Calculate maximum values of voltage, charge and force for the following capacitor.A parallel plate capacitor with capacitance C = 45*10^6 F has a...

We have a parallel plate capacitor with capacitance C =
45*10^-6 F and the plates are 1 mm apart. There is paper between the plates. The paper
has a Edis of 1.5*10^7 V/m. The maximum voltage field that the paper can withstand is
1.5*10^7 V/m. As the distance between the plates is just 1 mm, the maximum voltage that
can be applied between them is 1/1000 of 1.5*10^7

=>
Vmax = 1.50*10^4 V

We know that Q =
C*V

As C = 45*10^-6 F

Qmax =
C*Vmax

=> Qmax = 45*10^-6*1.50*10^4
C

=> Qmax = 67.50*10^-2
C

The electric field can be expressed in terms of N/C,
Qmax= 67.50*10^-2 C

Fmax =
1.5*10^7*67.5*10^-2

=> 10.12*10^5
N

=> 1.01
MN

The required values are Vmax = 1.50*10^4
V, Qmax = 67.50*10^-2 C and Fmax = 1.01 MN