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 10degree 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: Charge after insertion of dielectric:

Explanation / Answer

As long as the capacitor is connected to the charging voltage, its voltage will not change because of the dielectric.
Using that the voltage stays constant, you can write before & after dielectric as;
Vo = V
QoCo = QC

Then since "C" with dielectric is related to "Co" without dielectric by;
C = kCo

Then it follows that;
Q = Qo/k

Just divide the given charge transferred "7.03x10^-5" by the dielectric constant "7.43".

Q= (7.03x10^-5)/ 7.43

Q = 9.46x10^06 C

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