# Capacitor and work

• scholio

## Homework Statement

a 20 microfarad capacitor is charges to 50 volts. how much work would it require to add an additional charge of 0.001 coulombs to each plate?

## Homework Equations

capacitance C = Q/V where Q is charge, V is electric potential

electric potential energy U = (1/2)Q^2/C

work W = deltU = change in electric potential energy = U_f - U_i

## The Attempt at a Solution

i am not sure about the wording of this problem, when they say add 0.001 coulombs to EACH plate, do i quadruple the 0.001 charge for Q rather than only double it in the electric potential energy equation.

i tried only doubling the Q, but firstly:

solve for charge with C = 20*10^-6 farads, V = 50 volts use C = Q/V, solve for Q = 0.001 coulombs

solve for U_i , so Q = 0.001, C = 20*10^-6 --> U_i = 0.025 joules

solve for U_f, so Q = 0.002, C = 20*10^-6 --> U_f = 0.10 joules

solve for work W = U_f = U_i = 0.10 - 0.025 = 0.075 joules

did solve it correctly, did i make the correct assumption (double instead of quadruple)?

thanks