An initially uncharged air-filled capacitor is connected to a 2.69-V charging so
ID: 1550098 • Letter: A
Question
An initially uncharged air-filled capacitor is connected to a 2.69-V charging source. As a result, 7.91 times 10^-5 C of charge is transfered from one of the capacitor's plates to the other. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric Constant of this substance is 7.89. Find the Capacitor's potential difference and charge after the insertion. Potential difference after insertion of dielectric: Charge after insertion of dielectric:Explanation / Answer
Co = initial capacitance = 7.91 x 10-5 F
Vo = initial potential difference across the plates = 2.69 volts
initial charge stored on each plate is given as
Qo = Co Vo = (7.91 x 10-5) (2.69) = 0.000213 C
as the dielectric in inserted
Cf = final Capacitance = k Co = 7.89 (7.91 x 10-5) = 0.000624 F
the potential difference across the plates remains constant since the charging source has not changed.
hence Vf = final potential difference between plates = Vo = 2.69 Volts
charge stored after insertion of dielectric is given as
Qf = Cf Vf = (0.000624 ) (2.69) = 0.00168 C = 1.68 x 10-3 C