Question

a capacitor of 10 mew is charged to a potential V TV with a battery e the battery is is now disconnected and an additional charge 200 is give to the positive plate of capacitor new potential difference across the capacitor will be



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Answer 1

C = 10 × 10⁻⁶F

V = 50V

Q = CV = 500 × 10⁻⁶C

When an additional charge 200μC is given to the positive plate, it distributes on both sides of the plate equally. So, the charge on the inner side of each plate will be 100μC and -100μC.

i.e. charge on the capacitor will be 500μC + 100μC = 600μC

p.d. across the capacitor will be $\frac{600\times10^-^6}{10\times10^-^6}$ = 60V