Question

Two identical capacitors 1 and 2 are connected in series. The capacitor 2 contains a dielectric slab of constant $K$ as show. They are connected to as battery of emf ${\text{V}}_{\text{0}}$ volts. The dielectric slab is then removed. Let ${\text{Q}}_{\text{1}}$ and ${\text{Q}}_{\text{2}}$ be the charge stored in the capacitors before removing the slab and ${{\text{Q}}_{\text{1}}}^{\prime },$ and ${{\text{Q}}_{\text{2}}}^{\prime }$ be the values after removing the slab. Then:

• A. $\frac{{{\text{Q}}_1}^{\prime }}{Q_1}=\left(\frac{K+1}{K}\right)$
• B. $\frac{{{\text{Q}}_2}^{\prime }}{Q_2}=\frac{\left(K+1\right)}{2}$
• C. $\frac{{{\text{Q}}_2}^{\prime }}{Q_2}=\frac{K+1}{2K}$
• D. $\frac{{{\text{Q}}_1}^{\prime }}{Q_1}=\frac{K}{2}$

Answer 1

The correct option is C $\frac{{{\text{Q}}_2}^{\prime }}{Q_2}=\frac{K+1}{2K}$

Total capacitance of the circuit

$=\frac{1}{C}+\frac{1}{KC}$

$=\frac{1}{C}\left\{\frac{K+1}{K}\right\}$

$=\frac{CK}{K+1}$

Charge stared on Capacitor $1$,

$Q=\left(C_{eq}\right)\left(V_0\right)$

$Q_1=\frac{CKV}{K+1}{Q}_2=\frac{KV}{K+1}$

When dielectric slab is removed, Total capacitance of the circuit.

$\frac{1}{C_{eq}^1}=\frac{1}{C_1}+\frac{1}{C_2}$

$C_{eq}^1=\frac{C_1{C}_2}{C_1+C_2}$

$C_1=C_2\Rightarrow C_{eq}^1=\frac{1}{2}$

$\therefore$ Charge stared on each capacitor is given by,

$Q_1^1=C_{eq}^1{V}_0,Q_2^1=\frac{1}{2}V$

$=\frac{1}{2}V$

$\therefore \frac{Q_1}{Q_1^1}=\frac{V}{2}\left(\frac{KV}{K+1}\right)÷\left\{\frac{1}{2}V\right\}$

$\frac{Q_1}{Q_1^1}=\frac{2K}{K+1}$

Now, $\frac{Q_1}{Q_1^1}=\frac{Q_2}{Q_2^1}$

$\Rightarrow \frac{Q_2}{Q_2^1}=\frac{2K}{K+1}$

$\Rightarrow \frac{Q_2^1}{Q_2}=\frac{K+1}{2K}$

$\therefore$ Option (C) is the correct option.