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: Insertion of a dielectric slab between the plates of a charged capacitor enhances capacitance by a factor of the dielectric constant \( \kappa \). The stored energy is \( U = \frac{Q^2}{2C} \). With a constant charge \( Q \) (due to isolation), an increased capacitance \( C \) results in diminished stored energy. Step 2: The electric field within the capacitor is reduced. The dielectric material polarizes, counteracting the field generated by the plate charges, thereby lowering the net electric field. Step 3: Consequently, both the stored energy and the electric field diminish.