A Level Physics 19.1 Capacitors and Capacitance MCQs – 30 Practice Questions with Answers (Cambridge 9702)
These 30 original practice questions cover Section 19.1 of the Cambridge International AS & A Level Physics (9702) syllabus – an A Level (A2) topic. They build from the definition of capacitance, through the parallel-plate capacitor and the effect of a dielectric, to combining capacitors in series and in parallel.
Formula recap (see also the A Level Physics Formula Sheet):
- Capacitance: C = Q / V
- Parallel-plate capacitor: C = ε0A / d, and with a dielectric C = εrε0A / d, where ε0 = 8.85 × 10−12 F/m
- Capacitors in parallel: C = C1 + C2 + C3 + …
- Capacitors in series: 1/C = 1/C1 + 1/C2 + 1/C3 + …
Capacitance Basics
Q1. The capacitance of a capacitor is defined as
- (A) the charge stored on it multiplied by the p.d. across it
- (B) the charge stored on one plate divided by the p.d. across the plates
- (C) the energy stored divided by the charge stored
- (D) the p.d. across it divided by the charge stored on one plate
Show Answer
Answer: B – Capacitance is the charge stored per unit p.d., C = Q / V. A capacitor that stores a large charge for a small p.d. has a large capacitance.
Q2. One farad (1 F) is equal to
- (A) 1 coulomb per volt
- (B) 1 volt per coulomb
- (C) 1 joule per coulomb
- (D) 1 coulomb per metre
Show Answer
Answer: A – From C = Q / V, 1 F = 1 C V−1. The farad is a very large unit: a 1 F capacitor holding 1 C of charge at only 1 V is physically huge, so real capacitors are rated in microfarads, nanofarads or picofarads.
Q3. A capacitor stores a charge of 240 μC when the p.d. across it is 12 V. Its capacitance is
- (A) 2.0 μF
- (B) 12 μF
- (C) 20 μF
- (D) 2880 μF
Show Answer
Answer: C – C = Q / V = 240 μC / 12 V = 20 μF.
Q4. A 470 μF capacitor is connected to a 9.0 V supply. The magnitude of the charge stored on each plate is
- (A) 52 μC
- (B) 0.42 mC
- (C) 42 mC
- (D) 4.2 mC
Show Answer
Answer: D – Q = CV = 470 × 10−6 × 9.0 = 4.23 × 10−3 C ≈ 4.2 mC.
Q5. When a capacitor is charged by a battery, which statement is correct?
- (A) Only the plate connected to the positive terminal of the battery gains charge.
- (B) Both plates gain positive charge.
- (C) The two plates gain equal amounts of charge of opposite sign, so the capacitor as a whole remains neutral.
- (D) The net charge of the capacitor equals the charge on its positive plate.
Show Answer
Answer: C – Charging moves electrons from one plate to the other, so the plates carry +Q and −Q. The capacitor stores charge separation, not net charge: its total charge stays zero.
Q6. A capacitor is charged from a battery through a resistor. The charging current decreases as time passes because
- (A) charge leaks off the plates as fast as it arrives
- (B) the p.d. across the capacitor rises towards the supply p.d., so the p.d. across the resistor (which drives the current) falls
- (C) the capacitance of the capacitor increases as it charges
- (D) the internal resistance of the battery keeps increasing
Show Answer
Answer: B – The supply p.d. is shared between the resistor and the capacitor. As charge builds up, the p.d. across the capacitor grows, leaving less p.d. across the resistor, so the current falls (eventually to zero when the capacitor p.d. equals the supply p.d.).
Q7. The p.d. across a capacitor is increased from 6 V to 12 V. What happens to the stored charge and to the capacitance?
- (A) The charge doubles and the capacitance is unchanged.
- (B) The charge doubles and the capacitance doubles.
- (C) The charge is unchanged and the capacitance halves.
- (D) The charge halves and the capacitance is unchanged.
Show Answer
Answer: A – Capacitance is fixed by the construction of the capacitor (plate area, separation and dielectric); it does not depend on Q or V. Since Q = CV with C constant, doubling V doubles Q.
Q8. The capacitance of a parallel-plate capacitor depends on
- (A) the charge stored on the plates
- (B) the p.d. across the plates
- (C) the current used to charge it
- (D) the plate area, the plate separation and the material between the plates
Show Answer
Answer: D – C = ε0A / d (times εr if a dielectric is present). Charge and p.d. set how much charge is stored, not the capacitance itself.
Q9. Which capacitance is typical of a capacitor used in an electronic circuit?
- (A) 100 F
- (B) 100 kF
- (C) 100 μF
- (D) 100 C
Show Answer
Answer: C – The farad is an enormous unit, so practical capacitors have values in microfarads (μF), nanofarads (nF) or picofarads (pF). Option D is not even a unit of capacitance.
Q10. A charge of 50 μC is stored on a 10 μF capacitor. The p.d. across the capacitor is
- (A) 0.20 V
- (B) 5.0 V
- (C) 2.0 V
- (D) 500 V
Show Answer
Answer: B – V = Q / C = 50 μC / 10 μF = 5.0 V.
The Parallel-Plate Capacitor
Q11. For a parallel-plate capacitor with air (or vacuum) between the plates, area A and separation d, the capacitance is given by
- (A) C = ε0A / d
- (B) C = ε0d / A
- (C) C = A / (ε0d)
- (D) C = ε0Ad
Show Answer
Answer: A – C = ε0A / d: capacitance increases with plate area and decreases with plate separation. With a dielectric filling the gap, ε0 is replaced by εrε0.
Q12. Two parallel plates, each of area 0.010 m2, are separated by 1.0 mm of air. The capacitance is approximately
- (A) 8.9 pF
- (B) 89 pF
- (C) 890 pF
- (D) 8.9 nF
Show Answer
Answer: B – C = ε0A / d = 8.85 × 10−12 × 0.010 / (1.0 × 10−3) = 8.85 × 10−11 F ≈ 89 pF. Air-spaced plates give only picofarads, which is why real capacitors use very thin dielectrics and rolled-up or stacked plates.
Q13. The plate separation of an air-filled parallel-plate capacitor is doubled. Its capacitance becomes
- (A) four times as large
- (B) twice as large
- (C) unchanged
- (D) one half as large
Show Answer
Answer: D – C is inversely proportional to d, so doubling d halves C.
Q14. The plate area of a parallel-plate capacitor is doubled and the plate separation is halved. The capacitance is multiplied by
- (A) 1
- (B) 2
- (C) 4
- (D) 8
Show Answer
Answer: C – C ∝ A / d. Doubling A gives ×2 and halving d gives another ×2, so C increases by a factor of 4.
Q15. A capacitor has a capacitance of 60 pF with air between its plates. A dielectric of relative permittivity εr = 3.0 is inserted so that it completely fills the gap. The new capacitance is
- (A) 180 pF
- (B) 60 pF
- (C) 20 pF
- (D) 540 pF
Show Answer
Answer: A – With a dielectric, C = εrε0A / d = εr × Cair = 3.0 × 60 pF = 180 pF.
Q16. Parallel plates of area 0.020 m2 are separated by 0.50 mm. The gap is completely filled with a dielectric of relative permittivity 2.0. The capacitance is approximately
- (A) 177 pF
- (B) 354 pF
- (C) 710 pF
- (D) 1.42 nF
Show Answer
Answer: C – C = εrε0A / d = 2.0 × 8.85 × 10−12 × 0.020 / (0.50 × 10−3) = 2.0 × 3.54 × 10−10 = 7.08 × 10−10 F ≈ 710 pF.
Q17. Inserting a dielectric between the plates of a capacitor increases its capacitance because
- (A) the dielectric adds extra free charge to the plates
- (B) the molecules of the dielectric polarise, and the induced charges on its surfaces reduce the electric field (and so the p.d.) between the plates for the same stored charge
- (C) the dielectric increases the separation of the plates
- (D) the dielectric reduces the value of the permittivity of free space
Show Answer
Answer: B – In the field between the plates, dielectric molecules polarise so that bound charge of opposite sign appears on each face of the dielectric. This partly cancels the field inside, so a given charge Q on the plates produces a smaller p.d. V – and since C = Q / V, the capacitance rises by the factor εr.
Q18. Two metal plates, each 5.0 cm × 4.0 cm, are separated by 0.40 mm of air. The capacitance is approximately
- (A) 4.4 pF
- (B) 8.9 pF
- (C) 890 pF
- (D) 44 pF
Show Answer
Answer: D – A = 0.050 × 0.040 = 2.0 × 10−3 m2. C = 8.85 × 10−12 × (2.0 × 10−3) / (0.40 × 10−3) = 8.85 × 10−12 × 5.0 = 4.4 × 10−11 F ≈ 44 pF. (Careful: work in metres throughout.)
Q19. An air-filled parallel-plate capacitor has a capacitance of 200 pF. The plates are pulled apart so that the separation becomes three times larger. The new capacitance is
- (A) 600 pF
- (B) 67 pF
- (C) 100 pF
- (D) 22 pF
Show Answer
Answer: B – C ∝ 1 / d, so tripling d reduces C to 200 / 3 ≈ 67 pF.
Q20. A parallel-plate capacitor has a capacitance of 350 pF and is connected to a 24 V supply. The magnitude of the charge on each plate is
- (A) 8.4 nC
- (B) 0.35 nC
- (C) 84 nC
- (D) 8400 nC
Show Answer
Answer: A – Q = CV = 350 × 10−12 × 24 = 8.4 × 10−9 C = 8.4 nC.
Capacitors in Series & Parallel
Q21. Which statement about capacitors connected in parallel is correct?
- (A) The charge on each capacitor is the same.
- (B) The total capacitance is smaller than the smallest individual capacitance.
- (C) The p.d.s across the capacitors add up to the supply p.d.
- (D) The total capacitance is the sum of the individual capacitances, and each capacitor has the full supply p.d. across it.
Show Answer
Answer: D – In parallel every capacitor is connected directly across the supply, so each has the same p.d. V. The charges add: Qtotal = Q1 + Q2 = (C1 + C2)V, giving Ctotal = C1 + C2.
Q22. A 4.0 μF capacitor and a 6.0 μF capacitor are connected in parallel. The total capacitance is
- (A) 10 μF
- (B) 2.4 μF
- (C) 5.0 μF
- (D) 24 μF
Show Answer
Answer: A – In parallel, Ctotal = 4.0 + 6.0 = 10 μF.
Q23. Which statement about capacitors connected in series is correct?
- (A) The total capacitance is the sum of the individual capacitances.
- (B) Each capacitor carries the same charge, and the total capacitance is smaller than the smallest individual capacitance.
- (C) The p.d. across each capacitor is equal to the supply p.d.
- (D) The stored charge splits between the capacitors in proportion to their capacitances.
Show Answer
Answer: B – In series the same charge Q appears on every capacitor (the isolated middle plates can only gain induced charge), while the p.d.s add up to the supply p.d. This gives 1/Ctotal = 1/C1 + 1/C2, so Ctotal is always less than the smallest capacitor in the chain.
Q24. A 6.0 μF capacitor and a 3.0 μF capacitor are connected in series. The total capacitance is
- (A) 9.0 μF
- (B) 4.5 μF
- (C) 2.0 μF
- (D) 1.5 μF
Show Answer
Answer: C – 1/C = 1/6.0 + 1/3.0 = 0.167 + 0.333 = 0.50, so C = 2.0 μF – smaller than the smallest capacitor, as always for a series combination.
Q25. Two identical 10 μF capacitors are connected first in series and then in parallel. The total capacitances are
- (A) 5.0 μF in series and 20 μF in parallel
- (B) 20 μF in series and 5.0 μF in parallel
- (C) 10 μF in series and 10 μF in parallel
- (D) 5.0 μF in series and 10 μF in parallel
Show Answer
Answer: A – Series: 1/C = 1/10 + 1/10 = 0.20, so C = 5.0 μF (two equal capacitors in series give half of one). Parallel: C = 10 + 10 = 20 μF.
Q26. A 2.0 μF capacitor and a 4.0 μF capacitor are connected in series across a 12 V supply. The p.d. across the 2.0 μF capacitor is
- (A) 4.0 V
- (B) 6.0 V
- (C) 8.0 V
- (D) 12 V
Show Answer
Answer: C – Ctotal = (2.0 × 4.0) / (2.0 + 4.0) = 1.33 μF, so the charge on each capacitor is Q = 1.33 × 12 = 16 μC. Then V = Q / C = 16 / 2.0 = 8.0 V across the 2.0 μF capacitor (and 16 / 4.0 = 4.0 V across the other). In series, the smaller capacitor takes the larger share of the p.d.
Q27. A 3.0 μF capacitor and a 6.0 μF capacitor are connected in parallel across a 10 V supply. The charge stored on the 6.0 μF capacitor is
- (A) 20 μC
- (B) 30 μC
- (C) 45 μC
- (D) 60 μC
Show Answer
Answer: D – In parallel each capacitor has the full 10 V across it, so Q = CV = 6.0 × 10 = 60 μC. (The 3.0 μF capacitor stores 30 μC: in parallel, charge splits in proportion to capacitance.)
Q28. Two 4.0 μF capacitors are connected in parallel, and this combination is connected in series with an 8.0 μF capacitor. The total capacitance of the network is
- (A) 16 μF
- (B) 4.0 μF
- (C) 8.0 μF
- (D) 20 μF
Show Answer
Answer: B – Reduce the network step by step. The parallel pair gives 4.0 + 4.0 = 8.0 μF. This in series with 8.0 μF gives 1/C = 1/8.0 + 1/8.0 = 0.25, so C = 4.0 μF.
Q29. Several capacitors are connected in series across a supply. The charge stored on each capacitor is
- (A) the same on every capacitor
- (B) largest on the capacitor with the largest capacitance
- (C) largest on the capacitor with the smallest capacitance
- (D) different on each capacitor, in direct proportion to its capacitance
Show Answer
Answer: A – The plates between neighbouring capacitors form an isolated, neutral section, so charge can only appear on them by induction – every capacitor in the chain ends up with the same magnitude of charge, equal to Ctotal × Vsupply. (Sharing charge in proportion to capacitance is what happens in parallel, not in series.)
Q30. Capacitors of 1.0 μF, 2.0 μF and 3.0 μF are connected in series across a supply. The largest p.d. appears across the
- (A) 3.0 μF capacitor
- (B) 2.0 μF capacitor
- (C) All three capacitors have the same p.d. across them.
- (D) 1.0 μF capacitor
Show Answer
Answer: D – In series every capacitor carries the same charge Q, and V = Q / C, so the p.d. is largest across the smallest capacitance – here the 1.0 μF capacitor.
Keep practising: work through more topic-by-topic questions on the Physics MCQs page, and keep the key formulas beside you with the A Level Physics Formula Sheet.
These are original practice questions written for revision purposes; they are not Cambridge past-paper questions. Quaidianpoint.com is not affiliated with Cambridge Assessment International Education.