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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?
Homework Equations
The Attempt at a Solution
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