Question

1) An isolated capacitor with capacitance C = 4

Answer 1

a) The energy stored in a capacitor is:
U = Q^2/(2C) = (80*10^-6)^2/(2*4*10^-6) = 8 x 10^-4 J

b) Charge:
The charge on the original capacitor plates is unchanged.
So Q = 80 x 10^-6 C

c) Capacitance:
Because the charge and charge density on the capacitor plates are unchanged, the E-field intensity is also unchanged - except where the conducting material has intruded and "killed" the E-field. So now the E-field is non-zero only over 1 - 1/3 = 2/3 of its original distance, so the voltage drop is 2/3 of the orginal. Since V = Q/C, C = Q/V, and since Q is unchanged but V has changed by factor 2/3, C has changed by factor 3/2: C' = (3/2)C
So the new capacitance = (3/2) old_capacitance
= (3/2)*4*10^-6
= 6 x 10^-6 F

d) U' = Q^2/(2C') = (2/3)(Q^2/(2C)) = (2/3)*U = (2/3)*(8*10^-4) = 5.33 x 10^-4 J