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.
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