# Answer in Electric Circuits for Rashinda #159593

(i) Let “C_1=C_2=C=10cdot10^{-6}space F,” “U_1=U_2=U=50space 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=10cdot10^{-6}cdot50^2=25cdot10^{-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=66.7space 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{(10cdot10^{-6}+frac{10cdot10^{-6}}{2})66.7^2}{2}=33.3cdot10^{-3}space J.”

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