# A dielectric-filled parallel-plate capacitor has plate area

## Homework Statement

A dielectric-filled parallel-plate capacitor has plate area A = 15.0 cm2 , plate separation d = 9.00 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10^−12 C2/N⋅m2 .

U1 = 6.64×10^−10 J
U2 = 4.15×10−10J

A)The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

B)In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?

## The Attempt at a Solution it says that both my w and U_3 are incorrect so i tried it again by doing it differently such as:

Q=C2*V=3.16*10^-12*15 =4.74*10^-11 C

C3 =eoA/d =(8.85*10^-12)(15*10^-4)/(9*10^-3)

C3=1.475*10^-12 F

U3 =(1/2)(Q^2/C3) =(1/2)(4.74*10^-11)^2/(1.475*10^-12)

U3=7.616*10^-10 J

(but it still continues to tell me i am incorrect) so what am i doing wrong? what is the correct answer?
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and same goes for W i re did it and did:

W=U3-U2 =7.616-10^-10-4.15*10^-10

W=3.466*10^-10 J

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TSny
Homework Helper
Gold Member
Hello, and welcome to PF!

.... i tried it again by doing it differently such as:

Q=C2*V=3.16*10^-12*15 =4.74*10^-11 C

C3 =eoA/d =(8.85*10^-12)(15*10^-4)/(9*10^-3)

C3=1.475*10^-12 F
Shouldn't C2 be the total capacitance, Ceq, when the dielectric is half-way out? The value you have here does not look correct.

Your value of C3 looks good.