Question

A dielectric-filled parallel-plate capacitor has plate area
*A* = 30.0 cm2 , plate separation *d* = 8.00 mm and
dielectric constant *k* = 4.00. The capacitor is connected
to a battery that creates a constant voltage *V* = 5.00 V .
Throughout the problem, use *ϵ*0 = 8.85×10^{−12}
C2/N⋅m2 .

Part A- Find the energy *U*1 of the dielectric-filled
capacitor.

Part B- The dielectric plate is now slowly pulled out of the
capacitor, which remains connected to the battery. Find the energy
*U*2 of the capacitor at the moment when the capacitor is
half-filled with the dielectric.

Part C- 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,
*U*3.

Part D- 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?

Answer #1

Area A = 30.0 cm^2 = 0.003 m^2, d = 8.0 mm = 0.008 m

Part - A -

C = k * εo A /d = (4x 8.85x10^-12x 0.003) / 0.008 = 1.33 x 10^-11
F

U1 = Energy stored = 0.5CV^2 = 0.5x1.33x10^-11x5^2 = 1.662 x 10^-10
J

Part -B

U2 = Energy stored = [1+k] /2k* U1 = [(1+4) / (2x4)]x1.662 x 10^-10 = 1.04 x 10^-10 J

And, charge residing in this new capacitor =2* U2/ V =
2*1.04x10^-10 /5

Q= 4.16 x 10^-11 couloumb.

Part -C

After fully removing the dielectric,

Now the capacitance C /k = (1.33 x 10^-11)/4= 3.32 x 10^-12 F

U3 = Energy stored = Q^2 / (2 x3.32x10^-12) = (4.16x10^-11)^2 / (2
x3.32x10^-12) = 2.61 x 10^-10 J

Part -D

U3-U2 = 2.61x10^-10 - 1.04 x 10^-10 = 1.57 x 10^-10 J

So, work done by the external agent = 1.57 x 10^-10 J.

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