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  4. 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

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Q. 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

EAMCETEAMCET 2004Electrostatic Potential and Capacitance

Solution:

As capacitance of capacitor after introducing dielectric is given by
$ C = \frac{K \varepsilon_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.