Question

The oil dielectric to be used in a parallel plate capacitor has a

relative permittivity of 2.3 and the maximum working potential

gradient in the oil is not to exceed 104 V/m. Calculate the

Approximate plate area required for a capacitance of 0.0003μF,the

Maximum working voltage being 10,000 volts.

Answer 1

On connecting the two given capacitors, let the final voltage be V.

If capacity of capacitor without the dielectric is C, then the charge on this capacitor is q1​=CV

The other capacitor with dielectric has capacity εC.

Therefore, charge on it is q2​=εCV

As ε=αV, therefore q2​=ε=αV

The initial charge on the capacitor (Without dielectric)

that was charged is q0​=CV0​

From the conservation of charge, q0​=q1​+q2​