An initially uncharged air-filled capacitor is connected to a 2.27-V charging so

Question

An initially uncharged air-filled capacitor is connected to a 2.27-V charging source As a result, 7.03 Times 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 7.43. Find the capacitor's potential difference and charge after the insertion. Potential difference after insertion of dielectric:

Explanation / Answer

Q = CV

C = epsilon0 A/d

Insertion of dielectric will result in the increase of Capacitance,

so, charge increases

Q = 7.03*10^-5* 7.43

Q = 52.23*10^-5 C

Potential remains the same as 2.27 V

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