Finding the work done to remove a dielectric slab.

[Cisneros778]

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.

 

[gneill]

You may want to check your formula U = Uo/k. When the dielectric is pulled out, work will be done. Energy stored in the dielectric is going to end up back in the capacitor.

If C_o is the initial capacitance, then the new capacitance without the dielectric will be C_o/k. If the initial charge is Q = V_oC_o, then after the dielectric is removed the voltage will rise to V = kV_o. If you plug the expressions for the new capacitance and new voltage into the energy expression you should find that E = k(1/2)C_oV_o².