Question

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.20 and a dielectric strength of 1.25

Answer 1

We know that

                   E =V/d then separation between the plates is given by d =V/E =5400/1.25*10⁷ =4320*10-7m=0.432mm

Now the capacitance of the capacitor with dielectric is given by C =keoA/d

Then area between the paltes is A =C*d /kEo =(1.50*10^-9)(0.432*10^-3)/(3.20)(8.85*10^-12) =0.0228m²

Answer 2

12.5 MV/ m (12.5 kV/mm) and 5400 volts means:
5400 / 12.5 kV/mm =432 mm separation

Parallel plate cap
C = ???r(A/d)
?? is 8.85e-12 F/m
?r is dielectric constant (vacuum = 1)
A and d are area of plate in m