Question

An initially uncharged air-filled capacitor is connected to a 5.67-V charging source. As a result, 3.49 × 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 5.99. Find the capacitor\'s potential difference and charge after the insertion.

Please show all work! Thank you!

Answer 1

V = potential difference = 5.67 volts

Q_o = charge stored by the capacitor when without dielectric = 3.49 x 10^-5 C

C_o = Capacitance of capacitor without the dielectric

Using the formula

Q_o = C_o V

(3.49 x 10^-5 ) = C_o (5.67)

C_o = 6.2 x 10^-6 F

k = dielectric constant = 5.99

C = Capacitance with the dielectric = k C_o = 5.99 (6.2 x 10^-6 ) = 37.14 x 10^-6 F

Potential difference is due to the external charging source so it remains same

V = potential difference after insertion = 5.67 volts

New charge stored , Q = C V = (37.14 x 10^-6 )(5.67) = 0.000211 C