Question

An initially uncharged air-filled capacitor is connected to a 3.49 3.49 V charging source. As a result, 8.15× 10 −5 8.15×10−5 C of charge is transferred 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 ? κ of this substance is 4.49 4.49 . Find the voltage ? V across the capacitor and the charge ? f Qf stored by it after the dielectric is inserted and the circuit has returned to a steady state

Answer 1

Voltage across the capacitor will not change after inserting sheet of dielectric material .

so V = 3.49 V

New charge,

Q = kQo

Q = 4.49 * 8.15*10^(-5)

Q = 36.59*10^(-5) C