An initially uncharged air-filled capacitor is connected to a 4.51-V charging so

Question

An initially uncharged air-filled capacitor is connected to a 4.51-V charging source. As a result, 6.43 times 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 7.29. Find the capacitor's potential difference and charge after the insertion. Potential difference after insertion of dielectric: large after insertion of dielectric

Explanation / Answer

Capacitance of capacitor = charge/potential = 6.43 * 10-5 / 4.51 = 1.43 * 10-5 F

After inserting the dielectric material its capacitance become 7.29*(old capacitance) = 1.04 * 10-4 F

Since the capactior is still connected to the battery its potential difference will remain equal to the battey i.e. 4.51V

And since q = CV , charge becomes q = 1.04 * 10-4 *4.51 = 4.69 * 10-4 C

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