Question

A parallel plate capacitor is charged and then disconnected from the battery. If a dielectric slab is now inserted between the plates of the capacitor, the energy stored in the capacitor:

• A. Increases
• B. Decreases
• C. Remains same
• D. Becomes zero

Hint:

Remember:
Battery disconnected $\rightarrow$ Charge $Q$ is constant $\rightarrow$ $U = \frac{Q^2}{2C}$. Since $C$ increases, $U$ decreases.
Battery remains connected $\rightarrow$ Potential $V$ is constant $\rightarrow$ $U = \frac{1}{2}CV^2$. Since $C$ increases, $U$ increases.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to determine what happens to the electrostatic potential energy stored in a charged parallel plate capacitor when a dielectric slab is inserted after disconnecting the charging battery.

Step 2: Key Formula and Approach:
When the battery is disconnected, the charge $Q$ on the plates remains constant because it has no path to escape:
\[ Q = Q_0 \] The potential energy stored in a capacitor can be expressed as:
\[ U = \frac{Q^2}{2C} \] Inserting a dielectric slab of dielectric constant $K \gt 1$ increases the capacitance $C$.

Step 3: Detailed Explanation:

• Initial State:
Let $C_0$ be the initial capacitance and $Q_0$ be the charge.
The initial energy stored is:
\[ U_0 = \frac{Q_0^2}{2C_0} \]

• After inserting the dielectric slab ($K \gt 1$):
The new capacitance is:
\[ C = K C_0 \] Since the battery was disconnected, the charge remains:
\[ Q = Q_0 \]

• Calculate the new energy stored ($U$):
\[ U = \frac{Q^2}{2C} = \frac{Q_0^2}{2(K C_0)} = \frac{1}{K} \left(\frac{Q_0^2}{2C_0}\right) = \frac{U_0}{K} \] Since $K \gt 1$, we have:
\[ U \lt U_0 \] The energy decreases because the positive work done by the attractive electrostatic force pulling the dielectric slab into the plates reduces the stored electric energy.


Step 4: Final Answer:
The energy of the capacitor decreases, which corresponds to Option (B).