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## Homework Statement

how much work is done in increasing the charge on a 3 microfarad capacitor from 300 microcoulombs to 600 microcoulombs?

i am supposed to get work W = 0.045 joules

## Homework Equations

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

change in electric potential V = change in potential energy U/ q where q is charge

work W = change in potential energy U

## The Attempt at a Solution

first i calculated the electric potential at 300 microcoulombs using the first eq:

C = Q/V --> V = QC = (300*10^-6)(3*10^-6) = 9*10^-10 volts

then calculated the electric potential at 600 microcoulombs using same eq:

C = Q/V --> V = QC = (600*10^-6)(3*10^-6) = 1.8*10^-9 volts

since change in electric potential V = change in potential energy U/q ---> change in V = (1.8*10^-9 - 9*10^-10) = 9*10^-10 volts and since ---> change in U = change in V * q ---> i let change in V = 9*10^-10 and q = 3*10^-6 to get:

change in U = work W = (9*10^-10)(3*10^-6) = 2.7*10^-15 joules

i know i did something wrong because i am supposed to get work W = 0.045 joules

what did i do wrong?

thanks