} = Q_{C3} = +1.5 \mu\text{C}$
6
Final charge on Sphere B is $+ 1.5 μ C$ .

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 $k = 8.99 \times 1 0^{9} Nm^{2} / C^{2}$ , $q_{i}$ 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.

📐 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 $s$ ):
3
$U_1 = k \frac{(+q)(+q)}{s} = \frac{kq^2}{s}$
4
Pair 2 ( $+ q, - q$ , separation $s$ ):
5
$U_2 = k \frac{(+q)(-q)}{s} = -\frac{kq^2}{s}$
6
Pair 3 ( $+ q, - q$ , separation $s$ ):
7
$U_3 = k \frac{(+q)(-q)}{s} = -\frac{kq^2}{s}$
8
Sum the three potential energies:
9
$U_{\text{total}} = \frac{kq^2}{s} - \frac{kq^2}{s} - \frac{kq^2}{s} = -\frac{kq^2}{s}$
10
The negative sign confirms this is a bound system, as expected with two attractive interactions and one repulsive interaction.

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?

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, $Φ_{E} = 0 = \frac{Q _{enclosed}}{ϵ _{0}}$ , 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 zero.

5. Common Pitfalls

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 $Q_{enclosed}$ 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. Quick Reference Cheatsheet

Category	Formula	Notes
Conservation of Charge (Closed System)	$\sum Q_{initial} = \sum Q_{final}$	Applies when no charge crosses the system boundary
Conservation of Charge (Open System)	$Δ Q_{system} = Q_{in} - Q_{out}$	Applies when charge can enter/leave the system
Charge Redistribution (Identical Conductors)	$Q_{1} = Q_{2} = \frac{Q _{total}}{2}$	Only for identical conductors after contact at equilibrium
Multi-Charge Potential Energy	$U_{total} = k \sum_{i < j} \frac{q _{i} q _{j}}{r _{ij}}$	Count each unique pair only once; $k = 1/ (4 π ϵ_{0})$
Gauss's Law	$\oint E \cdot d A = \frac{Q _{enclosed}}{ϵ _{0}}$	Only charge inside the Gaussian surface contributes to net flux
Induced Charge (Hollow Conductor)	$Q_{inner} = - Q_{cavity}$	Applies for any hollow conductor with charge inside its cavity
Electric Field Outside Conducting Sphere	$E = \frac{k Q _{outer}}{r ^{2}}$	Matches the field of a point charge equal to the outer surface charge

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Charge redistribution on conductors
2023 · FRQ
Induced charge on conducting shells

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.