Two parallel plates, each having area A = 3755cm 2 are connected to the terminal

Question

Two parallel plates, each having area A = 3755cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.33cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate.

3.)

A dielectric having dielectric constant = 3 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3755 cm2 and thickness equal to half of the separation (= 0.165 cm) . What is the charge on the top plate of this capacitor?

4.)What is U, the energy stored in this capacitor

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The battery is now disconnected from the capacitor and then the dielectric is withdrawn.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

3)
A=3755 cm^2 = 0.3755 m^2
d = 0.33 cm = 0.0033 m
It can be simplified as 2 capacitor connected in series.

1st capacitor is with dielectric and thickness of d/2
C1 = K*ebsoleneo*A/(d/2)
= 3*(8.854*10^-12)*(0.3755 m^2) / (0.0033/2)
=6*10^-9 F

2nd capacitor is withot any dielectric and thickness of d/2
C2 = ebsoleneo*A/(d/2)
= (8.854*10^-12)*(0.3755 m^2) / (0.0033/2)
=2*10^-9 F

Net capacitance,C = C1*C2 / (C1+C2)
=(6*10^-9 F)*(2*10^-9 F) /(6*10^-9 F + 2*10^-9 F)
= 1.5*10^-9 F

charge on top plate will be positive which is equal to = C*V
=1.5*10^-9 F * 6V
= 9*10^-9 C

4)
Energy stored, U= 0.5 * C* V^2
= 0.5*(9*10^-9)*(6)^2
=1.62*10^-7 J
5)
Now if battery is disconnected and dielectric is removed,
C = ebsoleneo*A/(d)
= (8.854*10^-12)*(0.3755 m^2) / (0.0033)
=1*10^-9 F
Q will not change, Q = 9*10^-9 C

V = Q/C
= 9*10^-9 C /1*10^-9 F
= 9 V

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