An initially uncharged air-filled capacitor is connected to a 3.49-V charging so
ID: 1327942 • Letter: A
Question
An initially uncharged air-filled capacitor is connected to a 3.49-V charging source. As a result, 7.21 x 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 2.17. Find the capacitor^?s potential difference and charge after the insertion. Potential difference after insertion of dielectric: Charge after insertion of dielectric:Explanation / Answer
If the dielectric completely fills the space between the conductors of the capacitor ,the capacitance is increased by an factor K which is characterstics of the dielectric and This factor is known as the dielectric constan.
Dielectric constant of vaccum is unity.
the capacitor is filled by a dielectric the P.D decrease to a new value
V=V0/K.
=3.49/2.17
=1.608volts
if the dielectric is inserted while the battery is still connected then battery would have to supply some amount of charge to maintain the P.D between the plates and then total charge on the plates would be Q=KQ0.
Q=2.17*7.21*10-5
=15.64*10-5C