A parallel plate capacitor of capacitance 100 `mu` F Is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes

Options

A. 250 V
B. 500 V
C. 1000 V
D. 2000 V

Answer

B

Explanation

Here, ` " " C. = 2C`, since the charge remains the same ,
`q = C.V. = CV rArr V. = (CV)/(2C) = (500)/(2) = 250 `C