A parallel plate capacitor with a capacitance of 2 µF is connected to a 100 V battery. If the separation between the plates is doubled, the capacitance is:a) 1 µFb) 4 µFc) 0.5 µFd) 2 µF
Question
A parallel plate capacitor with a capacitance of 2 µF is connected to a 100 V battery. If the separation between the plates is doubled, the capacitance is:a) 1 µFb) 4 µFc) 0.5 µFd) 2 µF
Solution
The capacitance (C) of a parallel plate capacitor is given by the formula:
C = ε * (A/d)
where: ε is the permittivity of the material between the plates, A is the area of one of the plates, and d is the separation between the plates.
From this formula, we can see that the capacitance is inversely proportional to the separation between the plates. This means that if the separation (d) is doubled, the capacitance will be halved.
Therefore, if the original capacitance was 2 µF, when the separation is doubled, the new capacitance will be 1 µF.
So, the answer is a) 1 µF.
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