Practice a College Board-style free response question on Electric Force, Field, & Potential. Write your response, then reveal the model answer to see exactly what earns each point.
Free Response Question · Unit 2 · Electric Fields & Potential
A point charge of +Q = 4.0 × 10⁻⁶ C is fixed in place. Point P is located a distance of 0.30 m from the charge, and point S is located a distance of 0.60 m from the charge, along the same radial line.
Quantity
Value
Source charge, Q
+4.0 × 10⁻⁶ C
Distance to point P
0.30 m
Distance to point S
0.60 m
Coulomb's constant, k
8.99 × 10⁹ N·m²/C²
A
Calculate the magnitude of the electric field at point P, and state its direction relative to the source charge.
✓ Model answer (earns the point)
Using E = kQ/r²: E = (8.99 × 10⁹)(4.0 × 10⁻⁶) / (0.30)² ≈ 4.0 × 10⁵ N/C. Since the source charge is positive, the electric field at P points directly away from the charge (radially outward), in the direction a positive test charge placed at P would be pushed.
Why it scores: Uses the correct formula for the field of a point charge, substitutes values with correct units, arrives at the correct numerical magnitude, AND correctly states the direction (away from a positive source) with reasoning.
B
Without recalculating from scratch, determine the magnitude of the electric field at point S, using the relationship between field and distance.
✓ Model answer (earns the point)
Point S is twice as far from the charge as point P (0.60 m vs. 0.30 m). Since electric field follows an inverse-square relationship with distance (E ∝ 1/r²), doubling the distance reduces the field to 1/4 of its value at P. So E_S = (4.0 × 10⁵ N/C) / 4 = 1.0 × 10⁵ N/C.
Why it scores: Recognizes the inverse-square scaling relationship explicitly (not just recalculating the formula from scratch — though that would also be accepted with correct numbers), correctly applies the factor of 4 reduction for doubled distance, AND arrives at the correct final value.
C
A small test charge of +2.0 × 10⁻⁹ C is moved from point S to point P (closer to the source charge). Determine whether positive or negative work is done by the electric force on the test charge during this move, and justify your answer without calculating a numerical value.
✓ Model answer (earns the point)
The electric force does negative work on the test charge. Both charges are positive, so the source charge repels the test charge — the electric force naturally pushes the test charge away from the source, toward S. Moving the test charge from S to P instead brings it closer to the source charge, against the direction the electric force would naturally push it, so the electric force does negative work on it (an external agent would need to do positive work to cause this motion).
Why it scores: Identifies that both charges are positive (so the force is repulsive), recognizes that moving the test charge closer to P means moving it against the natural direction of the repulsive force, AND concludes that the electric force does negative work on the charge. Full credit also accepts justification using V = kQ/r: potential is higher at P than at S (since P is closer), and moving a positive charge to higher potential means positive work was done ON the charge by an external force, so the electric force itself did negative work.
How to score points on AP Physics 2 FRQs
Show your equation before you substitute numbers. Graders award points for correctly identifying the relevant relationship (E = kQ/r², W = qΔV, etc.), not just for a final number.
State direction explicitly with reasoning, not just a number. "Away from the charge, since the source is positive" earns more than a bare magnitude.
Use proportional reasoning when asked to avoid recalculating. Recognizing that doubling r quarters E (inverse-square) is often exactly what's being tested.
Be extremely careful with sign conventions on work and potential energy. Moving a positive charge toward another positive charge requires positive external work — meaning the electric force itself does negative work. Write out the logic, don't just guess the sign.
Watch your units throughout. N/C for field, V for potential, J for energy — keep them straight and convert before calculating.