Two parallel plates, each having area A = 2599cm 2 are connected to the terminal

Question

Two parallel plates, each having area A = 2599cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.37cm. 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?

0.00062 F

2)

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

0.00373 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 = 2599 cm2 and thickness equal to half of the separation (= 0.185 cm) . What is the charge on the top plate of this capacitor?

0.095 C

4)

What is U, the energy stored in this capacitor?

______ J

5)

The battery is now disconnected from the capacitor and then the dielectric is withdrawn. What is V, the voltage across the capacitor?

_______ V

Explanation / Answer

1) C = A*epsilon/d

= 2599*10^-4*8.854*10^-12/(0.37*10^-2)

= 6.2*10^-10 F

2) Q = C*V

= 6.2*10^-10*6

= 3.72*10^-9 C

3) Capacitance when dielectric introduced,

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

= 4.1*2599*10^-4*8.854*10^-12/( (0.37*10^-2 - 0.185*10^-2)*4.1 + 0.185*10^-2)

= 9.9*10^-10 F

Q' = C'*V

= 9.9*10^-10*6

= 5.9*10^-9 C or 0.059 micro C

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

= (1/2)*9.9*10^-10*6^2

= 1.78*10^-8 J

5) V' = 4.1*V

= 4.1*6

= 24.6 V

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