A-level Physics
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    5.4 Capacitance — Practise with help

    Work each problem first. Open a hint only if you are stuck, and the full working only to check. Write the unit at every step.

    Warm-up: remember

    1. A 6.0 Ω6.0\,\Omega resistor and a 3.0 Ω3.0\,\Omega resistor are connected in parallel. Find the equivalent resistance.
    2. A battery of emf 9.0 V9.0\,\text{V} drives a current of 1.5 A1.5\,\text{A} through an external circuit. Find the power delivered to the external circuit.
    3. Write 220 μF220\,\mu\text{F} in farads.
    Answers
    1. 1R=16.0+13.0=12.0\dfrac{1}{R} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{2.0}, so R=2.0 ΩR = 2.0\,\Omega.
    2. P=VI=9.0×1.5=13.5 WP = VI = 9.0 \times 1.5 = 13.5\,\text{W}.
    3. 220×10−6 F=2.20×10−4 F220 \times 10^{-6}\,\text{F} = 2.20 \times 10^{-4}\,\text{F}.

    If question 1 goes wrong, look again at Section 5.3 and remember that the reciprocals add. If question 2 goes wrong, use P=VIP = VI. If question 3 goes wrong, remember that micro means 10−610^{-6}.

    Practise with help

    1. A capacitor stores a charge of 6.0×10−4 C6.0 \times 10^{-4}\,\text{C} when the pd across it is 12 V12\,\text{V}. Find its capacitance.

    Hint

    Capacitance is the charge divided by the potential difference.

    Next step

    Write C=QVC = \dfrac{Q}{V} with Q=6.0×10−4Q = 6.0 \times 10^{-4} and V=12V = 12.

    Full working C=QV=6.0×10−412=5.0×10−5 FC = \frac{Q}{V} = \frac{6.0 \times 10^{-4}}{12} = 5.0 \times 10^{-5}\,\text{F}

    The capacitance is 5.0×10−5 F5.0 \times 10^{-5}\,\text{F}, which is 50 μF50\,\mu\text{F}.

    2. A capacitor of capacitance 470 μF470\,\mu\text{F} is charged to a pd of 300 V300\,\text{V}. Find the charge stored.

    Hint

    Change the capacitance to farads first, then use Q=CVQ = CV.

    Next step

    Write 470 μF=4.70×10−4 F470\,\mu\text{F} = 4.70 \times 10^{-4}\,\text{F}, then multiply by 300300.

    Full working Q=CV=4.70×10−4×300=0.141 C\begin{aligned} Q &= CV \\ &= 4.70 \times 10^{-4} \times 300 \\ &= 0.141\,\text{C} \end{aligned}

    The charge stored is 0.141 C0.141\,\text{C}.

    3. A capacitor of capacitance 220 μF220\,\mu\text{F} is charged to 50 V50\,\text{V}. Find the energy stored.

    Hint

    Use E=12CV2E = \tfrac{1}{2}CV^2. Change microfarads to farads and square the pd.

    Next step

    Substitute C=220×10−6C = 220 \times 10^{-6} and V=50V = 50 into E=12CV2E = \tfrac{1}{2}CV^2.

    Full working E=12CV2=12×220×10−6×502=12×220×10−6×2500=0.275 J\begin{aligned} E &= \frac{1}{2}CV^2 = \frac{1}{2} \times 220 \times 10^{-6} \times 50^2 \\ &= \frac{1}{2} \times 220 \times 10^{-6} \times 2500 \\ &= 0.275\,\text{J} \end{aligned}

    The energy stored is about 0.28 J0.28\,\text{J}.

    4. A capacitor of capacitance 100 μF100\,\mu\text{F} is discharged through a 200 Ω200\,\Omega resistor. Find the time constant.

    Hint

    The time constant is the product of the resistance and the capacitance.

    Next step

    Write τ=RC\tau = RC with R=200R = 200 and C=100×10−6C = 100 \times 10^{-6}.

    Full working τ=RC=200×100×10−6=2.0×10−2 s\tau = RC = 200 \times 100 \times 10^{-6} = 2.0 \times 10^{-2}\,\text{s}

    The time constant is 2.0×10−2 s2.0 \times 10^{-2}\,\text{s}, or 20 ms20\,\text{ms}.

    5. A capacitor of capacitance 50 μF50\,\mu\text{F} is charged to 100 V100\,\text{V} and then discharged through a 1.0 kΩ1.0\,\text{k}\Omega resistor. Find the time constant and the potential difference after one time constant.

    Hint

    Find τ=RC\tau = RC first. After one time constant the pd is 1e\dfrac{1}{e} of its starting value.

    Next step

    Work out τ=1000×50×10−6\tau = 1000 \times 50 \times 10^{-6}, then find 100 e−1100\,e^{-1}.

    Full working τ=RC=1000×50×10−6=0.050 sV=V0e−1=100×0.368=36.8 V\begin{aligned} \tau &= RC = 1000 \times 50 \times 10^{-6} = 0.050\,\text{s} \\ V &= V_0 e^{-1} = 100 \times 0.368 = 36.8\,\text{V} \end{aligned}

    The time constant is 0.050 s0.050\,\text{s}, and after one time constant the pd is about 37 V37\,\text{V}.

    6. A parallel plate capacitor has plates of area 0.020 m20.020\,\text{m}^2 separated by 5.0×10−5 m5.0 \times 10^{-5}\,\text{m}. The dielectric has a relative permittivity of 3.03.0. Find the capacitance.

    Hint

    Use C=ε0εrAdC = \dfrac{\varepsilon_0 \varepsilon_r A}{d} with ε0=8.85×10−12 F m−1\varepsilon_0 = 8.85 \times 10^{-12}\,\text{F}\,\text{m}^{-1}.

    Next step

    Substitute the four values. Keep the area on the top and the separation on the bottom.

    Full working C=ε0εrAd=8.85×10−12×3.0×0.0205.0×10−5=1.1×10−8 F\begin{aligned} C &= \frac{\varepsilon_0 \varepsilon_r A}{d} \\ &= \frac{8.85 \times 10^{-12} \times 3.0 \times 0.020}{5.0 \times 10^{-5}} \\ &= 1.1 \times 10^{-8}\,\text{F} \end{aligned}

    The capacitance is about 1.1×10−8 F1.1 \times 10^{-8}\,\text{F}, or 11 nF11\,\text{nF}.

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