A 1.63 nF parallel-plate capacitor is charged to an initial potential difference ΔVi = 100 V and then isolated. The dielectric material between the plates is mica, with a dielectric constant of 5.00.

Question

(a) How much work is required to withdraw the mica sheet?

Work = .5CV^2
C= k*E_0*A/d

But how can I solve for capacitance if I don't have area? And even if I find Capacitance how will I find work if I don't have volts?
THank You

Answer 1

When the dielctric material is withdrawn, the capacitance will decrease by a factor of 5, since it is proportional to the dielectric constant. Since C = Q/V and the charge Q remains the same, V will increase by a factor of 5.
The stored capacitor energy, (1/2) CV^2, will increase by a factor (1/5)*5^2 = 5

Initial stored energy
E1 = (1/2)*1.63*10^-9*100^2 = 8.15*10^-6 J
After withdrawing the mica,
E2 = 4.08*10^-4 J

The difference is the work required to remove the mica.

Answer 2

I understand now, thank you.