You have a parallelplate 677 106 F capacitor that is charge

You have a parallel-plate 6.77 × 10-6 F capacitor that is charged to 0.00833 C. While the capacitor is isolated, you change the plate separation so that the capacitance becomes 2.33 × 10-6 F. How much work do you perform in this process?

J?

Solution

Here ,

C1 = 6.77 *10^-6 F

C2 = 2.33 *10^-6 C

Q = 0.00833 C

work done in process = 0.5 * Q^2 * (1/C2 - 1/C1)

work done in process = 0.5 * 0.00833^2 * (1/(2.33 *10^-6 ) - 1/(6.77 *10^-6))

work done in process = 9.77 J

the work done in process is 9.77 J