A parallel plate capacitor of capacitance 100 μF is charged to 500 V.

1.	A parallel plate capacitor of capacitance 100 μF is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes(a) 250 V (b) 500 V (c) 1000V (d) 2000 V
Answer» (a) 250 V Here, C’ = 2C, since the charge remains the same. q = C’V’ = CV ⇒ V = \(\frac {CV}{2C}\,= \frac{500}{2}\) = 250V

A parallel plate capacitor of capacitance 100 μF is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes(a) 250 V (b) 500 V (c) 1000V (d) 2000 V

Answer»

(a) 250 V

Here, C’ = 2C, since the charge remains the same.

q = C’V’ = CV ⇒ V = \(\frac {CV}{2C}\,= \frac{500}{2}\) = 250V

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