Question

Two identical capacitors are connected as shown in given figure, having a charge $q_0$. A dielectric slab is introduced between the plates of the capacitor (I) so as to fill the gap, keeping the battery remain connected. The charge on each capacitors now will be :

• A. $\frac{2q_0}{[1+(1/k\left)\right]}$
• B. $\frac{q_0}{[1+(1/k\left)\right]}$
• C. $\frac{2q_0}{(1+k)}$
• D. $\frac{2q_0}{(1-k)}$

Answer 1

The correct option is A $\frac{2q_0}{[1+(1/k\left)\right]}$

Without dielectric : $C_{eq}=\frac{C_0{C}_0}{C_0+C_0}=\frac{C_0}{2}$ and $Q_{eq}=\frac{C_0}{2}\times 2V_0=C_0{V}_0$
As they are in series so the charge on each capacitors is equal to equivalent charge.i.e
$Q_{eq}=q_0=C_0{V}_0$
With dielectric in (I), the capacitance of it becomes $k{C}_0$
now $C_{eq}^{\prime }=\frac{k{C}_0{C}_0}{k{C}_0+C_0}=\frac{k}{k+1}{C}_0$ and
$Q_{eq}^{\prime }=\frac{k}{k+1}{C}_0\left(2V_0\right)=\frac{2k}{k+1}{q}_0$
Thus the on each capacitor is $q^{\prime }=Q_{eq}^{\prime }=\frac{2k}{k+1}{q}_0=\frac{2q_0}{1+1/k}$