Question

A dielectric slab of area A and thickness d is inserted between the plates of a capacitor of area 2A with constant speed v as shown in fig. Distance between the plates is d.

The capacitor is connected to a battery of e.m.f E. The current in the circuit varies with time as

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$C=\frac{K{ϵ}_{0}vtb}{d}+\frac{{ϵ}_0(1-vt)b}{d};q=CV$

$I=\frac{dq}{dt}=V\frac{dC}{dt}=V[\frac{K{ϵ}_{0}bv}{d}-\frac{{ϵ}_{0}bv}{d}]=\text{constant}$

So, till the dielectric is fully inserted, a constant current will flow. When the dielectric is fully inside, C will be constant, so I = 0. And when dielectric starts coming out, again a constant current will flow but in opposite direction.