CAIE A-Level Physics A2 19 Capacitance Questions

Practise analysing capacitance, series or parallel combinations, stored energy, discharge curves, time constants and exponential decay.

Exam points

define capacitance and calculate charge, voltage and equivalent series/parallel capacitance
calculate capacitor energy from graph area and energy formulae
analyse discharge graphs, time constants and exponential decay

Covered syllabus points

19.1 Capacitors and capacitance 19.2 Energy stored in a capacitor 19.3 Discharging a capacitor

[Maximum number: 2]

An isolated conducting sphere in a vacuum has radius r and is initially uncharged. It is then charged by friction so that it carries a final charge Q. This charge can be considered to be acting at the centre of the sphere.

By considering the electric potential at its surface, show that the capacitance C of the sphere is given by

C=4 \pi \varepsilon_{0} r

where $\varepsilon_{0}$ is the permittivity of free space.

C=Q / V
$V=Q /\left(4 \pi \varepsilon_{0} r\right)$ and so $C=Q /\left[Q /\left(4 \pi \varepsilon_{0} r\right)\right]=4 \pi \varepsilon_{0} r$

[Maximum number: 10]

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

Fig. 5.1

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

Fig. 5.2

The resistance of R is $25 \mathrm{k} \Omega$ and the resistance of S is $220 \mathrm{k} \Omega$ .
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.

[ 3 ]

(energy stored =) area under line or $\frac{1}{2}$ QV
$=1 / 2 \times 8.0 \times 1.2 \times 10^{-4}$ $=4.8 \times 10^{-4} \mathrm{~J}$

(b)

The switch is now moved to position Y.

[ 5 ]

(i)

Show that the time constant of the discharge circuit is 3.3 s .

[ 2 ]

(t=) R C
( t= ) $220 \times 10^{3} \times\left(1.2 \times 10^{-4} / 8.0\right)=3.3 \mathrm{~s}$

(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.

[ 3 ]

$E \propto V^{2}$
(so time to) $V_{\mathrm{o}} / 3$ $V=V_{0} e^{-t / R C}$ $\begin{aligned} \frac{V_{o}}{3}=V_{o} e^{-t / 3.3} \frac{1}{3}=e^{-t / 3.3} \end{aligned}$ $\mathrm{t}=3.6 \mathrm{~s}$

(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.

[ 2 ]

[Maximum number: 8]

Fig. 6.1 shows a capacitor of capacitance C connected in series with a resistor of resistance R.

Fig. 6.1

Initially the switch is open and there is a p.d. of 12 V across the capacitor.
At time t=0, the switch is closed so that there is a current I in the resistor.
Fig. 6.2 shows the variation of I with t.

Fig. 6.2

(a)

Explain the shape of the line in Fig. 6.2.

[ 3 ]

(b)

Use Fig. 6.2 to determine:

[ 3 ]

(i)

the time constant $\tau$ of the circuit in Fig. 6.1.

\begin{aligned} & \tau= \\ & \text { s [3] } \end{aligned}

[ 3 ]

correct read-off of at least one pair of values for I and t
attempted read-off of t when $I=0.048 \mathrm{~mA}$

or

substitution of a correct pair of values of I and t into $I=0.13 \exp (-t / \tau)$ $\tau=4.3 \mathrm{~s}$

(c)

Use your answers in (b) to determine capacitance C.

C =F [2]

[ 2 ]

$\tau=R C$ $\begin{aligned} C =\tau / R=4.3 /\left(9.2 \times 10^{4}\right) =4.7 \times 10^{-5} \mathrm{~F} \end{aligned}$

[Maximum number: 9]

A capacitor C is charged so that the potential difference (p.d.) V across its terminals is 8.0 V . The capacitor is connected into the circuit of Fig. 6.1.

Fig. 6.1

The switch is initially open. The switch is closed at time t=0.

(a)

Fig. 6.2 shows the variation of V with the charge Q on the plates of capacitor C as the capacitor discharges.

Fig. 6.2

[ 4 ]

(i)

Show that the energy stored in capacitor C at time t=0 is 1.8 mJ .

[ 2 ]

energy stored = area under graph
$=1 / 2 \times 450 \times 10^{-6} \times 8.0=1.8 \times 10^{-3} \mathrm{~J}$ or 1.8 mJ

(ii)

Determine the capacitance of capacitor C . Give a unit with your answer.

[ 2 ]

C=Q / V or $E=1 / 2 C V^{2}$ $\begin{aligned} C =\left(450 \times 10^{-6}\right) / 8.0 \text { or }\left(2 \times 1.8 \times 10^{-3}\right) / 8.0^{2} =5.6 \times 10^{-5} \mathrm{~F} \end{aligned}$

(b)

Fig. 6.3 shows the variation with t of $-\ln \left(\frac{V}{8.0 \mathrm{~V}}\right)$ .

Fig. 6.3

[ 5 ]

(i)

Show that, when t is equal to one time constant, the value of $-\ln \left(\frac{V}{8.0 \mathrm{~V}}\right)$ is equal to 1.0 .

[ 2 ]

$V=V_{0} \exp (-t / R C)$ and $\tau=R C$ $\begin{aligned} V=V_{0} \exp (-t / \tau) V_{0}=8.0 \mathrm{~V} \text {, and at one time constant, } t=\tau V / 8.0=\exp (-\tau / \tau), \text { so } \ln (V / 8.0)=-1.0 \text { or }-\ln (V / 8.0)=1.0 \end{aligned}$

(ii)

Determine the time constant $\tau$ of the circuit in Fig. 6.1.

\tau=

[ 1 ]

[ t read from graph at $-\ln (V / 8.0)=1.0]: \tau=3.2 \mathrm{~s}$

(iii)

Calculate the resistance of resistor R.
resistance = $\Omega$

[ 2 ]

$\tau=R C$ $\begin{aligned} R =3.2 /\left(5.6 \times 10^{-5}\right) =5.7 \times 10^{4} \Omega \end{aligned}$

All question bank results loaded

Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question 3

Question (a)

Question (b)

Question (i)

Question (c)

Question 4

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)

Question (iii)