Question

When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then:

• A. charge on capacitor increases
• B. potential difference across the capacitor increases
• C. energy stored increases
• D. capacity remains the same

Hint:

Battery connected: \(V\) constant \(\Rightarrow C\uparrow \Rightarrow Q\uparrow\).

Answer 1

The Correct Option is A

To understand the effect of introducing a dielectric slab between the plates of a capacitor while it is connected to a battery, we need to analyze the behavior of the capacitor's parameters such as charge, potential difference, energy, and capacitance.

Let's consider a parallel plate capacitor with capacitance \(C_0\) when the dielectric constant is 1 (i.e., in vacuum or air). The capacitance is given by:

\(C_0 = \frac{\varepsilon_0 A}{d}\)

where:

• \(\varepsilon_0\) is the permittivity of free space.
• \(A\) is the area of the plates.
• \(d\) is the distance between the plates.

When a dielectric slab with dielectric constant \(\kappa\) is introduced between the plates, the capacitance becomes:

\(C = \kappa C_0 = \kappa \frac{\varepsilon_0 A}{d}\)

Since the capacitor is connected to a battery, the potential difference \(V\) across the capacitor remains constant. Consequently, according to the relation between charge \(Q\), capacitance \(C\), and potential difference \(V\):

\(Q = CV\)

When the capacitance increases from \(C_0\) to \(C\), and since \(V\) is constant, the charge on the capacitor also increases. Thus, the charge \(Q\) becomes:

\(Q' = \kappa C_0 V = \kappa Q\)

This implies the charge on the capacitor increases.

Let’s examine the given options:

• **Charge on capacitor increases:** This is true due to the increased capacitance while maintaining a constant voltage.
• **Potential difference across the capacitor increases:** This is false, as the voltage remains constant due to the battery.
• **Energy stored increases:** The energy stored given by \(U = \frac{1}{2}CV^2\) does increase because capacitance increases with the introduction of the dielectric.
• **Capacity remains the same:** This is false because capacitance increases as it is directly proportional to the dielectric constant \(\kappa\).

Thus, the correct answer is charge on capacitor increases.