Question

A battery is kept connected to the plates of a parallel-plate capacitor. A dielectric slab of dielectric constant \(K\) is then introduced between the plates such that it covers the entire space between the plates. Choose the correct answer from the given alternatives.

• A. The electric field between the plates will increase.
• B. The charge on the plates will decrease.
• C. The energy stored in the capacitor will decrease.
• D. The electric field between the plates will remain the same.

Hint:

For a capacitor connected to a battery: \[ V=\text{constant}. \] When a dielectric is inserted, \[ C \uparrow,\qquad Q \uparrow,\qquad U \uparrow, \] but \[ E=\frac{V}{d} \] remains unchanged.

Answer 1

The Correct Option is D

Concept: Since the capacitor remains connected to the battery, the potential difference across the plates remains constant. \[ V=\text{constant} \] For a parallel-plate capacitor, \[ E=\frac{V}{d} \] where \(d\) is the separation between the plates.

Step 1: Find the effect on capacitance. When a dielectric of dielectric constant \(K\) completely fills the space, \[ C'=KC. \] Thus capacitance increases.

Step 2: Find the effect on charge. Since the battery keeps the voltage constant, \[ Q'=C'V \] \[ =KCV \] \[ =KQ. \] Hence charge increases, not decreases.

Step 3: Find the effect on electric field. Because \[ V=\text{constant} \] and \[ d=\text{constant}, \] \[ E'=\frac{V}{d}=E. \] Therefore the electric field remains unchanged.

Step 4: Check the stored energy. \[ U=\frac12 CV^2. \] Since \(V\) is constant and \(C\) increases, \[ U'=\frac12 (KC)V^2 \] \[ =KU. \] Thus stored energy increases. \[\begin{aligned} \boxed{E'=E} \end{aligned}\] Therefore, \[ \boxed{\text{The electric field between the plates remains the same.}} \] Hence, option \(\mathbf{(D)}\) is correct.