Question

When a dielectric slab is inserted between the plates of an isolated charged capacitor, the energy stored in it:

• A. \( \) increases and the electric field inside it also increases.
• B. \( \) decreases and the electric field also decreases.
• C. \( \) decreases and the electric field increases.
• D. \( \) increases and the electric field decreases.

Hint:

Inserting a dielectric increases the capacitance of a capacitor, which reduces the energy stored if the charge is constant. The dielectric also reduces the electric field inside the capacitor.

Answer 1

The Correct Option is B

Step 1: When a dielectric slab is inserted between the plates of a charged capacitor, the capacitance increases due to the dielectric constant \( \kappa \) of the material. The energy stored in a capacitor is given by: \[ U = \frac{Q^2}{2C} \] where \( Q \) is the charge and \( C \) is the capacitance. Since the capacitor is isolated, the charge remains constant, and increasing the capacitance leads to a decrease in the energy stored. Step 2: The electric field inside the capacitor also decreases because the dielectric reduces the effective field between the plates. The dielectric polarizes and partially cancels the electric field created by the charges on the plates. Step 3: Therefore, the energy stored decreases and the electric field also decreases.