An initially uncharged air-filled capacitor is connected to a 5.67-V charging source. As a result, 3.49 * 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 5.99. Find the capacitor/'s potential difference and charge after the insertion.

Question

Gwendolyn Mueller

Physics

24 October, 06:01

An initially uncharged air-filled capacitor is connected to a 5.67-V charging source. As a result, 3.49 * 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 5.99. Find the capacitor/'s potential difference and charge after the insertion.

+1

Answers (1)

Know the Answer?

Juliette Sloan · Answer 1

V = 5.67 V

Q' = 2.09 x 10^-4 C

Explanation:

Potential difference, V = 5.67 V

Charge, Q = 3.49 x 10^-5 C

dielectric constant, K = 5.99

Capacitance of the capacitor

C = Q / V = 3.49 x 10^-5 / 5.67 = 6.15 x 10^-6 F

Now the dielectric is inserted, so,

C' = K x C = 5.99 x 6.15 x 10^-6 = 3.687 x 10^-5 F

After insertion of dielectric, as the battery is connected, so potential difference remains same.

Charge, Q' = C' x V = 3.687 x 10^-5 x 5.67 = 2.09 x 10^-4 C