Two parallel plates, each having area A = 2522cm 2 are connected to the terminal

Question

Two parallel plates, each having area A = 2522cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.47cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate.

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

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

C)

A dielectric having dielectric constant = 2.7 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 2522 cm2 and thickness equal to half of the separation (= 0.235 cm) . What is the charge on the top plate of this capacitor?_____C

D) What is U, the energy stored in this capacitor?______J

E)

The battery is now disconnected from the capacitor and then the dielectric is withdrawn.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)The capacitance of a capacitor is given by C = eo*A/d (where A is the area and d is the separation between the plates)

so,

C=8.85*10^-12*2522*10^-4/(0.0047)

=4.75*10^-10 F

=4.75*10^-4 uF

2)Q=CV

=4.75*10^-4*6 uC

=28.5*10^-4 uC

3)C = eo*A/(d-t+t/K)

where t is the thickness of the dielectric and K is the dielectric constant.

so,

C = 8.85*10^-12*2522*10^-4/(0.0047 - 0.00235 + 0.00235/2.7)

=6.93*10^-10 F

=6.93*10^-4 uF

so charge = CV

=6.93*10^-4*6

=41.58 *10^-4 uC

4)Energy = 0.5CV^2

=0.5*6.93*10^-4*6^2

=1.247*10^-8 J

5)The charge on the capacitor does not change. So,

V = Q/C

=41.58/4.75

=8.75 V

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