1.

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 ?

Answer»

100 V
50 V
25 V
75 V

Solution :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 `


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