Electric Force, Field, and Potential: Unit Diagnostic (Physics 2 Unit 10)

What this diagnostic is for

Use this check to find which Unit 10 topics you should revisit. It has eight multiple-choice and three short-answer questions. These are original Marlbridge practice questions, not past exam questions. They are not calibrated against exam results, and your total does not predict a score. What matters is which questions you miss.

How to take it:

Work without notes, in about 30 minutes, and give a reason for each answer.
A scientific calculator is allowed. Data: k = 9.0 × 10⁹ N·m²/C²; e = 1.60 × 10⁻¹⁹ C; mₚ = 1.67 × 10⁻²⁷ kg. Potential is zero very far from all charges. Treat small spheres as point charges.
Open each answer only after committing to your own.

Question 1 (multiple choice · 10.1)

Small sphere A carries +6.0 nC and small sphere B carries +2.0 nC. The electric force exerted on B by A is 0.012 N, directed east. What is the electric force exerted on A by B?

(A) 0.036 N west
(B) 0.0040 N west
(C) 0.012 N west
(D) 0.012 N east

Answer and explanation

Answer: (C). The forces form an interaction pair: one magnitude, k|q_A q_B|/r², acts on both. Like charges repel, so A is pushed west.

(A) scales the force by the charge ratio.
(B) makes the same mistake the other way.
(D) has the wrong direction.

If you missed this: read “Coulomb’s law” in the Topic 10.1 study guide.

Question 2 (multiple choice · 10.2)

A neutral electroscope has a metal plate joined by a metal rod to two thin leaves. A negative rod is brought near the plate without touching it, and the leaves spread apart. Which statement is correct?

(A) Electrons jump from the rod to the electroscope, so it is now negative overall.
(B) The electroscope is still neutral. Electrons are pushed down to the leaves, so both leaves are negative and repel.
(C) Protons move up toward the rod, leaving the leaves negative.
(D) The rod pulls electrons up to the plate, so both leaves are positive and repel.

Answer and explanation

Answer: (B). The rod repels free electrons down to the leaves, leaving the plate positive. No charge enters or leaves, so the net charge stays zero.

(A) needs contact or a spark. Nothing touched.
(C) In a metal only electrons move.
(D) gets the direction wrong. A negative rod repels electrons.

If you missed this: read “Induced charge separation” in the Topic 10.2 study guide.

Question 3 (multiple choice · 10.3)

What is the electric field 0.20 m from a small sphere carrying −5.0 nC?

(A) 225 N/C, pointing toward the sphere
(B) 1125 N/C, pointing away from the sphere
(C) 225 N/C, pointing away from the sphere
(D) 1125 N/C, pointing toward the sphere

Answer and explanation

Answer: (D). E = k|q|/r² = (9.0 × 10⁹)(5.0 × 10⁻⁹) ÷ (0.20)² = 1125 N/C. A positive test charge would be attracted, so the field points toward the sphere.

(A) divides by r, not r².
(B) uses the “away” rule for a positive charge.
(C) makes both mistakes.

If you missed this: read “Field of a point charge” in the Topic 10.3 study guide.

Question 4 (multiple choice · 10.3)

A neutral solid metal sphere is placed in a uniform electric field that points to the right. Once no charges are moving, which statement is correct?

(A) The field inside the metal is zero. The left surface is negative and the right surface is positive.
(B) The field inside the metal is the same as the outside field, because the sphere is neutral.
(C) The field inside the metal is zero. The left surface is positive and the right surface is negative.
(D) Induced charge spreads through the volume, so the field inside is smaller but not zero.

Answer and explanation

Answer: (A). Free electrons move against the field, to the left, leaving the right surface positive. They stop when their field cancels the outside field inside the metal.

(B) A neutral conductor can still separate charge.
(C) reverses the sides. Electrons move opposite to the field.
(D) describes an insulator, not a conductor.

If you missed this: read “Charged conductors and insulators” in the Topic 10.3 study guide.

Question 5 (multiple choice · 10.4)

A +2.0 μC charge is fixed at x = 0 and a −1.0 μC charge is fixed at x = 0.20 m. An external force slowly brings a +1.0 μC charge from far away to x = 0.10 m. How much work does it do?

(A) 0.27 J
(B) 0.090 J
(C) 0.045 J
(D) 0 J

Answer and explanation

Answer: (B). The work equals the change in U_E, which comes from the two new pairs: W = k(1.0 × 10⁻⁶)[(2.0 × 10⁻⁶) + (−1.0 × 10⁻⁶)] ÷ 0.10 = 0.090 J.

(A) drops the sign of the −1.0 μC charge.
(C) uses 0.20 m. The new charge is 0.10 m from each.
(D) is the total final U_E (−0.090 J + 0.090 J), not the change.

If you missed this: read “Work done to rearrange charges” in the Topic 10.4 study guide.

Question 6 (multiple choice · 10.5)

A charge +Q is fixed at x = 0 and a charge +9Q is fixed at x = 4d. Point P is on the x-axis at x = d. Which statement about the net field E and the potential V at P is correct?

(A) E = 0 and V = 0
(B) E = 2kQ/d² and V = 4kQ/d
(C) E = 0 and V = 4kQ/d
(D) E = 0 and V = 10kQ/d

Answer and explanation

Answer: (C). The +Q charge gives kQ/d² at P, pointing in +x. The +9Q charge is 3d away, so it gives 9kQ/(3d)² = kQ/d², pointing in −x. The fields cancel. Potential is a scalar: V = kQ/d + 9kQ/(3d) = 4kQ/d.

(A) The fields cancel, but two positive potentials cannot.
(B) adds the field sizes without directions.
(D) uses d instead of 3d for the +9Q charge.

If you missed this: read “Potential of point charges” in the Topic 10.5 study guide.

Question 7 (multiple choice · 10.6)

A parallel-plate capacitor on a battery is swapped, on the same battery, for one with twice the plate area and the same separation. Compared with the first, which is correct?

(A) Q doubles, E doubles, U_C quadruples.
(B) Q is unchanged, E halves, U_C halves.
(C) Q doubles, E is unchanged, U_C quadruples.
(D) Q doubles, E between the plates is unchanged, U_C doubles.

Answer and explanation

Answer: (D). C = κε₀A/d doubles. The battery fixes ΔV, so Q = CΔV doubles. E = ΔV/d is unchanged. U_C = ½C(ΔV)² doubles.

(A) forgets that the extra charge is spread over twice the area.
(B) holds Q fixed, as for an isolated capacitor.
(C) uses U_C ∝ Q² but forgets that C also doubles: Q²/(2C) doubles.

If you missed this: read “What sets the capacitance” in the Topic 10.6 study guide.

Question 8 (multiple choice · 10.7)

An electron is released from rest at point P, where V = 0. Point X, on one side, is at +50 V and point Y, on the other side, is at −50 V. Only electric forces act. Which statement is correct?

(A) It moves toward X. U_E decreases and K increases.
(B) It moves toward Y. U_E decreases and K increases.
(C) It moves toward X. U_E increases and K increases.
(D) It stays at P, because the potential there is zero.

Answer and explanation

Answer: (A). The field points toward lower potential (Y), and the force on a negative charge is opposite to it. Moving to X, ΔU_E = qΔV = (−e)(+50 V) < 0, and the lost U_E becomes K.

(B) is true for a positive charge. Negative charges move toward higher potential.
(C) breaks energy conservation: both cannot rise.
(D) confuses zero potential with zero field.

If you missed this: read “Which way does a released charge move?” in the Topic 10.7 study guide.

Question 9 (short answer · 10.2)

Identical metal spheres P and Q sit on insulating stands. P has −4.8 nC and Q is neutral. P touches Q and they are separated. Q is then touched briefly by a grounded wire. Finally P touches Q again and they are separated.

(a) Find the final charge on each sphere. (b) How many electrons went to the ground? (c) The total charge on P and Q has changed. Explain why this does not break conservation of charge.

Worked answer

(a) First contact: identical spheres share equally, −2.4 nC each. Grounding takes Q to 0. Second contact: (−2.4 nC + 0) ÷ 2 = −1.2 nC each.
(b) Q lost 2.4 × 10⁻⁹ C: 2.4 × 10⁻⁹ ÷ 1.60 × 10⁻¹⁹ = 1.5 × 10¹⁰ electrons, from Q to the ground.
(c) During grounding, charge passed from Q into the Earth. The total for P, Q and the Earth is unchanged.

If you missed this: read “Grounding” and “Charging by contact” in the Topic 10.2 study guide.

Question 10 (short answer · 10.5)

A +5.0 nC charge is fixed at x = 0 and a −3.0 nC charge is fixed at x = 0.50 m.

(a) Calculate the potential at x = 0.10 m and at x = 0.30 m. (b) Calculate the work an external force does to move a +2.0 nC charge slowly from x = 0.10 m to x = 0.30 m. (c) Estimate the average field between these points, with its direction.

Worked answer

(a) Add signed scalars. At x = 0.10 m: 450 V + (−67.5 V) = 382.5 V. At x = 0.30 m: 150 V + (−135 V) = 15 V.
(b) W = qΔV = (2.0 × 10⁻⁹ C)(15 − 382.5) V = −7.35 × 10⁻⁷ J. Negative: the field pushes it that way.
(c) E = −ΔV/Δx = 367.5 V ÷ 0.20 m = 1.84 × 10³ V/m, in the +x direction, toward lower potential.

If you missed this: read “Potential and field” in the Topic 10.5 study guide.

Question 11 (short answer · 10.7)

A proton moves along the x-axis under electric forces only. The table gives the potential at five points (invented data).

x (m)	0	0.10	0.20	0.30	0.40
V (V)	0	−40	−100	−60	+20

The proton is released from rest at x = 0.

(a) Which way does it start to move? Explain. (b) Calculate its kinetic energy and speed at x = 0.20 m. (c) Does it reach x = 0.40 m? Describe its later motion.

Worked answer

(a) +x. V falls in that direction, so the field points that way, and a positive charge is pushed along the field.
(b) ΔK = −qΔV = −(1.60 × 10⁻¹⁹ C)(−100 V) = 1.60 × 10⁻¹⁷ J, so v = √(2K/mₚ) = 1.38 × 10⁵ m/s.
(c) No. K = e(0 − V) is never negative, so the proton can reach only points where V ≤ 0. V passes 0 V between x = 0.30 m and 0.40 m. The proton stops there, turns back, and moves back and forth between x = 0 and that point.

If you missed this: read “Energy conservation for a moving charge” in the Topic 10.7 study guide.

Your next step

Count a short answer as wrong if any part is missing, and any guess as wrong.

Topic	Question(s)	If you missed it, read
10.1 Electric charge and electric force	1	10.1 study guide
10.2 Conservation of charge and charging	2, 9	10.2 study guide
10.3 Electric fields	3, 4	10.3 study guide
10.4 Electric potential energy	5	10.4 study guide
10.5 Electric potential	6, 10	10.5 study guide
10.6 Capacitors	7	10.6 study guide
10.7 Conservation of electric energy	8, 11	10.7 study guide

How to use your result

Start with the topic where you missed the most. Read its study guide, then do its practice questions.
Look at why you missed it. A sign slip or r instead of r² needs a habit, not a full re-read.
Missed nothing? Go to the Unit 10 mixed review, where questions combine topics.
Retake this check after a week of revision.