Conductors and Capacitors — AP Physics C: E&M Phys C E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: Electricity & Magnetism.

Covers: Electrostatic properties of conductors, parallel plate capacitance rules, energy stored in capacitors, dielectric effects on capacitance, and series/parallel capacitor network analysis.

You should already know: Calculus (especially integration), Phys C Mechanics or equivalent.

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 C: E&M style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Conductors and Capacitors?

Conductors are materials with loosely bound charge carriers that rearrange rapidly to eliminate internal electric fields when exposed to electrostatic forces, while capacitors are two-conductor components designed to store charge and electrical energy for use in circuits. This topic makes up 14-18% of the AP Physics C E&M exam per the official course and exam description, with both multiple-choice and free-response questions testing conceptual understanding and quantitative problem-solving. Common synonyms you will encounter include dielectric constant (relative permittivity) and capacitance (charge storage capacity).

2. Conductors in electrostatic equilibrium

Electrostatic equilibrium is the state where no net motion of free charge occurs inside a conductor. This state is reached within nanoseconds of a conductor being exposed to an external electric field or given excess charge, and it follows four non-negotiable rules tested on every exam:

1. The electric field inside the bulk of the conductor is exactly 0, regardless of excess charge on the conductor.
2. All excess charge resides entirely on the surface of the conductor (for hollow conductors, charge is on the inner or outer surface depending on cavity charge).
3. The electric field just outside the conductor surface is perpendicular to the surface, with magnitude $E=\frac{\sigma }{{ϵ}_0}$ where $\sigma$ is the local surface charge density.
4. The entire conductor is an equipotential: the potential difference between any two points on or inside the conductor is 0.

Worked Example

A solid spherical conductor of radius 0.2 m carries an excess charge of 5 μC. Calculate the electric field at r = 0.1 m (inside the conductor) and r = 0.3 m (outside the conductor), and the electric potential at the surface of the conductor.

• Solution: At r = 0.1 m, $E=0$ per electrostatic equilibrium rules. At r = 0.3 m, use Gauss's law: $E=\frac{kQ}{r^2}=\frac{9\times 10^9\times 5\times 10^{-6}}{0.3^2}=5\times 10^5$ N/C directed radially outward. Surface potential: $V=\frac{kQ}{R}=\frac{9\times 10^9\times 5\times 10^{-6}}{0.2}=2.25\times 10^5$ V.

3. Capacitance and parallel plate capacitor

Capacitance $C$ is defined as the ratio of the magnitude of charge stored on either conductor of a capacitor to the potential difference between them: $$C=\frac{Q}{V}$$ Units are farads (F), where 1 F = 1 C/V. Most capacitors used in circuits have capacitances in the microfarad ($10^{-6}$ F) to picofarad ($10^{-12}$ F) range.

The parallel plate capacitor, the most common geometry tested on the exam, consists of two identical flat conducting plates separated by a small gap. We derive its capacitance using the electric field between plates ($E=\frac{\sigma }{{ϵ}_0}=\frac{Q}{{ϵ}_{0}A}$) and the relationship $V=Ed$: $$C_{parallelplate}=\frac{{ϵ}_{0}A}{d}$$ Where ${ϵ}_0=8.85\times 10^{-12}$ F/m is the permittivity of free space, $A$ is the area of one plate, and $d$ is the separation between plates. A critical exam note: capacitance depends only on the geometry of the capacitor, not on the charge stored or applied voltage.

Worked Example

Calculate the capacitance of a parallel plate capacitor with plate area 0.04 m² and plate separation 1 mm.

• Solution: $C=\frac{8.85\times 10^{-12}\times 0.04}{0.001}=3.54\times 10^{-10}$ F = 354 pF.

4. Energy stored in capacitor

When you charge a capacitor, you do work to move charge from the low-potential plate to the high-potential plate. The total work done equals the energy stored in the capacitor, derived by integrating the work to add an infinitesimal charge $dq$ against a potential $V(q)=\frac{q}{C}$: $$U={\int }_0^Q\frac{q}{C}dq=\frac{Q^2}{2C}=\frac{1}{2}C{V}^2=\frac{1}{2}QV$$ You can also describe the energy as being stored in the electric field between the plates, with energy density (energy per unit volume) $u=\frac{1}{2}{ϵ}_0{E}^2$, which works for all capacitor geometries, not just parallel plates.

Exam tip: Before solving energy problems, always note if the capacitor remains connected to a battery (fixed $V$) or is disconnected after charging (fixed $Q$), as this determines which form of the energy formula is most convenient.

Worked Example

A 10 μF capacitor is charged to 12 V, then disconnected from the battery. Calculate the stored energy, and the energy density if the plate separation is 0.5 mm.

• Solution: $U=\frac{1}{2}C{V}^2=0.5\times 10\times 10^{-6}\times 12^2=7.2\times 10^{-4}$ J. The electric field between plates is $E=\frac{V}{d}=\frac{12}{0.0005}=2.4\times 10^4$ N/C, so $u=0.5\times 8.85\times 10^{-12}\times (2.4\times 10^4{)}^2=2.55\times 10^{-3}$ J/m³.

5. Dielectrics

A dielectric is an insulating material inserted between the plates of a capacitor to increase its capacitance. When inserted, the dielectric polarizes, creating an induced electric field that opposes the original field between the plates, reducing the net electric field and potential difference (for fixed $Q$) or allowing more charge to be stored (for fixed $V$).

The dielectric constant $\kappa$ (unitless, always > 1 for real materials) is the ratio of the capacitance with the dielectric to the capacitance without: $$C=\kappa C_0=\frac{\kappa {ϵ}_{0}A}{d}$$ Every dielectric has a maximum electric field it can withstand before breaking down and conducting, called the dielectric strength, which determines the maximum voltage the capacitor can handle.

Worked Example

The 354 pF parallel plate capacitor from Section 3 has a dielectric with $\kappa =3.5$ inserted between its plates, while it remains connected to a 9 V battery. Calculate the new capacitance, and the additional charge stored on the plates after insertion.

• Solution: New capacitance $C=3.5\times 354$ pF = 1239 pF = 1.24 nF. Original charge $Q_0=C_{0}V=354\times 10^{-12}\times 9=3.19\times 10^{-9}$ C. New charge $Q=CV=1.24\times 10^{-9}\times 9=1.12\times 10^{-8}$ C. Additional charge: $Q-Q_0=8.01\times 10^{-9}$ C = 8 nC.

6. Capacitor networks

Capacitors in circuits are often combined in series and parallel configurations, which you can simplify using two core rules, opposite to the rules for resistors:

1. Parallel combination: All capacitors share the same potential difference. Total capacitance is the sum of individual capacitances: $$C_{parallel}=C_1+C_2+...+C_n$$ This makes intuitive sense: parallel capacitors effectively increase the total plate area, increasing total capacitance.
2. Series combination: All capacitors share the same stored charge. The reciprocal of total capacitance is the sum of reciprocals of individual capacitances: $$\frac{1}{C_{series}}=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$$ This makes intuitive sense: series capacitors effectively increase the total plate separation, decreasing total capacitance.

When solving network problems, always simplify the innermost combinations first, working outward to find total capacitance, then work backward to find charge and voltage across individual capacitors.

Worked Example

A 2 μF capacitor and 4 μF capacitor are connected in series, then this series combination is connected in parallel with a 3 μF capacitor. Calculate the total capacitance of the network, and total charge stored when connected to a 12 V battery.

• Solution: First find the series capacitance of the 2 μF and 4 μF: $\frac{1}{C_{12}}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$, so $C_{12}=1.33$ μF. Total capacitance is $C_{total}=1.33+3=4.33$ μF = $\frac{13}{3}$ μF. Total charge: $Q=C_{total}V=\frac{13}{3}\times 10^{-6}\times 12=52$ μC.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Assuming the electric field inside a conductor in electrostatic equilibrium is non-zero if the conductor has excess charge. Why: Students confuse conductors with insulators, where excess charge is distributed throughout the material. Correct: E inside the bulk of any conductor in electrostatic equilibrium is always exactly zero, regardless of excess charge.
• Wrong move: Using resistor series/parallel rules for capacitors. Why: The rules for capacitors are the inverse of resistor rules, leading to frequent mix-ups. Correct: Remember "capacitors parallel add directly, series add reciprocally" opposite to resistors, and double-check your logic with plate area/separation intuition.
• Wrong move: Forgetting the dielectric constant $\kappa$ when calculating capacitance with an insulator between plates. Why: Students memorize the parallel plate formula for air gaps but neglect the material factor. Correct: Always multiply the vacuum capacitance by $\kappa >1$ when a dielectric is present.
• Wrong move: Assuming charge $Q$ is constant when modifying a capacitor connected to a battery. Why: Students forget that a battery maintains a fixed potential difference across the capacitor, so charge changes as capacitance changes. Correct: Before solving capacitor modification problems, explicitly note if the battery is connected (fixed V) or disconnected (fixed Q).
• Wrong move: Calculating stored energy as $QV$ instead of $\frac{1}{2}QV$. Why: Students confuse the total work done by the battery with the energy stored in the capacitor, half of which is lost as heat or radiation during charging. Correct: Always include the ½ factor for energy stored in a capacitor.

8. Practice Questions (AP Physics C E&M Style)

Question 1

A hollow spherical conductor has inner radius $R_1=0.1$ m and outer radius $R_2=0.2$ m, with total excess charge $+8$ μC. A point charge of $-3$ μC is placed at the center of the cavity. (a) Find the charge on the inner and outer surfaces of the conductor. (b) Calculate the electric field magnitude at $r=0.05$ m, $r=0.15$ m, and $r=0.3$ m. (c) Find the potential difference between the outer surface of the conductor and $r=0.3$ m.

Solution

(a) To make E inside the conductor zero, the inner surface must have charge equal and opposite to the cavity charge: $+3$ μC. The remaining charge is on the outer surface: $8$ μC $-3$ μC = $+5$ μC. (b) $r=0.05$ m (inside cavity): $E=\frac{k|q|}{r^2}=\frac{9e9\times 3e-6}{0.05^2}=1.08\times 10^7$ N/C directed inward. $r=0.15$ m (inside conductor bulk): $E=0$. $r=0.3$ m (outside conductor): $E=\frac{k{Q}_{outer}}{r^2}=\frac{9e9\times 5e-6}{0.3^2}=5\times 10^5$ N/C directed outward. (c) $V={\int }_{R_2}^{0.3}Edr=kQ(\frac{1}{R_2}-\frac{1}{0.3})=9e9\times 5e-6\times (\frac{1}{0.2}-\frac{1}{0.3})=7.5\times 10^4$ V = 75 kV.

Question 2

A parallel plate capacitor with area $A=0.1$ m² and separation $d=2$ mm is connected to a 24 V battery. A dielectric slab of $\kappa =4$, thickness 1 mm, is inserted to fill half the horizontal area between the plates. Calculate: (a) Initial capacitance before inserting the dielectric. (b) New capacitance after insertion. (c) Total energy stored in the capacitor after insertion.

Solution

(a) $C_0=\frac{{ϵ}_{0}A}{d}=\frac{8.85e-12\times 0.1}{0.002}=4.425\times 10^{-10}$ F = 442.5 pF. (b) The capacitor acts as two parallel capacitors: one with half the area, all air: $C_1=\frac{{ϵ}_0(A/2)}{d}=2.2125\times 10^{-10}$ F. The second half area is two capacitors in series: 1 mm air gap and 1 mm dielectric: $C_{air}=\frac{{ϵ}_0(A/2)}{0.001}=4.425\times 10^{-10}$ F, $C_{dielectric}=\frac{4{ϵ}_0(A/2)}{0.001}=1.77\times 10^{-9}$ F. Series combination $C_2=\frac{C_{air}{C}_{dielectric}}{C_{air}+C_{dielectric}}=3.54\times 10^{-10}$ F. Total $C=C_1+C_2=5.75\times 10^{-10}$ F = 575 pF. (c) $U=\frac{1}{2}C{V}^2=0.5\times 5.75e-10\times 24^2=1.66\times 10^{-7}$ J.

Question 3

For the following network: $C_1=1$ μF, $C_2=2$ μF, $C_3=3$ μF, $C_4=4$ μF. $C_1$ and $C_2$ are in parallel, this combination is in series with $C_3$, then the entire group is in parallel with $C_4$, connected to a 10 V battery. Calculate: (a) Total capacitance of the network. (b) Charge stored on $C_3$. (c) Voltage across $C_2$.

Solution

(a) Parallel combination of $C_1$ and $C_2$: $C_{12}=1+2=3$ μF. Series with $C_3$: $C_{123}=\frac{3\times 3}{3+3}=1.5$ μF. Total capacitance: $C_{total}=1.5+4=5.5$ μF. (b) Voltage across $C_{123}$ is 10 V, so charge on $C_3$ is $Q=C_{123}V=1.5e-6\times 10=15$ μC. (c) Voltage across $C_{12}$: $V=\frac{Q}{C_{12}}=\frac{15\mu C}{3\mu F}=5$ V, which is the voltage across $C_2$.

9. Quick Reference Cheatsheet

10. What's Next

This topic is the foundation for two major subsequent units in the AP Physics C E&M syllabus: RC circuits and AC circuits. You will use capacitance rules to analyze time-dependent charge and voltage behavior in RC circuits, which make up another 17-23% of the exam, and to understand capacitive reactance in AC circuits with oscillating voltages and currents. Electrostatic conductor rules also underpin advanced Gauss's law applications and problems involving Faraday's law of electromagnetic induction, where changing magnetic fields induce currents in conductors.

To reinforce your mastery, practice with official College Board AP Physics C E&M past papers, paying close attention to free-response rubrics to ensure you include all required steps for full marks. If you struggle with any concept, practice problem, or past paper question, you can ask Ollie, our AI tutor, for personalized explanations and step-by-step walkthroughs at any time.