An initially uncharged air-filled capacitor is connected to a 3.29-V charging so

Question

An initially uncharged air-filled capacitor is connected to a 3.29-V charging source. As a result, 5.45 x 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 of this substance is 7.15. Find the capacitor's potential difference and charge after the insertion. Potential difference after insertion of dielectric: Charge after insertion of dielectric:

Explanation / Answer

Let's first find the initial capacitance

Q =CV

C = Q/V = 5.45E-5/3.29 = 16.59uF

After insertion of the dielectric the capacitor will become

C' = kC = 7.15 * 16.59 = 118.44 uF

Potential difference will be same as the battery is connected

Charge will increas on the capacitor

Q' = C'V = 118.44*3.29 = 38.9 *10-5 C

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