An initially uncharged air-filled capacitor is connected to a 2.63 v charging source. as a result, 6.27×10−5 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 κ of this substance is 3.55. find the capacitor's potential difference and charge after the insertion.

Charge stored after insertion will be

Explanation:

We have given potential difference V =2.63 V

charge stored by the capacitor when without dielectric =

We know that , here is capacitance without dielectric

We have given

k = dielectric constant = 5.99

C = Capacitance with the dielectric = k = 3.55× =

Potential difference is due to the external charging source so it remains same

V = potential difference after insertion = 2.63 volts

New charge stored ,

the amount of current in the coil

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg