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AP Physics 2: Algebra-Based, AP exams, 15–18%

Unit 10: Electric Force, Field, and Potential

Unit 10 builds electrostatics from charge and Coulomb's law up through field and potential. The recurring difficulty is keeping the vector quantity (field) distinct from the scalar quantity (potential), and force distinct from energy. Field-line and equipotential diagrams are common representation tasks.

Unit 10: Electric Force, Field, and Potential at a glance

Position
2 of 7 units
Published weighting
15–18% in the official framework
Topics covered
6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Physics 2: Algebra-Based. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Electric charge and conservation
Charging by friction, conduction, and induction, and conservation of charge in every process.
Coulomb's law
The inverse-square dependence on separation, superposition of forces, and comparison with gravitation.
Electric fields
Field as force per unit charge, field-line diagrams, and the field of point charges and parallel plates.
Electric charge distributions
How charge arranges itself on conductors and insulators, and the effect of a nearby charged object.
Electric potential energy
Work done moving a charge in a field, and the sign that follows from the charges involved.
Electric potential
Potential as energy per unit charge, equipotential surfaces, and the relationship between field and potential.

Study tips

How to practice Electric Force and Field.

Section-specific guidance that follows from how this material is usually tested.

  • Label every quantity as vector or scalar before starting; field-versus-potential confusion causes most errors here.
  • Practice sketching field lines and equipotentials for the standard configurations until it is automatic.
  • Draw the explicit analogy to gravitation, which makes the inverse-square structure feel familiar rather than new.

Where this fits

Unit 10: Electric Force, Field, and Potential is 2 of 7 units in the AP Physics 2: Algebra-Based guide, and it carries a published weighting of 15–18% in the official framework. Use the full guide to balance it against everything else you have to review.

Sample Electric Force and Field questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

  1. Question 1Foundational

    Two small charged spheres attract each other with a force of 36 N. Without changing either charge, the distance between their centers is made three times larger. What is the new magnitude of the force between them?

    1. A4.0 N
    2. B12 N
    3. C108 N
    4. D324 N
    Show answer

    Answer: A. 4.0 N

    Coulomb's law depends on the inverse square of the separation, so tripling the distance divides the force by 9: 36 N / 9 = 4.0 N. The 12 N choice divides by 3, treating the force as inversely proportional to distance rather than to distance squared.

    Key insight: Coulomb's law and Newton's law of gravitation share the same inverse-square structure.

  2. Question 2Foundational

    A small object carrying a charge of 2.0 x 10^-6 C experiences an electric force of 0.60 N when placed at a point in space. What is the magnitude of the electric field at that point?

    1. A1.2 x 10^-6 N/C
    2. B3.3 x 10^-6 N/C
    3. C3.0 x 10^5 N/C
    4. D3.0 x 10^6 N/C
    Show answer

    Answer: C. 3.0 x 10^5 N/C

    Electric field is force per unit charge: E = F/q = 0.60 N / (2.0 x 10^-6 C) = 3.0 x 10^5 N/C. The 1.2 x 10^-6 N/C choice multiplies force by charge instead of dividing.

    Key insight: The field describes the space itself, so it is unchanged if the test charge placed there is swapped for a different one.

  3. Question 3Intermediate

    Two large parallel conducting plates are separated by 0.020 m, and a 120 V potential difference is maintained across them. What is the magnitude of the uniform electric field in the region between the plates?

    1. A2.4 V/m
    2. B6.0 x 10^2 V/m
    3. C6.0 x 10^3 V/m
    4. D6.0 x 10^4 V/m
    Show answer

    Answer: C. 6.0 x 10^3 V/m

    For a uniform field between parallel plates, E = V/d = 120 V / 0.020 m = 6.0 x 10^3 V/m. The 2.4 V/m choice multiplies the potential difference by the separation instead of dividing.

    Key insight: A volt per meter and a newton per coulomb are the same unit, which is why field can be read as a potential gradient.

  4. Question 4Intermediate

    How much work must an external agent do to move a charge of +3.0 x 10^-6 C from a point at 0 V to a point at +200 V, if the charge starts and ends at rest?

    1. A1.5 x 10^-8 J
    2. B6.0 x 10^-4 J
    3. C1.5 x 10^-2 J
    4. D6.0 x 10^-2 J
    Show answer

    Answer: B. 6.0 x 10^-4 J

    With no change in kinetic energy the external work equals the change in electric potential energy: W = q(change in V) = (3.0 x 10^-6 C)(200 V) = 6.0 x 10^-4 J. The 1.5 x 10^-8 J choice divides the charge by the potential difference instead of multiplying.

    Key insight: Potential difference is energy per unit charge, so multiplying by the charge converts volts into joules.

  5. Question 5Advanced

    Two identical positive point charges are held a fixed distance apart. At the midpoint of the line joining them, what are the electric field and the electric potential, taking the potential to be zero infinitely far away?

    1. ABoth the electric field and the electric potential are zero
    2. BThe electric field is zero and the electric potential is not zero
    3. CThe electric field is not zero and the electric potential is zero
    4. DBoth the electric field and the electric potential are nonzero
    Show answer

    Answer: B. The electric field is zero and the electric potential is not zero

    Field is a vector, so the two equal contributions at the midpoint point in opposite directions and cancel. Potential is a scalar, so the two positive contributions add to a positive value, and it is possible to have zero field where the potential is large.

    Key insight: Field is a vector sum and potential is a scalar sum, so one can vanish while the other does not.

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