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 constant $K . Q_{1}$ and $Q_{2}$ are the charges stored in $1$ and $2$ . Now, the dielectric slab is removed and the corresponding charges are $Q_{1}'$ and $Q_{2}'$. Then

• A. $\frac{Q_{1}'}{Q_{1}}=\frac{K+1}{K}$
• B. $\frac{Q_{2}'}{Q_{2}}=\frac{K+1}{2}$
• C. $\frac{Q_{2}'}{Q_{2}}=\frac{K+1}{2 K}$
• D. $\frac{Q_{1}'}{Q_{1}}=\frac{K}{2}$

Answer 1

Answer: (C)

Before the slab is removed, if
$C_{1}=C, C_{2}=K C$
$C_{\text {net }}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}$
$=\frac{C K C}{C(1+K)}=\frac{K C}{K+1}$
$\therefore Q_{1}=Q_{2}=\frac{K C E}{K+1}$
After the slab is removed, $C_{1}=C_{2}=C$
$C_{\text {net }}=\frac{C}{2} $
$\therefore Q_{1}'=Q_{2}'=\frac{C E}{2}$
$\frac{Q_{2}'}{Q_{2}}=\frac{C E / 2}{K C E /(K+1)}=\frac{K+1}{2 K}$