Question

The potential energy of charged parallel plate capacitor is $v_{0}$. If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be

• A. $\frac{v_{0{K}$
• B. $v_{0}K^{2}$
• C. $\frac{v_{0{K^{2$
• D. $v_{0}^{2}$

Hint:

Logic Tip: Inserting a dielectric increases the "storage capacity" (capacitance), which reduces the energy per unit of stored charge if the charge remains constant.

Answer 1

The Correct Option is A

Concept:
The potential energy $U$ of a capacitor is given by $U = \frac{Q^2}{2C}$. When a dielectric is inserted, the capacitance increases.
Step 1: Determine the change in capacitance.
On inserting a slab of dielectric constant $K$, the new capacitance $C'$ becomes $KC$.
Step 2: Calculate the new potential energy.
Assuming the charge $Q$ remains constant, the new potential energy $v_{0}'$ is: $$v_{0}' = \frac{Q^2}{2C'} = \frac{Q^2}{2(KC)}$$ $$v_{0}' = \frac{1}{K} \left( \frac{Q^2}{2C} \right) = \frac{v_{0{K}$$