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A capacitor having capacitance of 5μF is charged to a potential difference of 10,000V. The energy stored in the capacitor is 250J. Find Q.

Physics

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Answer

250,000 J\text{250,000 J}

Step-by-step solution

4 steps

  1. Identify the given values.

    The capacitance of the capacitor is C=5 mFC = 5 \text{ mF}.
    The potential difference across the capacitor is V=10,000 VV = 10,000 \text{ V}.

  2. Convert the capacitance to Farads.

    C=5 mF=5×10−3 FC = 5 \text{ mF} = 5 \times 10^{-3} \text{ F}

  3. Use the formula for the energy stored in a capacitor.

    The energy EE stored in a capacitor is given by the formula:

    E=12CV2E = \frac{1}{2} C V^2

  4. Substitute the values and calculate the energy.

    E=12(5×10−3 F)(10,000 V)2E = \frac{1}{2} (5 \times 10^{-3} \text{ F}) (10,000 \text{ V})^2 E=12(5×10−3)(104)2E = \frac{1}{2} (5 \times 10^{-3}) (10^4)^2 E=12(5×10−3)(108)E = \frac{1}{2} (5 \times 10^{-3}) (10^8) E=12(5×108−3)E = \frac{1}{2} (5 \times 10^{8-3}) E=12(5×105)E = \frac{1}{2} (5 \times 10^5) E=2.5×105 JE = 2.5 \times 10^5 \text{ J} E=250,000 JE = 250,000 \text{ J}

    This can also be expressed as 250 kJ250 \text{ kJ}.

    The energy stored in the capacitor is 250,000 J\text{250,000 J}.

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