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What Capacitance is required to store an energy of 10 kWh at a potential difference of 1000 volt.

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3 parts
1

72 F\text{72 F}

2

9.96×10−10 J\text{9.96} \times 10^{-10} \text{ J}

3

5.03 F\text{5.03 F}

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Part 1· 3 steps

What Capacitance is required to store an energy of 10 kWh at a potential difference of 1000 volt.

  1. Convert the energy from kilowatt-hours (kWh) to Joules (J).

    1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J} U=10 kWh×3.6×106 J/kWh=3.6×107 JU = 10 \text{ kWh} \times 3.6 \times 10^6 \text{ J/kWh} = 3.6 \times 10^7 \text{ J}

  2. Use the formula for energy stored in a capacitor, U=12CV2U = \frac{1}{2}CV^2, and rearrange it to solve for capacitance CC.

    C=2UV2C = \frac{2U}{V^2}

  3. Substitute the given values into the formula and calculate CC.

    C=2×3.6×107 J(1000 V)2C = \frac{2 \times 3.6 \times 10^7 \text{ J}}{(1000 \text{ V})^2} C=7.2×107 J1×106 V2C = \frac{7.2 \times 10^7 \text{ J}}{1 \times 10^6 \text{ V}^2} C=72 FC = 72 \text{ F}

    The required capacitance is 72 F\boxed{\text{72 F}}.

Part 2

9.96×10−10 J\text{9.96} \times 10^{-10} \text{ J}

How much energy is store in 1m³ (cubic meter) of air do to the fair weather electric field of magnitude 15 volt per meter.

  1. Identify the given values and the permittivity of free space (ϵ0\epsilon_0).

    Given: Volume Vol=1 m3Vol = 1 \text{ m}^3, Electric field E=15 V/mE = 15 \text{ V/m}.
    The permittivity of free space (which is approximately the permittivity of air) is ϵ0=8.854×10−12 F/m\epsilon_0 = 8.854 \times 10^{-12} \text{ F/m}.

  2. Calculate the energy density (uu) using the formula u=12ϵ0E2u = \frac{1}{2}\epsilon_0 E^2.

    u=12×(8.854×10−12 F/m)×(15 V/m)2u = \frac{1}{2} \times (8.854 \times 10^{-12} \text{ F/m}) \times (15 \text{ V/m})^2 u=12×8.854×10−12 F/m×225 V2/m2u = \frac{1}{2} \times 8.854 \times 10^{-12} \text{ F/m} \times 225 \text{ V}^2/\text{m}^2 u=9.96075×10−10 J/m3u = 9.96075 \times 10^{-10} \text{ J/m}^3

  3. Calculate the total energy stored (UU) by multiplying the energy density by the volume.

    U=u×VolU = u \times Vol U=(9.96075×10−10 J/m3)×(1 m3)U = (9.96075 \times 10^{-10} \text{ J/m}^3) \times (1 \text{ m}^3) U=9.96075×10−10 JU = 9.96075 \times 10^{-10} \text{ J}

    The energy stored is 9.96×10−10 J\boxed{\text{9.96} \times 10^{-10} \text{ J}}.

Part 3

5.03 F\text{5.03 F}

Calculate the Capacitance of a Capacitors whose Area 1m² and distance between the plate is 8.8 x 10⁻¹² m dielectric between the plate is the factors of 5.

  1. Identify the given values and the permittivity of free space (ϵ0\epsilon_0).

    Given: Area A=1 m2A = 1 \text{ m}^2, Distance d=8.8×10−12 md = 8.8 \times 10^{-12} \text{ m}, Dielectric constant κ=5\kappa = 5.
    The permittivity of free space is ϵ0=8.854×10−12 F/m\epsilon_0 = 8.854 \times 10^{-12} \text{ F/m}.

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

    C=κϵ0AdC = \frac{\kappa \epsilon_0 A}{d}.

  3. Substitute the values into the formula and calculate CC.

    C=5×(8.854×10−12 F/m)×(1 m2)8.8×10−12 mC = \frac{5 \times (8.854 \times 10^{-12} \text{ F/m}) \times (1 \text{ m}^2)}{8.8 \times 10^{-12} \text{ m}} C=44.27×10−12 F8.8×10−12C = \frac{44.27 \times 10^{-12} \text{ F}}{8.8 \times 10^{-12}} C=5.03068 FC = 5.03068 \text{ F}

    The capacitance is 5.03 F\boxed{\text{5.03 F}}.

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