Question

An initially uncharged air-filled capacitor is connected to a 2.67 V2.67 V charging source. As a result, the capacitor acquires 6.39×10−5 C6.39×10−5 C of charge. 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 6.03.6.03. Find the voltage VV across the capacitor and the charge ????fQf stored by it after the dielectric is inserted and the circuit has returned to a steady state.

Answer 1

Answer:

Explanation:

capacitance of air  capacitor C = charge / volt

C = 6.39 x 10⁻⁵ / 2.67

= 2.39 x 10⁻⁵ Farad

After inserting dielectric between plates , the capacitance of capacitor becomes C₁ where

C₁ = 6.03 x 2.39 x 10⁻⁵

= 14.41 x 10⁻⁵ Farad

The voltage of the capacitor will remain the same as earlier that is 2.67 V because it remains connected with the charging source .

New charge on the capacitor = new capacitance x voltage

= 14.41 x 10⁻⁵ x 2.67

= 38.47 x 10⁻⁵ C .