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## Homework Statement

A 6.8-nF parallel plate capacitor with a sheet of Mylar (κ = 3.1) filling the space between the plates is charged to a potential difference of 160 V and is then disconnected. (The initial capacitance including the dielectric is 6.8 nF.)

(a) How much work is required to completely remove the sheet of Mylar from the space between the two plates?

## Homework Equations

Uo = (.5)*Qo^2/Co

U = Uo/k

Work = U - Uo

## The Attempt at a Solution

I solve for Qo using Co = Qo/Vo

Co*Vo = Qo

Uo = (.5) * (Co*Vo)^2/Co

this simplifies to

Uo = (.5) * Vo^2 *Co

Plugging in the values I get

Uo = (.5) * 160^2 * 6.8*10^-9

Uo = 8.7*10^-5

U = 2.81*10^-5

Work = (+ or -)5.896*10^-5 J

My answer is wrong when I enter it in the system. I'm not sure what to do from here.