A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2, plate separation d = 5.00mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 10.0V . Throughout the problem, use Eo = 8.85×10−12 C2/nxm2.

**Answer 1**

a)

C = k * εo A /d = 4* 8.85e-12* 0.003 / 0.005 = 2.124e-11 F

U1 = Energy stored = 0.5CV^2 = 0.5*2.124e-11*10^2 = 1.062e-9 J

U1 = 1.062e-9 J

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b)

U2 = Energy stored = [1+k] /2k* U1 = 0.625*1.062e-9

U2= 6.6375e-10 J

Charge residing in this new capacitor =2* U2/ V = 2*6.6375e-10 /10

Q= 1.3275e-10 couloumb.

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c)

After fully removing the dieelectric,

Now the capacitance C /k = 2.124e-11/4= 5.31e-12 F

U3 = Energy stored = Q^2 / (2 *5.31e-12) = 1.659375e-9 J

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d)

U3-U2 = 1.659375e-9 – 6.6375e-10 = -9.95625e-10 J

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**Answer 2**

Dielectric Constant K