Two parallel plates, each having area A 3605cm2 are connected to the terminals o

Question

Two parallel plates, each having area A 3605cm2 are connected to the terminals of a battery of voltage Vb-6 v as shown. The plates are separated by a distance d- 0.42cm. 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? You've reached the maximum number of submissions (3) for this question. 2) What is Q, the charge stored on the top plate of the this capacitor? C Submit You currently have O submissions for this question. Only 3 submission are allowed You can make 3 more submissions for this question. 3) A dielectric having dielectric constant K 3.4 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3605 cm2 and thickness equal to half of the sepation0.21 cm) What is the charge on the top plate of this capacitor? d/2 C Submit You currently have O submissions for this question. Only 3 submission are allowed. You can make 3 more submissions for this question. 4) What is U, the energy stored in this capacitor? J Submit You currently have 0 submissions for this question. Only 3 submission are altowed. You can make 3 more submissions for this question. 5) The battery is now disconnected from the capacitor and then the dielectric is withdrawn. What is V, the voltage across the capacitor? V Submit You currently have O submissions for this question. Only 3 submission are allowed. You can make 3 more submissions for this question.

Explanation / Answer

1) C = A*epsilon/d

= 3605*10^-4*8.854*10^-12/(0.42*10^-2)

= 7.60*10^-10 F

= 7.6*10^-4 micro F

2) Q = C*Vb

= 7.6*10^-10*6

= 4.56*10^-9 C

= 4.56*10^-3 micro C

3) C' = k*A*epsilon/( (d-t)*k + t )

= 3.4*3605*10^-4*8.854*10^-12/( (0.0042 - 0.0021)3.4 + 0.0021)

= 1.17*10^-9 F

Q' = C'*V

= 1.17*10^-9*6

= 7.02*10^-9 C

= 7.02*10^-3 micro C

4) U' = (1/2)*C'*V^2

= (1/2)*1.17*10^-9*6^2

= 2.10*10^-8 J

5) V = Q'/C

= 7.02*10^-9/ (7.60*10^-10)

= 9.24 V

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