Two parallel plates, each having area A = 3574cm 2 are connected to the terminal

Question

Two parallel plates, each having area A = 3574cm2 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?F

2)

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

3)

A dielectric having dielectric constant = 3.4 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3574 cm2 and thickness equal to half of the separation (= 0.21 cm) . What is the charge on the top plate of this capacitor?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The battery is now disconnected from the capacitor and then the dielectric is withdrawn.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?

Explanation / Answer

part 1 )

C = eo*A/d

A =3574 cm^2 = 3574 x 10^-4 m^2

d = 0.42 cm = 0.42 x 10^-2 m

C = 7.53 x 10^-10 F

part 2 )

Q = CV

v = 6V

Q = 4.52 x 10^-9 C = 4518.56 x 10^-6 = 4518 muC

part 3 )

such arrangement can be consider a series combination

C1 = eo*A/d

d = 0.21 cm

C1 = 1.506 x 10^-9 F

C2 = k*C1 = 5.12 x 10^-9 F

Ceq = 1.1637 x 10^-9 F

Q is remain same in series

Q = CeqV = 6.9822 x 10^-9 C = 6982 muC

part 4 )

U = Q^2/2Ceq

U = 2.095 x 10^-8 J

part 5 )

V = Q/C

charge remain constant

Q = 6.9822 x 10^-9 C

C = 7.53 x 10^-10 F

V = Q/C = 9.27 V

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