Electric Force, Field, and Potential — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Coulomb's law, electric field of point charges, superposition, electric potential and potential difference, equipotential surfaces, work done by/on charges in fields — AP Physics 2 Unit 3.

You should already know: Vector addition, Newton's laws (AP Phys 1), conservation of energy.

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 papers.

1. Why this Unit Matters

Unit 3 is the foundation of AP Physics 2 — every later unit (Circuits, Magnetism, Modern Physics) builds on the language of charge, field, and potential. About 13–17% of the AP Physics 2 score comes from Unit 3 directly, and Unit 3 concepts permeate Units 4 (Circuits), 5 (Magnetism), and 6 (Optics-EM).

Three big concepts:

1. Coulomb's law — the force between two charges.
2. Electric field — force per unit charge at every point in space, due to other charges.
3. Electric potential — potential energy per unit charge, measuring "voltage hills".

2. Coulomb's law

The electrostatic force between two point charges $q_1$ and $q_2$ separated by distance $r$: $$F=\frac{k|q_1{q}_2|}{r^2},k=8.99\times 10^9{\text{ Ncdotpm}}^2/{\text{C}}^2$$

Direction: along the line joining the charges. Repulsive if same sign, attractive if opposite.

For multiple charges, superposition applies — net force on a charge is the vector sum of forces from all other charges (not the magnitudes added).

3. Electric field

The electric field at a point is the force per unit positive test charge: $$\vec{E}=\vec{F}/q_{\text{test}},\text{units N/C or V/m}$$

For a point source charge $Q$ at distance $r$: $E=kQ/r^2$. Direction: away from positive source, toward negative.

Field lines: visual representation. Start on +, end on −. Density of lines = field strength. Lines never cross. For uniform field (e.g. between parallel plates), lines are parallel and equally spaced.

Force on a charge in a field: $\vec{F}=q\vec{E}$. Positive charges accelerate along field; negative charges accelerate against field.

4. Continuous charge distributions (qualitative)

For an extended distribution (rod, ring, sheet), break into infinitesimal pieces, find $dE$ from each, integrate. AP Physics 2 stays mostly qualitative on the calculus — you should be able to:

• Identify the symmetry (a ring of charge has zero $E$ at its centre by symmetry, but non-zero on its axis).
• Recognise that infinite parallel plate fields are uniform between the plates.
• Estimate field magnitude from Gauss's law–like arguments without integration.

5. Electric potential and potential difference

Electric potential $V$ at a point is the work needed to bring a unit positive charge from infinity to that point, divided by the charge: $$V=\frac{kQ}{r}\text{ (point charge)},\text{units volts (V) = J/C}$$

It's a scalar — adds algebraically (with signs), unlike $\vec{E}$ which adds as a vector.

Potential difference $\Delta V=V_B-V_A$. Work done by the electric force on charge $q$ moving from A to B: $$W=-q\Delta V=-q(V_B-V_A)$$

A positive charge naturally moves from high $V$ to low $V$; negative charge moves from low to high.

Energy conservation in a field: $$\Delta KE=-\Delta U=-q\Delta V$$

6. Equipotential surfaces

An equipotential is a surface where $V$ is constant. Key properties:

• No work is done moving a charge along an equipotential (since $\Delta V=0$).
• Equipotentials are perpendicular to electric field lines.
• For a point charge, equipotentials are concentric spheres.
• For a uniform field (parallel plates), equipotentials are flat planes parallel to the plates.

Relation to field: $E=-dV/dx$ along the field direction. Higher density of equipotentials = stronger field (steeper potential drop per unit distance).

7. Worked Example

Two point charges, $q_1=+3.0$ μC and $q_2=-2.0$ μC, are placed 0.40 m apart on the x-axis at $x=0$ and $x=0.40$ m respectively. (a) Calculate the magnitude and direction of the electric field at the midpoint between them. (b) Calculate the electric potential at the midpoint. (c) A test charge $q=+1.0$ nC is placed at the midpoint. Calculate the force on it and its electric potential energy.

Solution.

(a) Midpoint at $x=0.20$ m, distance from each charge = 0.20 m.

• Field from $q_1$ at midpoint: $E_1=(8.99\times 10^9)(3.0\times 10^{-6})/(0.20{)}^2=6.74\times 10^5$ N/C, pointing right (away from positive).
• Field from $q_2$ at midpoint: $E_2=(8.99\times 10^9)(2.0\times 10^{-6})/(0.20{)}^2=4.50\times 10^5$ N/C, pointing right (toward negative).
• Both rightward, so $E_{\text{net}}=6.74+4.50=11.24\times 10^5$ N/C, rightward.

(b) $V$ at midpoint: $V=k{q}_1/r_1+k{q}_2/r_2=(8.99\times 10^9)[(3.0\times 10^{-6})/0.20+(-2.0\times 10^{-6})/0.20]$ = $(8.99\times 10^9)(5.0\times 10^{-6})/0.20=2.25\times 10^5$ V (algebraic — note the sign of $q_2$).

(c) $F=qE=(1.0\times 10^{-9})(11.24\times 10^5)=1.12\times 10^{-3}$ N rightward. $U=qV=(1.0\times 10^{-9})(2.25\times 10^5)=2.25\times 10^{-4}$ J.

8. Common Pitfalls

• Field is vector, potential is scalar: when summing, $E$ requires vector addition (components or law of cosines), but $V$ adds with signs algebraically.
• Sign of $W=-q\Delta V$: work done by the electric force. If positive charge moves from high to low $V$, $\Delta V<0$, so $W>0$ (force does positive work).
• $E=0$ doesn't mean $V=0$: between two equal +charges, the midpoint has $E=0$ but $V>0$ (sum of two positives).
• Test charge sign matters for force direction, not field direction: the field points one way regardless of test charge sign; the force on a negative test charge is opposite to the field.

9. Practice Questions (CED Style)

1. Three point charges of +2 μC each are placed at the corners of an equilateral triangle of side 1.0 m. Calculate the magnitude of the electric field at the centroid (using symmetry).
2. A 9.0 V battery is connected to two parallel plates separated by 3.0 mm. Calculate the electric field between the plates and the force on an electron in this field.
3. A +5.0 nC charge is moved 0.20 m through a region where the potential changes from 200 V to 50 V. Calculate the work done by the electric force.

10. Quick Reference Cheatsheet

• Coulomb's law: $F=k{q}_1{q}_2/r^2$, $k=8.99\times 10^9$.
• Field of point charge: $E=kQ/r^2$, away from + / toward −.
• Force on charge: $\vec{F}=q\vec{E}$.
• Potential of point charge: $V=kQ/r$ (with sign).
• Potential difference: $\Delta V=V_B-V_A$. $W=-q\Delta V$.
• Field & potential: $E=-dV/dx$.
• Field lines: + to −, never cross, density = magnitude.
• Equipotentials: ⊥ to field lines.

11. What's Next

Unit 3 leads directly into Unit 4 (Electric Circuits) — voltage drives current through resistance ($V=IR$), and the energy delivered by a battery comes from the same potential framework. Unit 5 (Magnetism) builds on field concepts to handle moving charges. Use Ollie to step through superposition problems with multiple charges or to verify the work-energy logic of charged particle motion through a potential difference.