16PS. Capacitors (2026) tutorial solutions NJC

National Junior College Science Department | Physics 1 16. Capacitors Problem set (suggested solutions) 1 (a) apply potential difference between the plates M1 causes charge separation between the plates A1 (b) (i) straight line starting at the origin B1 line with positive gradient ending at (V, Q) B1 (ii) work done is the area under the graph B1 W = ½ QV B1 (c) (i) Electric field is constant / remains the same B1 E = VL = VNL!D M1 VN="L−DL$V (ii) Charge on each plate remains the same. B1 Charge on each plate Q = CV C1 CN= QVN = CV"L!DL#V = !LL!D" C A1 2 (a) Combined capacitance = 24 + (1/24 + 1/24 + 1/24)–1 = 24 + 8 = 32 µF (b) (i) parallel combination of two in series and a single capacitor (ii) one capacitor in series with two in parallel

National Junior College Science Department | Physics 2 3 (a) relationship between charges relationship between p.d.s series QS = Q1 = Q2 VS = V1 + V2 parallel QS = Q1 + Q2 VS = V1 = V2 (b) (i) E = ½ CV2 19 × 10–3 = ½ (470 × 10–6)V2 C1 V = 9.0 V A1 (ii) E = Q2 / 2C (or use C = Q / V) 19 × 10–3 = Q2 / (2 × 470 × 10–6) C1 Q = 4.2 × 10–3 C A1 (iii) 1. total capacitance = (470 + 180) × 10–6 = 650 × 10–6 C1 total charge unchanged C1 p.d. across capacitor = p.d. across total capacitance = Q / C = (4.226 × 10–3) / (650 × 10–6) = 6.5 V A1 2. Total energy in the two capacitors = Q2 / 2C = (4.226 × 10–3)2 / (2 × 650 × 10–6) = 13.74 mJ C1 Decrease in total energy = 19 – 13.74 = 5.3 mJ A1

National Junior College Science Department | Physics 3 4 combined capacitance = (1 / 22 + 1 / 47)–1 = 14.99 µF C1 initial charge on 22 µF = initial charge on combined capacitors = 14.99 × 12 = 179.88 µC AND charge on 22 µF when p.d. is 6 V = 22 × 6 = 132 µC C1 132 = 179.88 exp[–t / (2.7 × 106 × 14.99 × 10–6)] C1 Note: For the time constant, it is incorrect to use 22 𝜇F even though we compute the charge using it. This is because the circuit comprises both capacitors and the time constant is due to both capacitors. t = 12.5 s or 13 s A1 5 (a) current in resistor causes capacitor to lose charge, since p.d. proportional to charge so p.d. across capacitor decreases B1 rate of change of p.d. across capacitor decreases as p.d. (or charge) decreases B1 (since !!"(𝑉#𝑒$!"+=−%&𝑉#𝑒$!", gradient of p.d. decreases) since p.d. across resistor = p.d. across capacitor B1 so rate of change of p.d. across resistor decreases as p.d. (or charge) decreases (b) [Method 1] Q0 = 0.90 mC B1 At t = one time constant, Q = 0.90 e–1 = 0.33 mC M1 Construction on Fig. (b) showing when Q = 0.33 mC, 𝜏 = 5.5 s A1 [Method 2] Evidence of two correct sets of readings for Q and t from the graph B1 Correct substitution of Q1 = Q0 exp (–t1 / 𝜏) and Q2 = Q0 exp (–t2 / 𝜏) M1 Correct calculation Q2 / Q1 = exp [(t1 – t2) / 𝜏] to give 𝜏 = 5.5 s A1 [Method 3] Read-off of half-life as 3.75 s B1 Use