Question

A parallel plate capacitance of 4uF is
having a charge of 0.5 C. what will be
its capacity if the charge is increased
to 1 coulomb?​

Answer 1

Answer:

Using the formula,

Q=CV

u can find the value

initially u have capacitance as 4 micro farad which is equal to 10^-4 farad.

therefore 0.5=10^-4 * V

therefore v=5000 volts

taking voltage constant ,

when the charge will be 1 C

the capacitance= charge/voltage

which is 1/5000

= 0.0002 = 2 × 10^-4 F which is 2uF

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