plz shows the answers with steps as much as you can. Two parallel plates, each h

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plz shows the answers with steps as much as you can.

Two parallel plates, each having area A = 3579cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.52cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate.

1)

What is C, the capacitance of this parallel plate capacitor?

F

2)

What is Q, the charge stored on the top plate of the this capacitor?.

C

3)

A dielectric having dielectric constant = 4.1 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3579 cm2 and thickness equal to half of the separation (= 0.26 cm) . What is the charge on the top plate of this capacitor?

C

4)

What is U, the energy stored in this capacitor?

J

(Survey Question)

5)

Below is some space to write notes on this problem.

Explanation / Answer

1) C = e0*A/d = 8.85*10^-12*3579*10^-4/0.0052 = 6.09*10^-10 F

2) Q = CV = 6.09*10^-10*6 = 36.54*10^-10 C = 3.654 nC

3) Now C' =2C , So Charge Q' = 2Q = 7.308 nC on top plate

4) C on top plate =C' = 1.22 nF

   C on bottom plate = C'' = C' = 4.1*1.22 nF = 5 nF

Equivalent capacitance = C = C'C''/(C' + C'') = 0.98 nF

Energy stored = U =0.5CV^2 = 0.5*0.98*10^-9*36 = 17.64 nJ

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