Question

Column – I		Column – II
(A) Charge on A	(1)	Increases
(B) Potential difference across A	(2)	Decreases
(C) Potential difference across B	(3)	Remains constant
(D) Charge on B	(4)	Cannot say

• A. (A) → (1); (B) → (2); (C) → (2); (D) → (1)
• B. (A) → (1); (B) → (1); (C) → (2); (D) → (2)
• C. (A) → (2); (B) → (2); (C) → (2); (D) → (4)
• D. (A) → (2); (B) → (2); (C) → (1); (D) → (2)

Answer 1

Answer: (D)

(A) → (2); (B) → (2); (C) → (1); (D) → (2)

Answer 2

To solve this problem, we need to analyze the changes in capacitance, charge, and potential difference when a dielectric slab is removed from one of the capacitors in a series combination connected to a constant voltage source.

Let's denote:

• $C_A$: Capacitance of capacitor A.
• $C_{B,initial}$: Initial capacitance of capacitor B (with dielectric).
• $C_{B,final}$: Final capacitance of capacitor B (without dielectric).
• $V$: Constant voltage of the source.

Initial State: Capacitors A and B are in series. The capacitance of B with a dielectric slab is given by $C_{B,initial} = k C_{B,0}$, where $C_{B,0}$ is the capacitance of B without the dielectric and $k > 1$ is the dielectric constant. When the dielectric is removed, $C_{B,final} = C_{B,0}$. Therefore, $C_{B,final} < C_{B,initial}$.

The equivalent capacitance for series combination is: $C_{eq} = \frac{C_A C_B}{C_A + C_B}$

Analysis of Changes:

1. Change in Capacitance of B ($C_B$): When the dielectric slab is removed from capacitor B, its capacitance decreases. So, $C_{B,final} < C_{B,initial}$.

2. Change in Equivalent Capacitance ($C_{eq}$): The capacitors are in series. The formula for equivalent capacitance is $C_{eq} = \frac{C_A C_B}{C_A + C_B}$. Alternatively, $\frac{1}{C_{eq}} = \frac{1}{C_A} + \frac{1}{C_B}$. Since $C_B$ decreases, $\frac{1}{C_B}$ increases. Therefore, $\frac{1}{C_{eq}}$ increases, which means $C_{eq}$ decreases.

3. Change in Total Charge (Q): The total charge supplied by the source is $Q = C_{eq} V$. Since the voltage source $V$ is constant and $C_{eq}$ decreases, the total charge $Q$ decreases. In a series combination, the charge on each capacitor is the same as the total charge: $Q_A = Q_B = Q$. Therefore:

   • (A) Charge on A: Decreases (2)
   • (D) Charge on B: Decreases (2)

4. Change in Potential Difference across A ($V_A$): The potential difference across capacitor A is $V_A = \frac{Q_A}{C_A}$. Since $Q_A$ decreases (as determined above) and $C_A$ remains constant, $V_A$ decreases. Therefore:

   • (B) Potential difference across A: Decreases (2)

5. Change in Potential Difference across B ($V_B$): For series capacitors connected to a voltage source, the sum of potential differences across them equals the source voltage: $V_A + V_B = V$. Since $V$ is constant and $V_A$ decreases, for the sum to remain constant, $V_B$ must increase. (Alternatively, $V_B = \frac{Q_B}{C_B}$. Both $Q_B$ and $C_B$ decrease. This is an indeterminate form. However, using $V_B = V - V_A$ is more direct and conclusive.) Therefore:

   • (C) Potential difference across B: Increases (1)

Summary of Matches:

• (A) Charge on A → (2) Decreases
• (B) Potential difference across A → (2) Decreases
• (C) Potential difference across B → (1) Increases
• (D) Charge on B → (2) Decreases