Question

A parallel plate capacitor is charged by connecting a 2 Volt battery across it. It is then disconnected from the battery and a glass slab is introduced between plates. Which of the following pair of quantities decrease ?

• A. Energy stored and capacitance
• B. Charge and potential difference
• C. Capacitance and charge
• D. Potential difference and energy stored

Answer 1

The Correct Option is D

Let's analyze the scenario of a capacitor being charged, disconnected, and then subjected to a dielectric material (glass slab) between its plates. Based on this analysis, we need to determine which pair of quantities decreases.

Step 1: Charging the Capacitor

The capacitor is charged by connecting a 2 Volt battery across it. Initially, the potential difference \( V \) across the capacitor is 2 Volts. As the voltage is applied, the charge \( Q \) stored on the plates increases.

Step 2: Disconnecting the Battery

After disconnecting the battery, the potential difference \( V \) across the capacitor remains the same as it was when charged, i.e., 2 Volts. The charge \( Q \) remains the same because there is no current flow, and the capacitor is isolated from the battery.

Step 3: Introducing the Glass Slab (Dielectric Material)

When a dielectric material, such as a glass slab, is introduced between the plates of the capacitor, it increases the capacitance \( C \) of the capacitor. The capacitance with the dielectric is given by the formula:

\( C = k \cdot C_0 \)

Where:

• k: The dielectric constant of the material (for glass, typically \( k > 1 \))
• C₀: The original capacitance of the capacitor without the dielectric

As the capacitance increases due to the introduction of the glass slab, the charge \( Q \) stored on the plates also increases to maintain the same potential difference \( V \). This is because the relationship between charge, capacitance, and voltage is given by:

\( Q = C \cdot V \)

Conclusion:

Based on the analysis, we can conclude that the correct pair of quantities that decreases is:

• Potential difference and energy stored.

Correct answer: (D) Potential difference and energy stored.

Answer 2

Let the initial capacitance of the parallel plate capacitor be \( C_0 \). When connected to a battery of voltage \( V_0 = 2 \, \text{V} \), it acquires a charge \( Q_0 \):

\[ Q_0 = C_0 V_0 \] 

The initial energy stored in the capacitor is:

\[ U_0 = \frac{1}{2} C_0 V_0^2 = \frac{1}{2} \frac{Q_0^2}{C_0} \]

The capacitor is then disconnected from the battery. This is a crucial step. When disconnected, the charge \( Q \) on the capacitor plates remains constant because there is no path for the charge to flow.

\[ Q = Q_0 \]

Next, a glass slab is introduced between the plates. Glass is a dielectric material with a dielectric constant \( K > 1 \).

Introducing a dielectric slab between the plates increases the capacitance. The new capacitance \( C \) is given by:

\[ C = K C_0 \]

Since \( K > 1 \), the capacitance increases (\( C > C_0 \)).

Now let's analyze the other quantities:

• Charge (Q): As established, it remains constant because the capacitor is isolated. \( Q = Q_0 \).
• Potential Difference (V): The new potential difference across the plates is given by \( V = \frac{Q}{C} \). Substituting the values for \( Q \) and \( C \): \[ V = \frac{Q_0}{K C_0} = \frac{C_0 V_0}{K C_0} = \frac{V_0}{K} \] Since \( K > 1 \), the new potential difference \( V \) is less than the initial potential difference \( V_0 \). So, the potential difference decreases.
• Energy Stored (U): The new energy stored can be calculated using \( U = \frac{1}{2} \frac{Q^2}{C} \). Substituting the values for \( Q \) and \( C \): \[ U = \frac{1}{2} \frac{Q_0^2}{K C_0} = \frac{1}{K} \left( \frac{1}{2} \frac{Q_0^2}{C_0} \right) = \frac{U_0}{K} \] Since \( K > 1 \), the new energy stored \( U \) is less than the initial energy stored \( U_0 \). So, the energy stored decreases.

Summary of changes:

• Charge: Remains constant
• Capacitance: Increases
• Potential Difference: Decreases
• Energy Stored: Decreases

The pair of quantities that decrease are the Potential difference and energy stored.