Question

Two identical parallel-plate capacitirs, each with capacitance 10.0 x 10^(-6)F, are charged to potential difference 50.0V and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled.
(i) Find the total energy of the system of two capacitors before the plate separation is doubled.
(ii) Find the potential difference across each capacitor after the plate separation is doubled.
(iii)Find the total energy of the system after the plate separation is doubled.

Answer 1

(i) Let $C_1=C_2=C=10\cdot10^{-6}\space F,$ $U_1=U_2=U=50\space V.$

The total energy of the system of two capacitors before the plate separation is doubled:

$W=\frac{C_1{U_1}^2}{2}+\frac{C_2{U_2}^2}{2}=CU^2=10\cdot10^{-6}\cdot50^2=25\cdot10^{-3} J.$

(ii) Because the capacitors are connected in parralel, the potential differences across each capacitor are equal:

$U_1'=U_2'=U'.$

The total charge of system before and after the plate separation is doubled remains constant:

$q_1+q_2=q_1'+q_2',$

$C_1U_1+C_2U_2=C_1'U_1'+C_2'U_2'.$

Let $C_1'=C_1$ and $C_2'=\frac{C_2}{2}$ (due to the plate separation is doubled).

Then

$C_1U_1+C_2U_2=C_1U_1'+\frac{C_2}{2}U_2',$

$2CU=\frac{3}{2}CU',$

$U'=\frac{4}{3}U=\frac{4}{3}\cdot50=66.7\space V.$

(iii) The total energy of the system after the plate separation is doubled:

$W'=\frac{(C_1'+C_2')U'^2}{2}=\frac{(C+\frac{C}{2})U'^2}{2}=\frac{(10\cdot10^{-6}+\frac{10\cdot10^{-6}}{2})66.7^2}{2}=33.3\cdot10^{-3}\space J.$

Answer: (i) $25\cdot10^{-3} J$. (ii) 66.7 V. (iii) $33.3\cdot10^{-3}\space J.$