5.4 Capacitance — Practise

Work through these on your own. Write the unit of every answer, and change microfarads to farads before you substitute. In the last questions, explain your reasoning in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A capacitor stores a charge of $2.0 \times 10^{-5}\,\text{C}$ at a potential difference of $10\,\text{V}$ . Find its capacitance.
A capacitor of capacitance $100\,\mu\text{F}$ is charged to a potential difference of $12\,\text{V}$ . Find the charge stored.
A capacitor of capacitance $50\,\mu\text{F}$ is charged to a potential difference of $100\,\text{V}$ . Find the energy stored.
A capacitor of capacitance $10\,\mu\text{F}$ is discharged through a $10\,\text{k}\Omega$ resistor. Find the time constant.
A capacitor of capacitance $100\,\mu\text{F}$ is charged to $10\,\text{V}$ and discharged through a $1.0\,\text{k}\Omega$ resistor. Find the potential difference after $0.10\,\text{s}$ .
A parallel plate capacitor has plates of area $0.050\,\text{m}^2$ separated by $2.0 \times 10^{-4}\,\text{m}$ . The gap is filled with air, which has a relative permittivity of $1.0$ . Find the capacitance.
A capacitor holds a charge of $0.50\,\text{C}$ at a potential difference of $100\,\text{V}$ . Find the energy stored.
A capacitor discharges with a time constant of $0.050\,\text{s}$ . Find the half-life, the time for the charge to fall to half its starting value.
A capacitor stores $0.10\,\text{J}$ when the potential difference across it is $50\,\text{V}$ . Find its capacitance.
A capacitor is charged to $2000\,\text{V}$ and discharged through a total resistance of $50\,\Omega$ . Find the current at the first instant of the discharge.
A student plots $\ln V$ against time for a discharging capacitor. The points are $(0.00, 2.303)$ , $(0.010, 1.803)$ , $(0.020, 1.303)$ and $(0.030, 0.803)$ . Find the gradient of the line, and state its unit.
A capacitor holds a charge of $0.200\,\text{C}$ at a potential difference of $2000\,\text{V}$ . The charge–potential difference graph is a straight line through the origin. Find the energy stored from the area under the graph.
For a discharging capacitor, the points $(0.00, 2.303)$ , $(0.010, 1.803)$ , $(0.020, 1.303)$ and $(0.030, 0.803)$ lie on a straight-line graph of $\ln V$ against time. Find the intercept of the line, which is $\ln V_0$ .
Explain why the charge on a discharging capacitor falls exponentially with time.

Answers and concise solutions

1. The capacitance

Answer: $C = \dfrac{Q}{V} = \dfrac{2.0 \times 10^{-5}}{10} = 2.0 \times 10^{-6}\,\text{F} = 2.0\,\mu\text{F}$ .

2. The charge stored

Answer: $Q = CV = 100 \times 10^{-6} \times 12 = 1.2 \times 10^{-3}\,\text{C}$ .

3. The energy stored

Answer: $E = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 50 \times 10^{-6} \times 100^2 = 0.25\,\text{J}$ .

4. The time constant

Answer: $\tau = RC = 10 \times 10^{3} \times 10 \times 10^{-6} = 0.10\,\text{s}$ .

5. The potential difference

Answer: The time constant is $\tau = 1000 \times 100 \times 10^{-6} = 0.10\,\text{s}$ , so after $0.10\,\text{s}$ the pd is $V = 10 e^{-1} = 3.7\,\text{V}$ .

6. The capacitance

Answer: $C = \dfrac{\varepsilon_0 \varepsilon_r A}{d} = \dfrac{8.85 \times 10^{-12} \times 1.0 \times 0.050}{2.0 \times 10^{-4}} = 2.2 \times 10^{-9}\,\text{F}$ .

7. The energy stored

Answer: $E = \tfrac{1}{2}QV = \tfrac{1}{2} \times 0.50 \times 100 = 25\,\text{J}$ .

8. The half-life

Answer: The half-life is $t_{1/2} = RC\ln 2 = 0.050 \times 0.693 = 0.035\,\text{s}$ .

9. The capacitance

Answer: From $E = \tfrac{1}{2}CV^2$ , $C = \dfrac{2E}{V^2} = \dfrac{2 \times 0.10}{50^2} = 8.0 \times 10^{-5}\,\text{F} = 80\,\mu\text{F}$ .

10. The initial current

Answer: At the first instant the full pd is across the resistance, so $I_0 = \dfrac{V_0}{R} = \dfrac{2000}{50} = 40\,\text{A}$ .

11. The gradient

Answer: The gradient is $\dfrac{0.803 - 2.303}{0.030 - 0.00} = -50\,\text{s}^{-1}$ . The gradient is $-\dfrac{1}{RC}$ , so the time constant is $\dfrac{1}{50} = 0.020\,\text{s}$ .

12. The energy stored

Answer: The area under the charge–potential difference graph is a triangle, so $E = \tfrac{1}{2}QV = \tfrac{1}{2} \times 0.200 \times 2000 = 200\,\text{J}$ .

13. The intercept

Answer: The intercept is $2.303$ , so $\ln V_0 = 2.303$ and $V_0 = e^{2.303} = 10\,\text{V}$ . The starting potential difference is $10\,\text{V}$ .

14. Why the discharge is exponential

The current through the resistor is proportional to the potential difference across the capacitor, which is proportional to the charge left on it. So a larger charge gives a larger current.
The current is the rate at which the charge leaves the capacitor. With a larger current, the charge falls faster.
So the rate at which the charge falls is proportional to the charge remaining. That is the condition for exponential decay, and it gives Q equals Q zero times e to the minus t over RC.

Answer: The current is proportional to the pd, and the pd is proportional to the charge remaining. So the rate at which the charge falls is proportional to the charge that is left. That is exponential decay, with time constant $RC$ .