The capacitance of a capildtor is 10 `muF `. The potential difference on it is 50 V. If the distance between its platsis halved, what will he the potential difference now ?

Options

A. 100 V
B. 50 V
C. 25 V
D. 75 V

Answer

B

Explanation

V = Ed but E `= (sigma)/(epsilon_(0))`
`:. V = (sigma)/(epsilon_(0)).d `where `(sigma)/(epsilon_(0)) ` is constant
`:. V prop d`
`:. (V)/(V) = (d)/(d)`
`:. (V)/(50)= (d)/(2d) :. V = (50)/(2) = 25 V `