Question

When a dielectric slab is inserted between the plates of one of the two identical capacitors shown in the figure then match the following

• A. Charge on A
• B. Potential difference across A
• C. Potential difference across B
• D. Charge on B

Answer 1

A→ 1, B→ 1, C → 2, D→ 1

Answer 2

The problem describes a circuit with two identical capacitors, A and B, connected in series with a voltage source V. A dielectric slab is then inserted into capacitor B. We need to determine how the charge and potential difference across each capacitor change.

Let the initial capacitance of both capacitors be $C$. So, $C_A = C_B = C$.

Initial State:

1. Equivalent Capacitance: Since A and B are in series, the initial equivalent capacitance $C_{eq,i}$ is: $\frac{1}{C_{eq,i}} = \frac{1}{C_A} + \frac{1}{C_B} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$

   $C_{eq,i} = \frac{C}{2}$

2. Total Charge: The total charge supplied by the battery is $Q_i = C_{eq,i} V$: $Q_i = \frac{CV}{2}$

3. Charge on each capacitor: In a series combination, the charge on each capacitor is the same as the total charge: $Q_{A,i} = Q_{B,i} = Q_i = \frac{CV}{2}$

4. Potential Difference across each capacitor: $V_{A,i} = \frac{Q_{A,i}}{C_A} = \frac{CV/2}{C} = \frac{V}{2}$

   $V_{B,i} = \frac{Q_{B,i}}{C_B} = \frac{CV/2}{C} = \frac{V}{2}$

Final State (after inserting dielectric in B):

Let the dielectric constant be $K$. Since a dielectric slab is inserted into capacitor B, its capacitance increases to $C'_B = KC_B = KC$. Capacitor A remains unchanged, so $C'_A = C_A = C$.

1. Equivalent Capacitance: The new equivalent capacitance $C_{eq,f}$ is: $\frac{1}{C_{eq,f}} = \frac{1}{C'_A} + \frac{1}{C'_B} = \frac{1}{C} + \frac{1}{KC} = \frac{K+1}{KC}$

   $C_{eq,f} = \frac{KC}{K+1}$

2. Comparison of Equivalent Capacitance: Since $K > 1$ for a dielectric, let's compare $C_{eq,f}$ with $C_{eq,i}$:

   $C_{eq,f} = \frac{KC}{K+1}$ and $C_{eq,i} = \frac{C}{2}$

   We compare $\frac{K}{K+1}$ with $\frac{1}{2}$.

   $2K$ vs $K+1$. Since $K > 1$, $2K > K+1$.

   Therefore, $\frac{K}{K+1} > \frac{1}{2}$, which means $C_{eq,f} > C_{eq,i}$.

3. Total Charge: The total charge supplied by the battery is $Q_f = C_{eq,f} V$. Since $C_{eq,f} > C_{eq,i}$ and V is constant, $Q_f > Q_i$.

4. Charge on each capacitor: In a series combination, the charge on each capacitor is equal to the total charge: $Q_{A,f} = Q_{B,f} = Q_f$.

   Since $Q_f > Q_i$, both $Q_A$ and $Q_B$ increase.

   • A. Charge on A: Increases (1)
   • D. Charge on B: Increases (1)

5. Potential Difference across each capacitor:

   • Potential difference across A ($V_{A,f}$): $V_{A,f} = \frac{Q_{A,f}}{C'_A} = \frac{Q_f}{C} = \frac{\frac{KC}{K+1} V}{C} = \frac{K}{K+1} V$

     Comparing $V_{A,f}$ with $V_{A,i}$: $V_{A,f} = \frac{K}{K+1} V$ and $V_{A,i} = \frac{V}{2}$

     As established, $\frac{K}{K+1} > \frac{1}{2}$ for $K>1$.

     Therefore, $V_{A,f} > V_{A,i}$, meaning $V_A$ increases.

     • B. Potential difference across A: Increases (1)

   • Potential difference across B ($V_{B,f}$): $V_{B,f} = \frac{Q_{B,f}}{C'_B} = \frac{Q_f}{KC} = \frac{\frac{KC}{K+1} V}{KC} = \frac{1}{K+1} V$

     Comparing $V_{B,f}$ with $V_{B,i}$: $V_{B,f} = \frac{1}{K+1} V$ and $V_{B,i} = \frac{V}{2}$

     Since $K > 1$, $K+1 > 2$. Therefore, $\frac{1}{K+1} < \frac{1}{2}$.

     Thus, $V_{B,f} < V_{B,i}$, meaning $V_B$ decreases.

     • C. Potential difference across B: Decreases (2)

Summary of Matches:

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

Matching this with the given options, the correct option is: A→ 1, B→ 1, C → 2, D→ 1.