Question

Two identical capacitor A and B are connected in series to a battery of E.M.F, 'E' Capacitor B contains a slab of dielectric constant constant K. $Q_A$ and $Q_B$ are the charges stored in A and B. When the dielectric slab is removed, the corresponding charges are $Q'_A$ and $Q'_B$. Then

• A. $\frac{Q'_A}{Q_A} = \frac{K}{2}$
• B. $\frac{Q'_B}{Q_B} = \frac{K+1}{2}$
• C. $\frac{Q'_A}{Q_A} = \frac{K+1}{K}$
• D. $\frac{Q'_B}{Q_B} = \frac{K+1}{2k}$

Answer 1

Answer: (D)

D

Answer 2

1. Initial State (Before removing dielectric):

Let the capacitance of the identical capacitors A and B be $C_0$.

When capacitor B contains a dielectric slab of dielectric constant K, its capacitance becomes $C_B = KC_0$. Capacitor A has capacitance $C_A = C_0$.

The capacitors are connected in series to a battery of EMF 'E'. The equivalent capacitance ($C_{eq}$) for series combination is given by: $\frac{1}{C_{eq}} = \frac{1}{C_A} + \frac{1}{C_B} = \frac{1}{C_0} + \frac{1}{KC_0} = \frac{K+1}{KC_0}$ Therefore, $C_{eq} = \frac{KC_0}{K+1}$

In a series combination, the charge stored on each capacitor is the same and equal to the total charge supplied by the battery. The total charge ($Q$) is given by $Q = C_{eq}E$. So, the charges stored in A and B are: $Q_A = Q_B = Q = \frac{KC_0E}{K+1}$

2. Final State (After removing dielectric):

When the dielectric slab is removed from capacitor B, its capacitance reverts to $C_0$. Now, both capacitors A and B have capacitance $C_A' = C_0$ and $C_B' = C_0$. They are still connected in series to the same battery of EMF 'E'. The new equivalent capacitance ($C'_{eq}$) for series combination is: $\frac{1}{C'_{eq}} = \frac{1}{C_0} + \frac{1}{C_0} = \frac{2}{C_0}$ Therefore, $C'_{eq} = \frac{C_0}{2}$

The new total charge ($Q'$) supplied by the battery is $Q' = C'_{eq}E$. So, the new charges stored in A and B are: $Q'_A = Q'_B = Q' = \frac{C_0E}{2}$

3. Calculate the Ratios:

We need to find the ratios $\frac{Q'_A}{Q_A}$ and $\frac{Q'_B}{Q_B}$. Since $Q'_A = Q'_B = Q'$ and $Q_A = Q_B = Q$, both ratios will be the same: $\frac{Q'_A}{Q_A} = \frac{Q'_B}{Q_B} = \frac{Q'}{Q}$ Substitute the expressions for $Q'$ and $Q$: $\frac{Q'}{Q} = \frac{\frac{C_0E}{2}}{\frac{KC_0E}{K+1}}$ $\frac{Q'}{Q} = \frac{C_0E}{2} \times \frac{K+1}{KC_0E}$ $\frac{Q'}{Q} = \frac{K+1}{2K}$

Thus, the correct option is D.