Question

An initially uncharged air-filled capacitor is connected to a 4.65 V charging source. As a result, 3.35 × 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 7.05 . Find the voltage V across the capacitor and the charge Q f stored by it after the dielectric is inserted and the circuit has returned to a steady state.

Answer 1

The law of capacitance states that,

C = Q/V,

Where C is the capacitance,

Q is the charge stored by the capacitor,

And V is the potential difference across it.

Therefore,

C = 3.35*10^-5/4.65 = 7.20*10^-6F

So now, as the question states that the charging source is not disconnected, hence we can say that the potential difference across capacitor will not change, and the V across capacitor will be 4.65 volts, even after the dielectric is inserted.

V = 4.65 volts

Now but the capacitance of the system changes, the new capacitance will be equal to the product of original capacitance (with air filled in it) and the dielectric constant (of the dielectric).

Hence C' (new capacitance) will be given by

C' = 7.05*7.20*10^-6 = 50.76*10^-6

And finally to find the final charge stored on the capacitor, we again use law of capacitance,

Q_{​​​​​​f} = (C')*V = 50.76*10^-6*4.65

Q_{​​​​​​f} = 23.60*10^-5 Coulumb