It takes 101 J of work to move 2 C of charge from the negative plate to the positive plate of a parallel plate capacitor. What voltage difference exists between the plates

10.05volts

Explanation:

Work done to move a charge from the positive side of a plate to the negative side is represented mathematically as 1/2CV² which is also the energy stored by the capacitor.

Work done = 1/2CV² where

C is the charge

V is voltage difference that exists between the plates.

Given Work done = 101Joules

C = 2Coulombs

Substituting the values in the formula to get V, we will have;

101 = 1/2 (2) V²

101 = V²

V = √101

V = 10.05volts

Therefore the voltage difference between the plates is 10.05colt's