Question

A parallel-plate capacitor has capacitance C = 15.7 pF when the volume between the plates is filled with air. The plates are circular, with radius 2.50 cm. The capacitor is connected to a battery and a charge of magnitude 28.0 pC goes into each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 pC.

(a) What is the dielectric constant K of the dielectric?

(b) What is the potential difference between the plates before and after the dielectric has been inserted?

(c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

Answer 1

Capacitance =Co = 1.57*10^-11 F

Area = A = (pi)r²

= 0.0019625 m²

As the potential difference between the plates 'V' remains same,the increase in capacity due to insertion of dielectric results in increase in charge

The dielectric constant =45/28
(A) The dielectric constant =1.607


(b)Potential difference between the plates before the dielectric has been inserted = Vo =qo/co

=28pC/15.7pF

=1.783V

Potential difference between the plates before the dielectric has been inserted = Vo = 1.783 V
Potential difference between the plates after the dielectric has been inserted remains same as battery remains connected

Potential difference between the plates after the dielectric has been inserted = 1.783 V

(c)Electric field at a point midway between the plates before the dielectric has been inserted = Eo=qo/Aeo

Eo =25*10^-12 /0.0019625 * 8.85*10^-12

Electric field at a point midway between the plates before the dielectric has been inserted = Eo=1439.42 N/C

The electric field at a point midway between the plates after the dielectric has been inserted= Ed=Eo / dielectric constant

Ed=1439.42/1.607

=895.72 N/C