Question

A dielectric-filled parallel-plate capacitor has plate area A = 20.0cm2 , plate separation d = 8.00mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0V . Throughout the problem, use ?0 = 8.85

Answer 1

B. The capacitance of the dielectric half of the capacitor C = 0.5*area*e0*k/d.

The additional capacitance of the non-dielectric half ?C = 0.5*area*e0/d.

Add ?C to C to obtain total capacitance C1.

Then repeat the U formula with C1, call the result U1.

(A shortcut for this step is to multiply U from step A by (k+1)/(2k). You should be able to prove this.)

C. For this you have to think about conservation of charge, since the battery is disconnected.

First compute the Q in C1 of step B. (Q = VC.)

The new capacitance C2 = 2?C.

Then the energy U2 = Q^2/(2C2).

D. Work = change in stored energy = U2-U1.

......................................

C = k * ?o A /d = 3* 8.85e-12* 0.002 / 0.008 = 6.63e-12 F

U1 = Energy stored = 0.5CV^2 = 0.5 * 6.63e-12 * 15^2 = 7.458e-10 J
U1 = 7.458e-10 J
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b)
U2 = Energy stored = [1+k] /2k* U1 = 0.625*7.458e-10
U2= 4.66e-10 J
Charge residing in this new capacitor =2* U2/ V = 2*4.66e-10 /15
Q= 6.21e-11 couloumb.
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c)
After fully removing the dieelectric,
Now the capacitance C /k = 6.63e-12/3= 2.21e-12 F
U3 = Energy stored = Q^2 / (2 *2.21e-12) = 8.71e-10 J
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d)
U3-U2 = 8.71e-10 -4.66e-10 = 4.05e-10 J