Question

Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in figure. Capacitor $2$ contains a dielectric slab of dielectric constant $k$ as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q_1^{\prime }$ and $Q_2^{\prime }$.
Then

• A. $\frac{Q_1^{\prime }}{Q_1}=\frac{k+1}{k}$
• B. $\frac{Q_2^{\prime }}{Q_2}=\frac{k+1}{2}$
• C. $\frac{Q_2^{\prime }}{Q_2}=\frac{k+1}{2k}$
• D. $\frac{Q_1^{\prime }}{Q_1}=\frac{k}{2}$

Answer 1

Answer: (C)

$\frac{Q_2^{\prime }}{Q_2}=\frac{k+1}{2k}$

The capacitor 1 has capacitance $C$ and as dielectric contains in capacitor 2 so its capacitance becomes $kC$.

The net capacitance $C_{eq}=\frac{C.kC}{C+kC}=\frac{Ck}{1+k}$

and $Q_{eq}=C_{eq}E$

When dielectric is reemoved from capacitor 2, its capacitance becomes $C$.

now net capacitance $C_{eq}^{\prime }=\frac{C.C}{C+C}=\frac{C}{2}$

and $Q_{eq}^{\prime }=C_{eq}^{\prime }E=\frac{CE}{2}$

When the capacitors are connected in series, they each have the same charge as the net capacitance. Thus, $Q_{eq}=Q_1=Q_2$ and $Q_{eq}^{\prime }=Q_1^{\prime }=Q_2^{\prime }$

$\therefore \frac{Q_1^{\prime }}{Q_1}=\frac{Q_2^{\prime }}{Q_2}=\frac{2k}{k+1}$