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## Homework Statement

A 4.0-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 120 V and is then disconnected. (The initial capacitance including the dielectric is 4.0 nF.)

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

(ii) What is the potential difference between the plates of the capacitor once the Mylar is completely removed?

## Homework Equations

Q=CV

C=kA(Epsilon_0)/d

-W=Uf-Ui

U = (Q^2)/2C = C(V^2)/2 = QV/2

## The Attempt at a Solution

Capacitance w/ dielectric = 4.0 nF

Capacitance w/o dielectric = (4.0 nF)/κ = (4.0 nF)/3.1

Potential Energy w/ dielectric = [(4.0 nF)(120 V)^2]/2 = 28,880 nJ

And...I don't know how to calculate the Potential Energy w/o the dielectric. Can someone point me to the right direction?

Thanks in advance!