Question

A parallel plate capacitor filled with a material of dielectric constant K is charged to a certain voltage. The dielectric material is removed. Then
(A) the capacitance decreases 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 increases by a factor K

• A. A and B are true
• B. A and C are true
• C. B and C are tru
• D. B and D are true

Answer 1

Answer: (B)

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 V = $\frac{Q}{C}.$
Hence, if C decreases in factor K, then V will increase by factor K.