The potential difference across the plates of a capacitor is 200 V and the charge stored is 4 µC. The capacitance is:a) 20 µFb) 2 µFc) 0.02 µFd) 0.2 µF
Question
The potential difference across the plates of a capacitor is 200 V and the charge stored is 4 µC. The capacitance is:a) 20 µFb) 2 µFc) 0.02 µFd) 0.2 µF
Solution
The capacitance (C) of a capacitor can be calculated using the formula:
C = Q/V
where: Q is the charge stored in the capacitor, and V is the potential difference across the capacitor.
Given in the problem, Q = 4 µC = 4 x 10^-6 C (since 1 µC = 10^-6 C) and V = 200 V.
Substituting these values into the formula, we get:
C = (4 x 10^-6 C) / (200 V) = 2 x 10^-8 F
Since 1 F = 10^6 µF, we can convert the capacitance to microfarads (µF):
C = 2 x 10^-8 F x 10^6 µF/F = 0.02 µF
So, the capacitance of the capacitor is 0.02 µF, which corresponds to option c).
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