An initially uncharged air-filled capacitor is connected to a 5.21-V charging so

Question

An initially uncharged air-filled capacitor is connected to a 5.21-V charging source. As a result, 9.61 × 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 2.85. Find the capacitor's potential difference and charge after the insertion.

Potential difference after insertionof dielectric:

______ V

Explanation / Answer

using Q = CV

9.61 x 10^-5 = C x 5.21

C = 1.84 x 10^-5 F

so capacitance of capcitor is 1.84 x 10^-5 F.

if we insert dielectric material between plates this increases the capacitance of
capacitor.

so nEw capacitance C' = k C = 2.85 x 1.84 x 10^-5 = 5.26 x 10^-5 F

capcitor is still connected to source.

so V will be 5.21 V.

Potential diff = 5.21 V ........Ans

charge Q = C'V = 5.26 x 10^-5 x 5.21 = 27.39 x 10^-5 C ...Ans

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