The energy stored in a parallel plate capacitor is 24 J. What is the potential difference be if the capacitance of the capacitor is 3 µF?
Correct answer: A. 4 KV
- A. 4 KV
- B. 16 kV
- C. 54 kV
- D. 8 kV
Explanation
The expression relating the energy stored in a capacitor with respect to potential difference and capacitance is : E = ½ CV² 24 = ½ (3×10^-6) V² V² = 16 × 10⁶ V = 4 × 10³ V = 4 kV
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About Capacitors
Capacitors store electric charge and energy in an electric field between conductors. Work includes capacitance, the relation Q = CV, dielectric materials, charging and discharging, energy formulas, and combinations of capacitors in series and parallel. Electric potential difference is needed to understand why charge and stored energy change.
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