A parallel plate capacitor has a capacity 80 10⁻⁶ ~F , when air is present between its plates. The space between the plates is filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of 30 ~V by wires. The dielectric slab is then removed. Then, the charge passing through the wire is

Step-by-step solution

C_ air =80 F C_ dielectric = _r C_ air =1600 F Charge stored in the presence of air, q_ air =C_ air V=80 30 10⁻⁶=2400 C Charge stored in presence of dielectric medium, aligned q_d=C_ dielectric V & =1600 30 10⁻⁶ & =48000 C aligned When dielectric is removed, effective charge remained is aligned q & =q_d-q_ air & =(48-2.4) 10⁻³ C q & =45.6 10⁻³ C aligned