When an additional charge of 1 μ C is given to a capacitor, its potential rises by 1mV. Then the capacitance of the capacitor in μ F is
0.001
1
100
1000
Q=cv dQ=cdv
=10−6=c10−3⇒c=10−3F
A capacitor of capacitance 10μF is charged to a potential 50 V with a battery. The battery is now disconnected and an additional charge 200 μC is given to the positive plate of the capacitor. The potential difference across the capacitor will be :