16ES. Capacitors (2026) notes NJC exercise solutions

National Junior College Science Department | Physics 1 Topic 16: Capacitors Solutions to Exercises Exercise 1 1.1 An uncharged capacitor is connected between earth Z and a terminal W. A positively charged rod is brought close to W. Which of the following describes the movement of charge? A Electrons move from W to X and from Y to Z. B Electrons move from W to X but not from Y to Z. C Electrons move from X to W and from Z to Y. D Electrons move from X to W but not from Z to Y. Answer: C Electrons near W attracted to positive rod causing electrons to move towards W. X becomes net positive. X being positive attracts electrons to move towards Y, drawing electrons from the earth. 1.2 A 20 µF capacitor is charged by a constant current of 10 mA. If the capacitor is initially uncharged, how long does it take for the potential difference across the capacitor to reach 300 V? A 6.0 × 10–4 s B 0.60 s C 15 s D 6.0 × 105 s Answer: B Charge on capacitor when p.d. is 300 V, Q = CV = (20 × 10–6)(300) = 0.006 C Since current = charge / time, 10 × 10–3 = 0.006 / t t = 0.6 s

National Junior College Science Department | Physics 2 1.3 Four identical capacitors are connected as shown. Which of the following lists the arrangements in order of decreasing capacitance? A PQRS B PSRQ C QRSP D QSRP Answer: B Let C be the capacitance of each capacitor. Circuit P: Ctotal = C + C + C + C = 4C Circuit Q: Ctotal = [1/C + 1/C + 1/C + 1/C]–1 = C/4 Circuit R: Ctotal = [1/C + 1/(C + C) + 1/C]–1 = 2C/5 Circuit S: Ctotal = [1/C + 1/(C + C + C)]–1 = 3C/4 Decreasing capacitance is PSRQ. 1.4 Two capacitors are connected in series as shown. What is the charge carries by each of these capacitors? charge on C1 / μC charge on C2 / μC A 4.0 4.0 B 4.0 8.0 C 8.0 4.0 D 8.0 8.0 Answer: D Ctotal = (1/1.0 + 1/2.0)–1 = 2/3 μF p.d. across equivalent capacitor V = 12 – 0 = 12 V Charge on equivalent capacitor = Ctotal V = (2/3) (12) = 8 μC Capacitors in series have the same amount of charge.

National Junior College Science Department | Physics 3 1.5 The energy stored in a capacitor of capacitance C, carrying charge Q with potential difference V between its plates. may be obtained by calculating the area under an appropriate graph. Which graph shows the correct relationship between a pair of the quantities C, Q and V, and in addition shows a shaded area which corresponds to the energy stored in the capacitor? Answer: A C = Q / V and energy stored in capacitor =!"𝑄𝑉=!"𝐶𝑉"=!"#!$. So Option A: Graph of V against Q gives a straight line with gradient = 1/C. Area under graph = !"𝑄𝑉 = energy stored Option B: Graph of C against Q gives a straight line with gradient = 1/V. Area under graph = !"𝑄𝐶 ≠ energy stored Option C: Graph of C against 1/V gives a straight line with gradient = Q. Area under graph = !"𝐶'!%(=!"$% ≠ energy stored Option D: Graph of Q against V cannot be a