What is the first step in most AP Physics 2 electricity problems?

Identify whether the quantity is a vector or a scalar, and draw the right diagram. Electric force and field are vectors, so you must add contributions component by component and show directions; electric potential and potential energy are scalars, so you add signed values arithmetically. A field-line sketch or a clear circuit schematic with the current direction marked turns a confusing scenario into a set of equations you can solve, and AP graders award points for it.

How do you decide between the electric field and the electric potential?

Use the field (a vector, $E = kQ/r^2$) when you need a force or a direction, since $\vec{F} = q\vec{E}$. Use the potential (a scalar, $V = kQ/r$) when you need energy, since the work to move a charge is $W = q\Delta V$ and the kinetic energy gained is $\Delta KE = q\Delta V$. Many problems mix the two: find the field for the force, and the potential for the energy.

How do you analyze a series-parallel circuit?

Reduce it step by step. Combine parallel groups (reciprocals for resistors), then series groups (sums), until the network is a single equivalent resistance. Find the total current from the source with $I = V/R$, then work back outward: the current is the same through series elements, and the voltage is the same across parallel elements. Track the shared quantity, current in series, voltage in parallel.

When do you need Kirchhoff's rules instead of series and parallel?

Use Kirchhoff's loop and junction rules when a circuit cannot be reduced by series and parallel alone, for example circuits with more than one battery or bridge configurations. Assign a current to each branch, write junction equations (conservation of charge) and loop equations (conservation of energy), and solve the simultaneous equations. A negative current just means your assumed direction was reversed.

How is the magnetic force different from the electric force?

The electric force acts on any charge, $F = qE$, along the field. The magnetic force acts only on a moving charge, $F = qvB\sin\theta$, perpendicular to both the velocity and the field, so it does no work and only bends the path. A charge at rest feels no magnetic force, and a charge moving along the field feels none either. Use the right-hand rule for direction (reverse it for a negative charge).

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AP Physics 2 solving electricity and magnetism problems: a complete skills guide to Coulomb's law, electric fields and potential, circuit analysis, magnetic forces and electromagnetic induction

A deep-dive AP Physics 2 skills guide to the problem-solving core of Units 10, 11 and 12: Coulomb's law, electric fields and potential, circuit analysis with Ohm's law and Kirchhoff's rules, magnetic forces, and electromagnetic induction. Includes worked examples and the exam technique that earns full free-response marks.

Generated by Claude Opus 4.820 min read10.1-12.4Updated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Why this is the spine of AP Physics 2
Coulomb's law and the inverse-square law
Electric fields: a vector
Electric potential and energy: a scalar
Circuit analysis: Ohm's law, series and parallel
Kirchhoff's rules for harder circuits
Magnetic forces and induction
Check your knowledge

Why this is the spine of AP Physics 2

Units 10, 11 and 12 supply the problem-solving engine of AP Physics 2. Electrostatics gives the field and potential ideas; circuits apply them to flowing charge; magnetism extends the field concept to moving charges. The recurring moves are the same: decide whether a quantity is a vector or a scalar, draw the right diagram, apply a core law, and reason with conservation of charge and energy. This guide ties together the matching dot-point pages, each of which has its own practice questions: electric charge and Coulomb's law, electric fields, electric potential and voltage, resistance and Ohm's law, resistors in series and parallel, and magnetism and moving charges.

Coulomb's law and the inverse-square law

The electric force between two point charges is $F = \dfrac{k|q_1||q_2|}{r^2}$ , an inverse-square law: doubling the separation quarters the force, tripling it cuts the force to a ninth. Like charges repel, unlike attract, and the force acts along the line joining them. For more than two charges, the net force is the vector sum of the pairwise forces, computed component by component. The single most common error is using a linear distance dependence; always square the distance.

Electric fields: a vector

The electric field is the force per unit charge, $\vec{E} = \vec{F}/q$ , and a point charge produces $E = \dfrac{kQ}{r^2}$ , pointing outward from a positive charge and inward toward a negative one. The force on a charge placed in a field is $\vec{F} = q\vec{E}$ , in the field's direction for a positive charge and opposite for a negative one. Because the field is a vector, add contributions from several charges as vectors. Between parallel plates the field is uniform, $E = V/d$ , and inside a conductor in equilibrium the field is zero.

Electric potential and energy: a scalar

Potential is energy per unit charge, $V = \dfrac{kQ}{r}$ (note the $1/r$ , not $1/r^2$ ), and it is a scalar, so contributions add arithmetically with their signs. The work to move a charge through a potential difference is $W = q\Delta V$ , and if only the electric force acts, this equals the kinetic energy gained, $\Delta KE = q\Delta V$ . The electric force is conservative, so $KE + U$ is conserved: a released charge speeds up as it moves to lower potential energy. Use the field for forces and directions, and the potential for energy.

Circuit analysis: Ohm's law, series and parallel

Current is the rate of charge flow, $I = \Delta q/\Delta t$ , and Ohm's law $V = IR$ links it to voltage and resistance. To analyze a network, reduce it step by step:

Series: resistances add, $R = R_1 + R_2 + \dots$ ; the current is the same through each, and the voltage divides.
Parallel: reciprocals add, $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$ ; the voltage is the same across each, and the current divides.

Find the total resistance, then the source current with $I = V/R_{total}$ , then work back outward. Power is $P = IV = I^2 R = V^2/R$ ; choose $I^2 R$ at fixed current and $V^2/R$ at fixed voltage.

Kirchhoff's rules for harder circuits

When series and parallel are not enough (multiple batteries, bridges), use Kirchhoff's rules. The junction rule ( $\sum I_{in} = \sum I_{out}$ ) is conservation of charge; the loop rule ( $\sum \Delta V = 0$ ) is conservation of energy. Assign a current to each branch, write the junction and loop equations, and solve. Crossing a resistor along the current is a voltage drop ( $-IR$ ); crossing a battery from $-$ to $+$ is a rise ( $+\mathcal{E}$ ).

Magnetic forces and induction

The magnetic force acts only on moving charges: $F = qvB\sin\theta$ , perpendicular to both the velocity and the field, so it does no work and bends the path into a circle ( $qvB = mv^2/r$ ). A current-carrying wire feels $F = BIL\sin\theta$ . A changing magnetic flux induces an emf (Faraday's law, $\mathcal{E} = N\Delta\Phi/\Delta t$ ), and the induced current opposes the change (Lenz's law). Use the right-hand rule for directions, reversing it for negative charges.

Two point charges and the field at a point

Charges $+4.0$ microcoulombs at the origin and $-4.0$ microcoulombs at $x = 0.30$ m. Find the net electric field at $x = 0.15$ m (the midpoint). Take $k = 8.99 \times 10^9$ N m squared per C squared.

step 1 Field from the positive charge

At $0.15$ m: $E_1 = \dfrac{kQ}{r^2} = \dfrac{(8.99 \times 10^9)(4.0 \times 10^{-6})}{(0.15)^2} = 1.60 \times 10^6$ N/C, pointing in the $+x$ direction (away from the positive charge).

step 2 Field from the negative charge

At $0.15$ m: $E_2 = 1.60 \times 10^6$ N/C, pointing in the $+x$ direction (toward the negative charge, which lies in the $+x$ direction).

step 3 Add as vectors

Both fields point in $+x$ , so they add: $E_{net} = 1.60 \times 10^6 + 1.60 \times 10^6 = 3.2 \times 10^6$ N/C in the $+x$ direction.

step 4 Interpret

Between a positive and a negative charge, the two fields reinforce (both point from $+$ toward $-$ ), giving a strong field midway, unlike between two like charges where they cancel.

AP Physics 2 Units 10 to 12 share one problem-solving routine: classify the quantity (vector force and field versus scalar potential and energy), draw the right diagram, apply a core law, and reason with conservation of charge and energy. Electrostatics uses Coulomb's law ( $F = kq_1q_2/r^2$ , inverse-square), the field ( $E = kQ/r^2$ , a vector, with $\vec{F} = q\vec{E}$ ) and the potential ( $V = kQ/r$ , a scalar, with $\Delta KE = q\Delta V$ ). Circuits use Ohm's law ( $V = IR$ ), series sums and parallel reciprocals, power ( $P = IV$ ), and Kirchhoff's rules (junction = charge conservation, loop = energy conservation) for harder networks. Magnetism uses $F = qvB\sin\theta$ (perpendicular, no work, circular motion) and $F = BIL\sin\theta$ , with Faraday's and Lenz's laws for the emf and direction of induction.

Check your knowledge

A mix of recall, calculation and application questions covering Units 10 to 12. Attempt them under timed conditions, then check against the solutions.

State the inverse-square law for the force between two point charges, and what happens to the force if the distance triples. (2 marks)
Two charges of $+2.0$ microcoulombs and $+6.0$ microcoulombs are $0.20$ m apart. Calculate the force between them ( $k = 8.99 \times 10^9$ ). (2 marks)
State the direction of the electric field around an isolated negative charge. (1 mark)
A charge of $+3.0 \times 10^{-6}$ C is accelerated through $400$ V. Calculate the kinetic energy it gains. (2 marks)
Find the equivalent resistance of $6.0$ ohms and $3.0$ ohms in parallel. (2 marks)
A $12$ V battery drives a $4.0$ -ohm resistor. Calculate the power dissipated. (2 marks)
State the conservation law expressed by Kirchhoff's junction rule. (1 mark)
A proton moves at $2.0 \times 10^6$ m/s perpendicular to a $0.30$ T field (charge $1.6 \times 10^{-19}$ C). Calculate the magnetic force on it. (2 marks)
Explain why a magnetic force does no work on a moving charge. (2 marks)
State what must change for an emf to be induced in a coil, and which law gives the direction of the induced current. (2 marks)

Solutions

Q1: The force is $F = \dfrac{k|q_1||q_2|}{r^2}$ , an inverse-square law. Tripling the distance reduces the force to one ninth ( $3^2 = 9$ ).
Q2: $F = \dfrac{(8.99 \times 10^9)(2.0 \times 10^{-6})(6.0 \times 10^{-6})}{(0.20)^2} = \dfrac{0.108}{0.04} = 2.7$ N (repulsive, both positive).
Q3: Radially inward, toward the negative charge.
Q4: $\Delta KE = q\Delta V = (3.0 \times 10^{-6})(400) = 1.2 \times 10^{-3}$ J.
Q5: $R = \dfrac{(6.0)(3.0)}{6.0 + 3.0} = \dfrac{18}{9.0} = 2.0$ ohms.
Q6: $P = \dfrac{V^2}{R} = \dfrac{12^2}{4.0} = \dfrac{144}{4.0} = 36$ W.
Q7: Conservation of charge.
Q8: $F = qvB\sin 90^\circ = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.30)(1) = 9.6 \times 10^{-14}$ N.
Q9: The magnetic force is always perpendicular to the velocity, so $W = Fd\cos 90^\circ = 0$ ; it changes the direction of motion but not the speed or kinetic energy.
Q10: The magnetic flux through the coil must change (Faraday's law); the direction of the induced current is given by Lenz's law (it opposes the change in flux).

Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)