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A parallel-plate capacitor has plates 0.8 mm apart and area 4 cm2. With a potential difference of 100 V, find its capacitance and the charge stored.

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2 parts
b(i)

4.425×10−12 F\text{4.425} \times 10^{-12}\text{ F}

b(ii)

4.425×10−10 C\text{4.425} \times 10^{-10}\text{ C}

Step-by-step solution

5 steps
  1. (a) Define capacitance. Capacitance is the measure of a capacitor's ability to store an electric charge. It is defined as the ratio of the magnitude of the charge stored on either conductor to the potential difference between the conductors.

Part b(i)· 3 steps

Find the capacitance of the capacitor (take ϵ0=8.85×10−12 FM−1\epsilon_0 = 8.85 \times 10^{-12} \text{ FM}^{-1}).

Two plates of a parallel-plate capacitor are 0.8mm0.8\text{mm} apart and each has an area 4cm24\text{cm}^2. Given that the potential difference between the plates is 100V100\text{V},

  1. Convert given values to SI units.

    Distance between plates, d=0.8 mm=0.8×10−3 md = 0.8\text{ mm} = 0.8 \times 10^{-3}\text{ m}
    Area of plates, A=4 cm2=4×(10−2 m)2=4×10−4 m2A = 4\text{ cm}^2 = 4 \times (10^{-2}\text{ m})^2 = 4 \times 10^{-4}\text{ m}^2
    Permittivity of free space, ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12}\text{ F/m}

  2. Use the formula for the capacitance of a parallel-plate capacitor.

    C=ϵ0AdC = \frac{\epsilon_0 A}{d}

  3. Substitute the values and calculate CC.

    C=(8.85×10−12 F/m)×(4×10−4 m2)0.8×10−3 mC = \frac{(8.85 \times 10^{-12}\text{ F/m}) \times (4 \times 10^{-4}\text{ m}^2)}{0.8 \times 10^{-3}\text{ m}} C=35.4×10−160.8×10−3 FC = \frac{35.4 \times 10^{-16}}{0.8 \times 10^{-3}}\text{ F} C=44.25×10−13 FC = 44.25 \times 10^{-13}\text{ F} C=4.425×10−12 FC = 4.425 \times 10^{-12}\text{ F}

    The capacitance of the capacitor is 4.425×10−12 F\boxed{\text{4.425} \times 10^{-12}\text{ F}}.

Part b(ii)

4.425×10−10 C\text{4.425} \times 10^{-10}\text{ C}

Calculate the charge stored in the capacitor.

  1. Use the formula relating charge, capacitance, and potential difference.

    Q=CVQ = CV

    Given potential difference, V=100 VV = 100\text{ V}
    Capacitance, C=4.425×10−12 FC = 4.425 \times 10^{-12}\text{ F} (from part b(i))

  2. Substitute the values and calculate QQ.

    Q=(4.425×10−12 F)×(100 V)Q = (4.425 \times 10^{-12}\text{ F}) \times (100\text{ V}) Q=442.5×10−12 CQ = 442.5 \times 10^{-12}\text{ C} Q=4.425×10−10 CQ = 4.425 \times 10^{-10}\text{ C}

    The charge stored in the capacitor is 4.425×10−10 C\boxed{\text{4.425} \times 10^{-10}\text{ C}}.

Part c

State two applications of a capacitor.

  • Capacitors are used for energy storage, such as in camera flashes and defibrillators, where they release a large amount of energy quickly.
  • They are used in filtering circuits to smooth out voltage fluctuations in power supplies or to block DC current while allowing AC current to pass.
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