Question

An initially uncharged air-filled capacitor is connected to a 3.33 V charging source. As a result, 8.13×10−5 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 6.27.

Find the voltage V across the capacitor and the charge Qf stored by it after the dielectric is inserted and the circuit has returned to a steady state.

Answer 1

initially capacitance

C = q/v = 8.13 / 3.33 = 2.441 * 10^-5 F

capcitance after inserting dielectric

C = k Co = 6.27* 2.441* 10^-5

C = 15.308* 10^-5 F

=====

potential difference even after insertion remains constant = 3.33 V

=====

charge

Q = CV = 15.308* 3.33 = 50.975* 10^-5 C

======

Comment before rate in case any doubt, will reply for sure... goodluck