A parallel plate capacitor filled with a material of dielectric constant K is charged to a certain voltage. The battery is disconnected. The dielectric material is removed. Then
a) The capacitance decreased by a factor K.
b) The electric field reduces by a factor K.
c) The voltage across the capacitor increases by a factor K.
d) The charge stored in the capacitor increased by a factor K.

As capacitance of capacitor after introducing dielectric is given by

$C=\frac{K{\epsilon }_{0}A}{d}$

Here, if dielectric is introduced the capacitance increases by K, but when dielectric is removed, then

capacitance will decrease by factor K and since $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}} \mathrm{as} \mathrm{charge} \mathrm{remains} \mathrm{constant}.$

Hence, if C decreases in factor K, then V will increase by factor K.

Electric field will increase by a factor K as potential is increased and distance between plates remains same.