Question on: SS2 Physics - Electromagnetism
A parallel plate capacitor with a capacitance of 8 pF is connected to a 12 V battery. What is the energy stored in the capacitor?
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Given: C = 8 pF = 8 × 10(-12) F, V = 12 V
The energy stored in the capacitor is given by the formula: E = 1/2CV2
E = 1/2 × (8 × 10(-12) F) × (12 V)2
E = 1/2 × 8 × 10(-12) F × 144 V2
E = 576 × 10(-12) F × V2
E = 576 × 10(-12) J
Therefore, the energy stored in the capacitor is 576 × 10(-12) J.
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