College Physics Quiz: Redistribution Of Charge Between Conductors
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Redistribution Of Charge Between ConductorsQuestion 1 of 20

A conducting sphere A with charge +8 μC is connected by a conducting wire to an initially uncharged conducting sphere B. The spheres have radii R_A = 2 cm and R_B = 6 cm respectively. After equilibrium is reached, what is the charge on sphere A?

Sphere A has a final charge of +2 μC
Sphere A has a final charge of +4 μC
Sphere A has a final charge of +6 μC
Sphere A has a final charge of +1 μC
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College Physics Quiz

College Physics Quiz: Redistribution Of Charge Between Conductors

Practice Redistribution Of Charge Between Conductors in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Redistribution Of Charge Between Conductors, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A conducting sphere A with charge +8 μC is connected by a conducting wire to an initially uncharged conducting sphere B. The spheres have radii R_A = 2 cm and R_B = 6 cm respectively. After equilibrium is reached, what is the charge on sphere A?

  1. Sphere A has a final charge of +2 μC (correct answer)
  2. Sphere A has a final charge of +4 μC
  3. Sphere A has a final charge of +6 μC
  4. Sphere A has a final charge of +1 μC
Explanation: When connected by a conducting wire, the spheres reach the same electric potential: V_A = V_B, which means kQ_A/R_A = kQ_B/R_B, so Q_A/R_A = Q_B/R_B. This gives Q_B = Q_A × (R_B/R_A) = Q_A × (6/2) = 3Q_A. Since charge is conserved: Q_A + Q_B = 8 μC, and Q_B = 3Q_A, we have Q_A + 3Q_A = 8 μC, so 4Q_A = 8 μC, therefore Q_A = 2 μC. Choice B incorrectly assumes equal charge distribution. Choice C assumes charge distributes by radius ratio directly. Choice D uses an incorrect ratio calculation.

Question 2

A +3nC+3\,\text{nC} rod touches an identical neutral rod, then separates. What is the resulting charge on each rod?

  1. Each: +1.5nC+1.5\,\text{nC} (correct answer)
  2. Charged rod: +3nC+3\,\text{nC}, neutral rod: 0nC0\,\text{nC}
  3. Each: +3nC+3\,\text{nC}
  4. Each: 0nC0\,\text{nC}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. Contact shares with neutral, conserve. Identical equal. Choice A is correct: total +3 nC, each +1.5. Choice B is incorrect, no share. To help students: Include neutrals. Reinforce halves.

Question 3

A positive rod at +12nC+12\,\text{nC} touches a negative rod at 4nC-4\,\text{nC}, then separates. If rods are identical, what charge is on each?

  1. Each: +4nC+4\,\text{nC} (correct answer)
  2. Each: +8nC+8\,\text{nC}
  3. Each: 4nC-4\,\text{nC}
  4. Each: 0nC0\,\text{nC}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. Identical rods share equally, conserve total. Mixed signs average. Choice A is correct: total +8 nC, each +4. Choice B is incorrect, doubling. To help students: Handle signs carefully. Practice opposites.

Question 4

Identical spheres: A has 12nC-12\,\text{nC}, B has +4nC+4\,\text{nC}. After touching then separating, what is each charge?

  1. A: 4nC-4\,\text{nC}, B: 4nC-4\,\text{nC} (correct answer)
  2. A: 12nC-12\,\text{nC}, B: +4nC+4\,\text{nC}
  3. A: +4nC+4\,\text{nC}, B: 12nC-12\,\text{nC}
  4. A: 0nC0\,\text{nC}, B: 8nC-8\,\text{nC}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. Contact equalizes charge for identical spheres. Conservation applies. Choice A is correct: total -8 nC, each -4 nC. Choice B is incorrect, not conserving total. To help students: Focus on signs. Use conservation equations.

Question 5

Capacitors in parallel: C1=1μFC_1=1\,\mu\text{F} with Q1=+2μCQ_1=+2\,\mu\text{C}, C2=4μFC_2=4\,\mu\text{F} with Q2=+6μCQ_2=+6\,\mu\text{C}. After connection, what are charges?

  1. Q1=+1.6μCQ_1=+1.6\,\mu\text{C}, Q2=+6.4μCQ_2=+6.4\,\mu\text{C} (correct answer)
  2. Q1=+4.0μCQ_1=+4.0\,\mu\text{C}, Q2=+4.0μCQ_2=+4.0\,\mu\text{C}
  3. Q1=+2μCQ_1=+2\,\mu\text{C}, Q2=+6μCQ_2=+6\,\mu\text{C}
  4. Q1=+6.4μCQ_1=+6.4\,\mu\text{C}, Q2=+1.6μCQ_2=+1.6\,\mu\text{C}
Explanation: This question tests understanding of charge redistribution in capacitors in introductory college physics. Connection conserves total Q, equal V. Q proportional to C. Choice A is correct: total 8 μC, V=1.6V, Q1=1.6, Q2=6.4. Choice B is incorrect, equalizing. To help students: Emphasize proportions. Practice numbers.

Question 6

Identical spheres have 0C0\,\text{C} and +14μC+14\,\mu\text{C}; touch then separate. Final charge on each?

  1. Each sphere ends with +7.0μC+7.0\,\mu\text{C} (correct answer)
  2. Each sphere ends with +14μC+14\,\mu\text{C}
  3. Neutral stays 0C0\,\text{C}; charged stays +14μC+14\,\mu\text{C}
  4. Each sphere ends with +3.5μC+3.5\,\mu\text{C}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, charge from charged to neutral is shared equally. Choice A is correct because total +14 μC splits to +7.0 μC each for identicals. Choice C is incorrect as transfer happens. To help students: Treat neutral as zero charge in average. Reinforce with calculations.

Question 7

Two different-size conducting spheres connect by wire; which statement best describes the final charge distribution?

  1. Charge redistributes until both spheres reach the same potential (correct answer)
  2. Both spheres end with equal charge because they are connected
  3. Only the smaller sphere can hold charge after contact
  4. Total charge changes because the wire supplies electrons
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, equilibrium is at equal potential, not equal charge. Choice A is correct because potential equalizes, charge by capacitance. Choice B is incorrect for different sizes. To help students: Recall V same implies Q proportional to C. Use sphere radius for capacitance intuition.

Question 8

Identical spheres start at 12μC-12\,\mu\text{C} and +4μC+4\,\mu\text{C}; touch then separate. Final charges?

  1. Each sphere ends with 8μC-8\,\mu\text{C}
  2. Each sphere ends with 4μC-4\,\mu\text{C} (correct answer)
  3. Each sphere ends with +4μC+4\,\mu\text{C}
  4. Each sphere ends with 16μC-16\,\mu\text{C}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, net negative charge is shared equally. Choice B is correct because total -8 μC gives -4 μC each. Choice A is incorrect as it ignores averaging. To help students: Sum with signs carefully. Practice mixed charge scenarios.

Question 9

A +7.0μC+7.0\,\mu\text{C} rod touches a 7.0μC-7.0\,\mu\text{C} rod; after separation, total charge becomes?

  1. +14μC+14\,\mu\text{C}
  2. 14μC-14\,\mu\text{C}
  3. 0C0\,\text{C} (correct answer)
  4. Cannot be determined because charge is not conserved
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, equal opposite charges neutralize overall. Choice C is correct because +7.0 μC + (-7.0 μC) = 0 C total. Choice A is incorrect as it ignores the negative. To help students: Emphasize algebraic summing. Discuss neutralization in equal magnitudes.

Question 10

Identical spheres have +9.0μC+9.0\,\mu\text{C} and 3.0μC-3.0\,\mu\text{C}; touch then separate. Final charge on each?

  1. Each sphere ends with +6.0μC+6.0\,\mu\text{C}
  2. Each sphere ends with +3.0μC+3.0\,\mu\text{C} (correct answer)
  3. Each sphere ends with 0C0\,\text{C}
  4. Each sphere ends with +12.0μC+12.0\,\mu\text{C}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, the spheres equalize by averaging the positive and negative charges. Choice B is correct because total +6.0 μC divides to +3.0 μC each. Choice C is incorrect as it suggests neutralization to zero, which isn't the case. To help students: Remember to add algebraic sums for total charge. Practice with mixed signs to master conservation.

Question 11

Two identical spheres have initial charges Q1Q_1 and Q2Q_2 and then touch. After separation, what is the final charge on each sphere?

  1. Each sphere has (Q1+Q2)/2(Q_1+Q_2)/2 due to equal sharing and charge conservation (correct answer)
  2. Each sphere has Q1Q2Q_1-Q_2 because opposite charges subtract on contact
  3. Sphere 1 keeps Q1Q_1 and sphere 2 keeps Q2Q_2 because charges are fixed
  4. Each sphere becomes neutral because conductors always discharge when touched
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. Identical spheres share equally. Conservation key. Choice A is correct: each (Q1+Q2)/2. Choice D is incorrect, not neutralizing. To help students: Generalize formulas. Apply to specifics.

Question 12

Two capacitors in parallel: C1=2μFC_1=2\,\mu\text{F} with Q1=+6μCQ_1=+6\,\mu\text{C}, C2=1μFC_2=1\,\mu\text{F} with Q2=0Q_2=0. After connection, what are Q1Q_1 and Q2Q_2?

  1. Q1=+4μCQ_1=+4\,\mu\text{C}, Q2=+2μCQ_2=+2\,\mu\text{C} (correct answer)
  2. Q1=+3μCQ_1=+3\,\mu\text{C}, Q2=+3μCQ_2=+3\,\mu\text{C}
  3. Q1=+6μCQ_1=+6\,\mu\text{C}, Q2=0μCQ_2=0\,\mu\text{C}
  4. Q1=+2μCQ_1=+2\,\mu\text{C}, Q2=+4μCQ_2=+4\,\mu\text{C}
Explanation: This question tests understanding of charge redistribution in capacitors, analogous to conductors in introductory college physics. Parallel connection equalizes voltage, conserving total charge. Charges redistribute per capacitance. Choice A is correct: total 6 μC, V=2V, Q1=4, Q2=2. Choice B is incorrect, equalizing wrongly. To help students: Use Q=CV. Vary capacitances.

Question 13

Two identical spheres have +1.2μC+1.2\,\mu\text{C} and +0.8μC+0.8\,\mu\text{C}. After contact and separation, what charges remain?

  1. A: +1.0μC+1.0\,\mu\text{C}, B: +1.0μC+1.0\,\mu\text{C} (correct answer)
  2. A: +2.0μC+2.0\,\mu\text{C}, B: 0C0\,\text{C}
  3. A: +1.2μC+1.2\,\mu\text{C}, B: +0.8μC+0.8\,\mu\text{C}
  4. A: +0.2μC+0.2\,\mu\text{C}, B: +1.8μC+1.8\,\mu\text{C}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. Identical spheres average charges on contact. Total charge conserved. Choice A is correct: total +2.0 μC, each +1.0 μC. Choice B is incorrect, assuming no change. To help students: Practice averaging. Discuss equilibrium.

Question 14

Identical spheres start +1.0μC+1.0\,\mu\text{C} and 5.0μC-5.0\,\mu\text{C}; touch then separate. Final charges?

  1. Each sphere ends with 2.0μC-2.0\,\mu\text{C} (correct answer)
  2. Each sphere ends with +2.0μC+2.0\,\mu\text{C}
  3. They keep +1.0μC+1.0\,\mu\text{C} and 5.0μC-5.0\,\mu\text{C}
  4. Each sphere ends with 4.0μC-4.0\,\mu\text{C}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, net negative charge is equally shared. Choice A is correct because total -4.0 μC gives -2.0 μC each. Choice C is incorrect as redistribution occurs. To help students: Calculate total first. Use electron flow visuals for negatives.

Question 15

Two capacitors connect in parallel; which statement best describes the final charges on them?

  1. Total charge is conserved; charge redistributes until voltages match (correct answer)
  2. Total charge increases because connecting wires create charge
  3. Each capacitor keeps its initial charge because plates are isolated
  4. Both capacitors end with equal charge regardless of capacitance
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When two capacitors connect in parallel, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, voltages equalize while total Q is preserved. Choice A is correct because it describes conservation and voltage matching. Choice C is incorrect as charge does transfer. To help students: Recall V = Q/C for equilibrium. Simulate with numbers to see redistribution.

Question 16

Two identical rods have +4.0μC+4.0\,\mu\text{C} and 2.0μC-2.0\,\mu\text{C}; touch then separate. Final charge on each?

  1. Each rod ends with +1.0μC+1.0\,\mu\text{C} (correct answer)
  2. Each rod ends with +2.0μC+2.0\,\mu\text{C}
  3. They keep +4.0μC+4.0\,\mu\text{C} and 2.0μC-2.0\,\mu\text{C}
  4. Each rod ends with 0C0\,\text{C}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, identical rods share net positive charge equally. Choice A is correct because total +2.0 μC splits to +1.0 μC each. Choice C is incorrect as contact causes sharing. To help students: Average the charges for identicals. Mix signs in practice problems.

Question 17

Two identical metal rods have +5nC+5\,\text{nC} and 9nC-9\,\text{nC}. After touching then separating, what is the resulting charge on each?

  1. Each: 2nC-2\,\text{nC} (correct answer)
  2. Each: +2nC+2\,\text{nC}
  3. One: +5nC+5\,\text{nC}, other: 9nC-9\,\text{nC}
  4. Each: 7nC-7\,\text{nC}
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. Identical rods average charges. Total conserved. Choice A is correct: total -4 nC, each -2. Choice C is incorrect, no change. To help students: Compute net first. Use negatives.

Question 18

Two conductors are touched, allowing charge flow, then separated. What happens to the total charge of the two-conductor system?

  1. Total charge is conserved; it only redistributes between conductors (correct answer)
  2. Total charge increases because contact generates additional free charge
  3. Total charge decreases because some charge is lost during separation
  4. Total charge becomes zero because charges always neutralize on contact
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. System isolated, total Q conserved. Redistribution only. Choice A is correct: conserved, redistributes. Choice B is incorrect, no generation. To help students: Emphasize isolation. Discuss principles.

Question 19

Two identical spheres touch and separate; which statement best describes the total charge of the pair?

  1. It stays the same; charge only transfers between spheres (correct answer)
  2. It increases because electrons are created during contact
  3. It decreases because some charge disappears as heat
  4. It becomes zero because charges always neutralize
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, total charge doesn't change post-separation. Choice A is correct because charge only transfers, not created or destroyed. Choice D is incorrect as neutralization isn't guaranteed. To help students: Stress isolation of system. Analyze before/after totals.

Question 20

A small sphere has +6nC+6\,\text{nC} and a larger neutral sphere is wired to it. After equilibrium, which statement best describes charges?

  1. Both end with +3nC+3\,\text{nC} because charge always splits equally
  2. Total stays +6nC+6\,\text{nC}, but larger sphere holds more charge than smaller (correct answer)
  3. Neutral sphere remains 0nC0\,\text{nC} because it starts uncharged
  4. Total charge increases above +6nC+6\,\text{nC} when the wire is attached
Explanation: This question tests understanding of charge redistribution between conductors in introductory college physics. When conductors of different sizes are connected, charge redistributes proportionally to their capacitances, conserving total charge. In this scenario, the larger sphere has greater capacity, so it holds more charge at equilibrium. Choice B is correct because it states the total charge remains +6 nC and the larger sphere gets more, reflecting proportional sharing. Choice A is incorrect as it assumes equal sharing regardless of size, a common misconception. To help students: Analogize to capacitors in parallel. Practice with size ratios to understand uneven distribution.