S5 Physics Practice Paper: Capacitors, Electric Field and Potential (100 marks)

2026-11-10

S5 Physics – Capacitors and electric fields

S5 Physics paper: 6 questions, 22 parts, 100 marks, 60 min. Coulomb law, electric field, capacitors, networks, charge sharing, RC discharge; mark schemes included.

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Section A – Coulomb law, field and potential

Take k = 8.99 x 10^9 N m^2 C^-2. Show formula, substitution and unit.

#	Question	Answer and marking notes
1	Point charges of +2.0 uC and +5.0 uC are 0.30 m apart. Find the magnitude and direction of the electrostatic force between them.	F = kQ1Q2 / r^2 = 8.99 x 10^9 x 2.0 x 10^-6 x 5.0 x 10^-6 / 0.30^2 = 1.00 N; both charges are positive, so the force acts outwards along the line joining them (repulsion). (3 marks: formula 1; substitution 1; value and direction 1)
2	A point charge of +4.0 uC is isolated. Find the electric field strength 0.15 m from it.	E = kQ / r^2 = 8.99 x 10^9 x 4.0 x 10^-6 / 0.15^2 = 1.60 x 10^6 N/C, directed radially away from the charge. (3 marks: formula 1; substitution 1; value and direction 1)
3	Find the electric potential at the same point, 0.15 m from the +4.0 uC charge.	V = kQ / r = 8.99 x 10^9 x 4.0 x 10^-6 / 0.15 = 2.40 x 10^5 V (positive around a positive charge). (3 marks: formula 1; substitution 1; answer with unit 1)
4	A charge of +3.0 uC is moved from 0.15 m to 0.30 m away from the +4.0 uC charge. Find the work done by the electric field.	V(0.15 m) = 2.40 x 10^5 V; V(0.30 m) = 1.20 x 10^5 V; W = q dV = 3.0 x 10^-6 x (2.40 – 1.20) x 10^5 = 0.360 J, positive because the field pushes the charge outwards. (4 marks: two potentials 2; dV 1; answer with sign 1)

Section B – The parallel-plate capacitor

Take the permittivity of free space e0 = 8.85 x 10^-12 F/m.

#	Question	Answer and marking notes
5	A parallel-plate capacitor has plates of area 0.020 m^2 separated by 2.0 mm of air. Calculate its capacitance.	C = e0 A / d = 8.85 x 10^-12 x 0.020 / 2.0 x 10^-3 = 88.5 pF. (4 marks: formula 1; area 1; converting mm to m 1; answer 1)
6	The capacitor is connected to a 12 V supply. Find the charge on each plate.	Q = CV = 88.5 x 10^-12 x 12 = 1.06 nC, equal and opposite on the two plates. (3 marks: formula 1; substitution 1; answer 1)
7	Find the electric field strength between the plates, assuming a uniform field.	E = V / d = 12 / 2.0 x 10^-3 = 6000 V/m, from the positive plate to the negative plate. (3 marks: formula 1; substitution 1; direction 1)
8	Calculate the energy stored in the capacitor.	U = 1/2 CV^2 = 0.5 x 88.5 x 10^-12 x 144 = 6.37 x 10^-8 J (or 1/2 QV). (4 marks: formula 1; substitution 1; squaring V 1; answer with unit 1)

Section C – Series and parallel combinations

Three capacitors: C1 = 6.0 uF, C2 = 12 uF and C3 = 4.0 uF.

#	Question	Answer and marking notes
9	If all three capacitors are connected in parallel, find the equivalent capacitance.	Capacitances add in parallel: 6.0 + 12 + 4.0 = 22 uF. (6 marks: parallel rule 2; three additions 3; answer with unit 1)
10	If C1 and C2 are connected in series, find the equivalent capacitance.	1/C = 1/6.0 + 1/12 = 0.250, so C = 4.0 uF, smaller than either capacitor. (6 marks: reciprocal rule 2; substitution 2; answer 1; check 1)
11	C1 and C2 are first connected in parallel, and the pair is then connected in series with C3. Find the equivalent capacitance of the whole network.	Parallel pair: 6.0 + 12 = 18 uF; in series with 4.0 uF: C = (18 x 4.0) / 22 = 3.27 uF. (6 marks: parallel pair 2; series formula 2; working 1; answer 1)
12	The network is connected across 24 V. Find the total charge stored and the voltage across C3 and across the parallel pair.	Q = CV = 3.27 x 10^-6 x 24 = 78.5 uC; voltage across C3 = Q / 4.0 = 19.6 V; across the parallel pair = Q / 18 = 4.4 V, and the two add to 24 V. (8 marks: total charge 2; equal charge in series 2; each voltage 2)

Section D – Charge sharing and energy loss

A 10 uF capacitor is charged to 20 V and then connected in parallel with an uncharged 15 uF capacitor.

#	Question	Answer and marking notes
13	Find the common voltage across the two capacitors after connection.	Charge is conserved: Q = 10 x 10^-6 x 20 = 200 uC; V = Q / (10 + 15) uF = 8.0 V. (5 marks: initial charge 2; dividing by total capacitance 2; answer 1)
14	Find the final charge on each capacitor.	10 uF: 10 x 8.0 = 80 uC; 15 uF: 15 x 8.0 = 120 uC; total 200 uC, the same as before. (4 marks: each charge 1.5; conservation check 1)
15	Calculate the total energy stored before and after connection, and explain where the missing energy goes.	Before: 1/2 x 10 x 10^-6 x 20^2 = 2.0 mJ; after: 1/2 x 25 x 10^-6 x 8.0^2 = 0.80 mJ; loss 1.2 mJ (60%). The lost energy is dissipated as heat, sparking and electromagnetic radiation at the instant of connection. (6 marks: two energies 2 each; difference 1; explanation 1)

Section E – RC discharge

A 47 uF capacitor is charged to 12 V and then discharged through a 100 k-ohm resistor.

#	Question	Answer and marking notes
16	Find the time constant of the circuit.	tau = RC = 100 x 10^3 x 47 x 10^-6 = 4.7 s. (4 marks: formula 2; substitution 1; answer with unit 1)
17	Find the voltage across the capacitor 5.0 s after discharge begins.	V = V0 e^(-t/tau) = 12 x e^(-5.0/4.7) = 4.14 V, about one third of the initial value. (5 marks: formula 2; substitution 2; answer 1)
18	Find the time taken for the voltage to fall to 3.0 V.	3.0 = 12 e^(-t/4.7), so t = 4.7 x ln 4 = 6.52 s. (6 marks: taking logs 2; rearranging 2; using tau 1; answer 1)
19	Find the energy dissipated in the resistor during the first 4.7 s of discharge.	Initial energy 1/2 x 47 x 10^-6 x 144 = 3.38 mJ; at t = tau, V = 12/e = 4.41 V, energy 0.46 mJ; energy dissipated = 2.93 mJ. (6 marks: initial energy 2; voltage at tau 2; difference 2)

Section F – Dielectrics and applications

Use the capacitor of Section B (area 0.020 m^2, spacing 2.0 mm, C = 88.5 pF in air).

#	Question	Answer and marking notes
20	With the 12 V supply still connected, a dielectric of relative permittivity 3.0 fills the space between the plates. Find the new capacitance.	C = kappa C0 = 3.0 x 88.5 = 265.5 pF. (4 marks: relation 2; substitution 1; answer 1)
21	Continuing from the previous part with the supply still connected, find the factor by which the stored energy changes and state the role of the supply.	U = 1/2 CV^2 with V fixed, so the energy becomes 3 times larger; the supply pushes extra charge onto the plates, providing the extra energy. (4 marks: V fixed 1; factor 3 two; role of supply 1)
22	If the capacitor is charged to 12 V, disconnected from the supply, and only then filled with the same dielectric, find the new voltage and the ratio of the new stored energy to the original.	Charge is fixed: V = Q / C falls by a factor of 3, so V = 4.0 V; U = Q^2 / 2C, so the stored energy becomes one third of the original. (3 marks: V = 4.0 V 1.5; energy ratio 1.5)

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