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MCAT Capacitors Practice Questions: Capacitance, Dielectrics, Series and Parallel

By StudyStore · Published 2026-11-04 · 8 original questions · about 12 min

Eight free MCAT-style physics practice questions on parallel-plate capacitors, stored charge and energy, dielectrics, and capacitors in series and parallel, with worked explanations.

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Quick concept primer

A capacitor stores charge: Q = CV. For a parallel-plate capacitor, C = kappa epsilon_0 A / d, where kappa is the dielectric constant, A the plate area and d the separation. Capacitance rises with larger area, smaller separation or an inserted dielectric. Stored energy is U = (1/2) C V^2 = (1/2) Q V = Q^2 / (2C).

Parallel capacitors share the same voltage and capacitances add: C = C1 + C2. Series capacitors carry the same charge and their reciprocals add: 1/C = 1/C1 + 1/C2. This is the reverse of resistors. For dielectric problems, ask whether the battery is still connected (voltage constant) or disconnected (charge constant).

8 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

The separation between the plates of a parallel-plate capacitor is doubled while the plate area stays the same. What happens to the capacitance?

  1. It is halved
  2. It is doubled
  3. It is unchanged
  4. It is quadrupled
Show answer and explanation

Answer: A. It is halved

Capacitance is inversely proportional to plate separation (C = epsilon_0 A / d), so doubling d halves C. Capacitance increases only if the area grows, the plates move closer, or a dielectric is added.

Question 2

A 4 microfarad capacitor is connected across a 10 V battery. How much charge is stored on it?

  1. 2.5 microcoulombs
  2. 0.4 microcoulombs
  3. 14 microcoulombs
  4. 40 microcoulombs
Show answer and explanation

Answer: D. 40 microcoulombs

Q = CV = 4 microfarad x 10 V = 40 microcoulombs. Dividing instead of multiplying would give 0.4. Always multiply capacitance by voltage to find charge.

Question 3

How much energy is stored in a 4 microfarad capacitor charged to 10 V?

  1. 200 microjoules
  2. 400 microjoules
  3. 40 microjoules
  4. 20 microjoules
Show answer and explanation

Answer: A. 200 microjoules

U = (1/2) C V^2 = 0.5 x 4 microfarad x (10 V)^2 = 0.5 x 4 x 100 = 200 microjoules. Forgetting the factor of one half gives 400, and forgetting to square the voltage gives 20.

Question 4

Two capacitors of 6 microfarad and 3 microfarad are connected in series. What is the equivalent capacitance?

  1. 9 microfarad
  2. 4.5 microfarad
  3. 18 microfarad
  4. 2 microfarad
Show answer and explanation

Answer: D. 2 microfarad

For series capacitors, 1/C = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so C = 2 microfarad. Series capacitance is always smaller than the smallest individual capacitor. The value 9 is the parallel result.

Question 5

Two capacitors of 2 microfarad and 3 microfarad are connected in parallel across a battery. What is the equivalent capacitance?

  1. 5 microfarad
  2. 1.2 microfarad
  3. 6 microfarad
  4. 2.5 microfarad
Show answer and explanation

Answer: A. 5 microfarad

Parallel capacitances add, so C = 2 + 3 = 5 microfarad. The value 1.2 is the series result (6/5). Parallel plates effectively increase the total area, which raises capacitance.

Question 6

A capacitor is connected to a battery and stays connected while a dielectric is inserted between the plates. What happens to the charge on the plates?

  1. It decreases
  2. It stays the same
  3. It increases
  4. It becomes zero
Show answer and explanation

Answer: C. It increases

The battery holds voltage constant. The dielectric raises capacitance by the factor kappa, and with Q = CV and V fixed, the charge increases as the battery supplies more. Stored energy (1/2 C V^2) also increases.

Question 7

A charged capacitor is disconnected from its battery and a dielectric is then inserted. What happens to the voltage across the capacitor?

  1. It increases
  2. It stays the same
  3. It becomes zero
  4. It decreases
Show answer and explanation

Answer: D. It decreases

With the battery disconnected the charge Q cannot change. The dielectric raises C, so V = Q/C falls. The stored energy Q^2/(2C) also falls, because the field pulls the dielectric in and does work.

Question 8

A 6 microfarad capacitor and a 3 microfarad capacitor are connected in series across a 9 V battery. What is the voltage across the 3 microfarad capacitor?

  1. 3 V
  2. 6 V
  3. 9 V
  4. 4.5 V
Show answer and explanation

Answer: B. 6 V

The equivalent capacitance is 2 microfarad, so the total charge is Q = 2 x 9 = 18 microcoulombs, and the same charge sits on each capacitor in series. Then V = Q/C = 18 / 3 = 6 V across the 3 microfarad capacitor, and 3 V across the 6 microfarad one.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.

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