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...

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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**