A dielectric slab of area A and thickness d is inserted

SOLUTION

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

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

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.