# CIE AS & A Level Physics : 19.1 Capacitors and capacitance – Exam style question – Paper 4

### Question

The variation with potential difference V of the charge Q on one of the plates of a capacitor is shown in Fig. 5.1.

The capacitor is connected to an 8.0V power supply and two resistors R and S as shown in Fig. 5.2.

The resistance of R is $$25$$ kΩ and the resistance of S is $$220$$ kΩ.

The switch can be in either position X or position Y.

(a)     The switch is in position X so that the capacitor is fully charged.

Calculate the energy E stored in the capacitor.

(b)    The switch is now moved to position Y.

(i)      Show that the time constant of the discharge circuit is $$3.3$$s.

(ii)     The fully charged capacitor in (a) stores energy E.

Determine the time t taken for the stored energy to decrease from E to E/9

(c)     A second identical capacitor is connected in parallel with the first capacitor.

State and explain the change, if any, to the time constant of the discharge circuit.

Ans:

(a)   (energy stored =) area under line or $$1/2$$ QV

= $$1/2 \times 8.0 \times 1.2 \times 10^{-4}$$

= $$4.8 \times 10^{–4}$$ J

(b)     (i)      (τ=) RC

(τ=) $$220 \times 10^{3} \times (1.2 \times 10^{-4}/8.0)$$ = $$3.3$$ s

(ii)     $$E$$ ∝ $$V^{2}$$

(so time to) $$V_{0}/3$$

$$V$$ = $$V_{0}e^{-t/RC}$$

$$\frac{V_{0}}{3}$$ = $$V_{0}e^{-t/3.3}$$

$$\frac{1}{3}$$ = $$e^{-t/3.3}$$

$$t$$ = $$3.6$$ s

(c)     (total) capacitance is doubled

### Question

(a) A capacitor consists of two parallel metal plates, separated by air, at a variable distance x apart, as shown in Fig. 6.1. The capacitance C is inversely proportional to x.

The capacitor is charged by a supply so that there is a potential difference (p.d.) V between the plates.  State expressions, in terms of C and V, for the charge Q on one of the plates and for the
energy E stored in the capacitor.
Q = ……………………………………… E = ……………………………………… [1]

(b) The charged capacitor in (a) is now disconnected from the supply. The plates of the capacitor are initially separated by distance L. They are then moved closer together by a distance D, as
shown in Fig. 6.2.

State expressions, in terms of C, V, L and D, for:

(i) the new capacitance CN

CN = ………………………………………………… [1]

(ii) the new charge QN on one of the plates
QN = ………………………………………………… [1]
(iii) the new p.d. VN between the plates.
VN = ………………………………………………… [1]
(c) Explain whether reducing the separation of the plates in (b) results in an increase or decrease in the energy stored in the capacitor.          [1]                                                                                                                                                                                                                                                                                             [Total: 5]

Ans

(a) Q = CV and E = ½CV2

(b) (i) CN = CL / (L – D)

(b) (ii) (charge is unchanged by moving the plates so) QN = CV

(b) (iii) VN = QN / CN
= (CV) / [CL / (L – D)]
=V(L – D) / L

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