Question

Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be :(charge on each condenser is $q_0$; $k$ = dielectric constant )

• A. $\frac{2q_0}{1+1/k}$
• B. $\frac{q_0}{1+1/k}$
• C. $\frac{2q_0}{1+k}$
• D. $\frac{q_0}{1+k}$

Answer 1

The correct option is A $\frac{2q_0}{1+1/k}$

When capacitors are connected in series the charge present on them must be equal.
$\frac{Q}{c_1}+\frac{Q}{c_2}=2$ ------------ (1)
Now when $k$ is introduced $c_2$ becomes $k{c}_2$

Let $q_o$ be the charge on capacitor
given $c_1=c_2$
$\Rightarrow 2\frac{q_o}{c}=2,\frac{q_o}{c}=1$ from (1)
$c=q_o$

After introducing the dielectric, let charge on each plate become $Q_1$

$\Rightarrow \frac{Q_1}{q_o}+\frac{Q_1}{k{q}_0}=2$
$\Rightarrow \frac{k{Q}_1+Q_1}{k{q}_o}=2$
$\Rightarrow Q_1=\frac{2k{q}_o}{k+1}$
$\Rightarrow Q_1=\frac{2q_o}{1+\frac{1}{k}}$