Question

A parallel plate capacitor, filled with a material of dielectric constant $K$, is charged to a certain voltage and is isolated. The dielectric material is removed. Then:
(a) capacitance decreases by a factor $K$
(b) electric field reduces by a factor $K$
(c) voltage across the capacitor increases by a factor $K$
(d) 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 true
• D. (b) and (d) are true

Answer 1

The correct option is B (a) and (c) are true

Initial capacitance $c=\frac{k{\epsilon }_{0}A}{d}$
When the system is isolated, charge ($Q$) present on it is constant.
Now, when dielectric is removed $C=\frac{{\epsilon }_{0}A}{d}$
$\Rightarrow$ $C$ decreases by a factor of $K$.

We know that $Q=C\times V$

As $C$ decreases by factor of $K$ and $Q$ remains constant, $V$ increases by a factor of $K$.