Question

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.

A voltmeter reads 44.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 14.5 V . What is the dielectric constant of this material?

What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

Answer 1

let Co is the initial capacitnace and Qo is the charge on the capacitor.
Co = A*epsilon/d (A is the area of each plate and d is the distance betwen the plates)
Vo = 44 V

we know, Qo = Co*Vo

let C' is the capacitance when dielctric is fully inserted.
V' = 14.5 V

Q' = C'*V'

= k*Co*V'

But we know,

Q' = Qo

k*Co*V' = Co*Vo

k = Vo/V'

= 44/14.5

= 3.03 <<<<<<--------------Answer

when dielctric is partially inserted.

C'' = (2*A/3)*epsilon/d + k*(A/3)*epsilon/d

= A*epsilon/d*(2/3 + k/3)

= Co*(2/3 + 3.03/3)

= 1.676*Co

now use,

Q'' = Qo

C''*V'' = Vo*Co

1.676*Co*V'' = Vo*Co

V'' = Vo/1.676

= 44/1.676

= 26.2 V <<<<<<<<<<---------------------Answer