A Level Physics 19.2 Energy Stored in a Capacitor MCQs – 30 Practice Questions with Answers (Cambridge 9702)
These 30 original practice questions cover Cambridge AS & A Level Physics (9702) Section 19.2, Energy Stored in a Capacitor, an A Level (A2) topic. They build from the three energy formulae, through full combination and discharge calculations, to the classic connected-versus-disconnected capacitor changes.
Formula recap (see also the A Level Physics Formula Sheet):
- Energy stored: E = ½QV = ½CV2 = Q2 / (2C)
- The factor ½ appears because the p.d. rises from 0 to V during charging, so the average p.d. is V/2; the stored energy is the triangular area under the V–Q graph.
- Capacitance: C = Q / V. If a charged capacitor is disconnected, Q stays constant; if it stays connected to a battery, V stays constant.
Energy Stored: The Formulae
Q1. Which expression correctly gives the energy E stored in a capacitor of capacitance C charged to a potential difference V, storing charge Q?
- (A) E = CV2
- (B) E = ½QV
- (C) E = QV
- (D) E = ½CV
Show Answer
Answer: B – The stored energy is E = ½QV, which is also equal to ½CV2 and to Q2/(2C). The factor ½ is essential: E = QV would be the energy the battery supplies in total, not the energy stored.
Q2. The energy stored in a capacitor is ½QV rather than QV. The main reason for the factor ½ is that
- (A) half of the charge leaks away during charging
- (B) half of the stored energy is held in the dielectric
- (C) the p.d. across the capacitor rises gradually from 0 to V as it charges, so the average p.d. during charging is V/2
- (D) the battery only ever supplies half of its energy to the capacitor
Show Answer
Answer: C – Each small packet of charge is pushed onto the plates against the p.d. already there. The p.d. grows linearly from 0 to V, so the work done per unit charge averages V/2, giving total stored energy Q × (V/2) = ½QV. This is the triangular area under the V–Q graph.
Q3. A graph of potential difference V against charge Q for a charging capacitor is a straight line through the origin. The energy stored when the charge is Q0 and the p.d. is V0 is given by
- (A) the area of the triangle under the line up to Q0, equal to ½Q0V0
- (B) the gradient of the line
- (C) the area of the rectangle Q0 × V0
- (D) the value V0 / Q0
Show Answer
Answer: A – Energy is the area under the V–Q graph. Because the graph is a straight line through the origin, that area is a triangle: ½ × base × height = ½Q0V0. (The gradient is 1/C, and V0/Q0 is also 1/C.)
Q4. A 4.0 µF capacitor is charged to a potential difference of 12 V. The energy stored is
- (A) 576 µJ
- (B) 288 µJ
- (C) 144 µJ
- (D) 48 µJ
Show Answer
Answer: B – E = ½CV2 = ½ × 4.0 × 10−6 × 122 = ½ × 4.0 × 10−6 × 144 = 2.88 × 10−4 J = 288 µJ. (Option A is CV2, forgetting the ½; option D is the charge, 48 µC, not an energy.)
Q5. A capacitor stores a charge of 200 µC when the potential difference across it is 50 V. The energy stored is
- (A) 10 mJ
- (B) 2.5 mJ
- (C) 5.0 mJ
- (D) 20 mJ
Show Answer
Answer: C – E = ½QV = ½ × 200 × 10−6 × 50 = ½ × 1.0 × 10−2 = 5.0 × 10−3 J = 5.0 mJ. (Option A, 10 mJ, is QV with the ½ missing.)
Q6. A 20 µF capacitor holds a charge of 60 µC. Using E = Q2/(2C), the energy stored is
- (A) 180 µJ
- (B) 45 µJ
- (C) 30 µJ
- (D) 90 µJ
Show Answer
Answer: D – E = Q2/(2C) = (60 × 10−6)2 / (2 × 20 × 10−6) = 3.6 × 10−9 / 4.0 × 10−5 = 9.0 × 10−5 J = 90 µJ. Check: V = Q/C = 60/20 = 3.0 V, so E = ½QV = ½ × 60 × 3.0 = 90 µJ.
Q7. A capacitor is charged from a supply and the potential difference across it is then doubled, the capacitance staying the same. The energy stored
- (A) becomes four times as large
- (B) becomes twice as large
- (C) becomes eight times as large
- (D) is unchanged
Show Answer
Answer: A – E = ½CV2, so E is proportional to V2. Doubling V multiplies the stored energy by 22 = 4. The charge also doubles (Q = CV), which is consistent with E = ½QV growing by 2 × 2.
Q8. The charge stored on a capacitor of fixed capacitance is doubled (so the p.d. doubles as well). The energy stored
- (A) doubles
- (B) increases by a factor of the square root of 2
- (C) halves
- (D) quadruples
Show Answer
Answer: D – At fixed C, E = Q2/(2C), so E is proportional to Q2. Doubling Q multiplies E by 4. Equivalently, both Q and V in ½QV double, giving 2 × 2 = 4 times the energy.
Q9. The energy stored in a charged capacitor is held
- (A) as chemical energy in the plates
- (B) in the electric field between the plates
- (C) in the connecting wires
- (D) only in the dielectric, never in an air gap
Show Answer
Answer: B – Work done in separating the charges sets up an electric field in the space between the plates, and the energy is stored in that field. On discharge, the field collapses and the energy is released, whether the gap contains a dielectric or not.
Q10. A 10 µF capacitor is charged to 200 V. The energy stored is
- (A) 0.40 J
- (B) 2.0 J
- (C) 0.20 J
- (D) 0.020 J
Show Answer
Answer: C – E = ½CV2 = ½ × 10 × 10−6 × 2002 = ½ × 10 × 10−6 × 4.0 × 104 = 0.20 J. (Option A drops the ½; option D mishandles the powers of ten.)
Energy Calculations
Q11. A 470 µF capacitor is charged to 25 V. The energy stored is approximately
- (A) 0.29 J
- (B) 0.059 J
- (C) 0.074 J
- (D) 0.15 J
Show Answer
Answer: D – E = ½CV2 = ½ × 470 × 10−6 × 252 = ½ × 470 × 10−6 × 625 = 0.147 J, about 0.15 J. (Option A is CV2 without the ½.)
Q12. A capacitor stores 1.2 mC of charge at a potential difference of 40 V. The energy stored is
- (A) 24 mJ
- (B) 48 mJ
- (C) 12 mJ
- (D) 96 mJ
Show Answer
Answer: A – E = ½QV = ½ × 1.2 × 10−3 × 40 = ½ × 4.8 × 10−2 = 2.4 × 10−2 J = 24 mJ. (Option B is QV, forgetting the ½.)
Q13. Two capacitors, 2.0 µF and 6.0 µF, are each charged to the same potential difference of 10 V. How do their stored energies E2 (2.0 µF) and E6 (6.0 µF) compare?
- (A) They store equal energies
- (B) E6 is nine times E2
- (C) E6 is three times E2
- (D) E2 is three times E6
Show Answer
Answer: C – At the same V, E = ½CV2 is proportional to C. E2 = ½ × 2.0 × 10−6 × 100 = 100 µJ and E6 = ½ × 6.0 × 10−6 × 100 = 300 µJ, so the 6.0 µF capacitor stores three times as much.
Q14. The same charge Q is placed on a 10 µF capacitor and on a 40 µF capacitor. Which statement about the stored energies is correct?
- (A) The 40 µF capacitor stores four times as much energy
- (B) The 10 µF capacitor stores four times as much energy
- (C) Both store the same energy
- (D) The 10 µF capacitor stores sixteen times as much energy
Show Answer
Answer: B – At fixed Q, E = Q2/(2C), so E is inversely proportional to C. The smaller capacitor reaches the higher p.d. (V = Q/C) and stores more energy: E10/E40 = 40/10 = 4.
Q15. Two identical 10 µF capacitors are connected in series across a 20 V supply. The total energy stored in the combination is
- (A) 1.0 mJ
- (B) 2.0 mJ
- (C) 4.0 mJ
- (D) 0.50 mJ
Show Answer
Answer: A – In series, Ctotal = (10 × 10)/(10 + 10) = 5.0 µF. E = ½CtotalV2 = ½ × 5.0 × 10−6 × 400 = 1.0 × 10−3 J = 1.0 mJ. Check: each capacitor has 10 V across it and stores ½ × 10 µF × 100 = 0.50 mJ, totalling 1.0 mJ.
Q16. Two identical 10 µF capacitors are connected in parallel across a 20 V supply. The total energy stored is
- (A) 1.0 mJ
- (B) 8.0 mJ
- (C) 2.0 mJ
- (D) 4.0 mJ
Show Answer
Answer: D – In parallel, Ctotal = 10 + 10 = 20 µF, and each capacitor has the full 20 V across it. E = ½CtotalV2 = ½ × 20 × 10−6 × 400 = 4.0 × 10−3 J = 4.0 mJ, four times the series total in Q15.
Q17. A 200 µF capacitor is charged to 100 V and then discharged completely through a flash lamp. The total energy converted to light and heat in the lamp is
- (A) 2.0 J
- (B) 0.50 J
- (C) 1.0 J
- (D) zero, because the energy returns to the capacitor
Show Answer
Answer: C – E = ½CV2 = ½ × 200 × 10−6 × 1002 = ½ × 200 × 10−6 × 1.0 × 104 = 1.0 J. On complete discharge all of this stored energy is dissipated in the lamp as light and heat.
Q18. A camera flash unit uses a 1000 µF capacitor charged to 300 V, which discharges in a time of 1.0 ms. The average power of the flash is approximately
- (A) 45 W
- (B) 45 kW
- (C) 4.5 kW
- (D) 450 kW
Show Answer
Answer: B – E = ½CV2 = ½ × 1000 × 10−6 × 3002 = ½ × 1.0 × 10−3 × 9.0 × 104 = 45 J. Average power = E/t = 45 / (1.0 × 10−3) = 4.5 × 104 W = 45 kW. The huge power comes from releasing a modest energy in a very short time.
Q19. A 100 µF capacitor is charged to 12 V from a battery. The battery supplies a total energy of CV2 = 14.4 mJ during charging, while the capacitor stores ½CV2 = 7.2 mJ. The remaining 7.2 mJ is
- (A) dissipated as thermal energy in the resistance of the charging circuit
- (B) stored in the battery for later use
- (C) radiated away as light by the capacitor plates
- (D) an error: energy conservation means no energy is unaccounted for inside the circuit
Show Answer
Answer: A – The battery does work QV = CV2 moving the full charge through the fixed battery p.d. Only half ends up stored in the field; the other half is dissipated as thermal energy in whatever resistance the charging circuit has, however small that resistance is.
Q20. Two identical capacitors of capacitance C are connected first in parallel and then in series across the same supply of voltage V. The ratio (total energy stored in parallel) / (total energy stored in series) is
- (A) 1
- (B) 2
- (C) 8
- (D) 4
Show Answer
Answer: D – Parallel: Ctotal = 2C, so Ep = ½(2C)V2 = CV2. Series: Ctotal = C/2, so Es = ½(C/2)V2 = CV2/4. Ratio = CV2 / (CV2/4) = 4.
Changing a Charged Capacitor
Q21. A parallel-plate capacitor is charged and then disconnected from the battery. The plate separation is doubled. What happens to the stored energy?
- (A) It halves
- (B) It is unchanged
- (C) It doubles
- (D) It quadruples
Show Answer
Answer: C – Disconnected means Q is fixed. Doubling the separation halves C, so use E = Q2/(2C): halving C doubles E. The extra energy comes from the work done pulling the plates apart against the attraction between the oppositely charged plates.
Q22. A parallel-plate capacitor is charged and then disconnected from the battery. The plate separation is doubled. The potential difference between the plates
- (A) doubles
- (B) halves
- (C) is unchanged
- (D) quadruples
Show Answer
Answer: A – With Q fixed and C halved (C is inversely proportional to separation), V = Q/C doubles. This is the same chain as Q21: C halves, V doubles, and E = ½QV doubles because Q is constant.
Q23. A charged capacitor is disconnected from the battery, and a dielectric that multiplies the capacitance by 3 is then slid fully between the plates. What happens to the p.d. V and the stored energy E?
- (A) V triples and E triples
- (B) V is unchanged and E triples
- (C) V falls to one third and E triples
- (D) V falls to one third and E falls to one third
Show Answer
Answer: D – Disconnected, so Q is fixed. C becomes 3C, so V = Q/C falls to one third, and E = Q2/(2C) also falls to one third. The field does work pulling the dielectric in, and that work comes from the stored energy.
Q24. A parallel-plate capacitor remains connected to a battery while the plate separation is doubled. The stored energy
- (A) doubles
- (B) halves
- (C) is unchanged
- (D) quarters
Show Answer
Answer: B – Connected means V is fixed by the battery. Doubling the separation halves C, so E = ½CV2 halves, and Q = CV also halves as charge flows back into the battery. Contrast Q21: the same change with the capacitor disconnected doubles the energy instead.
Q25. A capacitor stays connected to a battery while a dielectric that doubles the capacitance is inserted between the plates. What happens to the charge Q and the stored energy E?
- (A) Q halves and E halves
- (B) Q is unchanged and E is unchanged
- (C) Q doubles and E doubles
- (D) Q doubles and E quadruples
Show Answer
Answer: C – Connected, so V is fixed. With C doubled, Q = CV doubles (extra charge flows from the battery) and E = ½CV2 doubles. The battery supplies both the extra stored energy and the work done pulling the dielectric in. Contrast Q23, where the disconnected capacitor loses energy on insertion.
Q26. An 8.0 µF capacitor is connected to a 10 V battery and stays connected while its plate separation is doubled, halving its capacitance to 4.0 µF. The stored energy changes from 400 µJ to
- (A) 200 µJ
- (B) 400 µJ
- (C) 800 µJ
- (D) 100 µJ
Show Answer
Answer: A – V stays at 10 V. New E = ½CV2 = ½ × 4.0 × 10−6 × 100 = 2.0 × 10−4 J = 200 µJ. Halving C at fixed V halves the stored energy.
Q27. An 8.0 µF capacitor is charged to 10 V and then disconnected, so it holds a charge of 80 µC and stores 400 µJ. Its plate separation is then doubled, halving the capacitance to 4.0 µF. The stored energy is now
- (A) 400 µJ
- (B) 200 µJ
- (C) 100 µJ
- (D) 800 µJ
Show Answer
Answer: D – Disconnected, so Q = 80 µC stays fixed. E = Q2/(2C) = (80 × 10−6)2 / (2 × 4.0 × 10−6) = 6.4 × 10−9 / 8.0 × 10−6 = 8.0 × 10−4 J = 800 µJ. The p.d. has risen to V = Q/C = 20 V. Same change as Q26, opposite result, because here Q (not V) is held constant.
Q28. An isolated (disconnected) charged capacitor stores 0.30 mJ. The plates are pulled further apart until the stored energy is 0.60 mJ. The work done by the external agent pulling the plates apart is
- (A) 0.60 mJ
- (B) 0.30 mJ
- (C) zero
- (D) 0.15 mJ
Show Answer
Answer: B – The capacitor is isolated, so no battery exchanges energy with it. By conservation of energy, the work done against the attraction between the plates equals the increase in stored energy: 0.60 − 0.30 = 0.30 mJ.
Q29. A charged capacitor is disconnected from the battery, and the overlap area of its plates is then doubled, doubling the capacitance. What happens to the p.d. V and the stored energy E?
- (A) V doubles and E doubles
- (B) V is unchanged and E doubles
- (C) V halves and E halves
- (D) V halves and E doubles
Show Answer
Answer: C – Disconnected, so Q is fixed. Doubling C halves V = Q/C, and E = ½QV therefore halves (equivalently, E = Q2/(2C) halves when C doubles).
Q30. When the plates of an isolated charged capacitor are pulled apart, the stored energy increases. The extra energy comes from
- (A) the work done by the external agent against the attraction between the oppositely charged plates
- (B) charge created by the friction between the plates
- (C) the battery, which stays connected through the air
- (D) a decrease in the charge on the plates
Show Answer
Answer: A – The plates carry opposite charges and attract each other. Separating them requires mechanical work against that attraction, and with Q fixed and no battery involved, that work is exactly the increase in the energy stored in the field.
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These are original practice questions written for revision. They are not Cambridge past-paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.