Question

A parallel-plate air capacitor of area A=12.0cm² and plate separation d=3.80mm is charged by a battery to a voltage 66.0V. If a dielectric material with kappa=3.60 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate?

Answer 1

A parallel-plate air capacitor of area (A) = 12 cm² =12*10^-4m²

And plate separation of ( d) = 3.80 mm=3.80*10^-3m

The batterry is charged to a voltage of (V) = 66.0 V

What is the charge on the capacitor (Q) =?

The capacitance of the capacitor is C =eoA/d =8.85*10-12*10^-4/3.80*10^-3m =27.94*10^-13F=2.794pF

Now the charge on the capacitor is Q =CV =2.794pF*66 =184.40pC =

Now if the dielectric material with ? = 3.60 is inserted so that it fills the volume between the plates, the additional charge flow is given by

            C =keoA/d =2.794pF*3.60=377.982*10-13F =10.05pF

Now charge on the plates is given by Q =CV =10.05pF*66=663.85pC

Now the additional charge will flow from the battery onto the positive plate is

                           663.85pC -184.40pC =479.45pC