Question

Two identical sized capacitors $C_1$ and $C_2$ are connected to a battery in series. A dielectric slab of dielectric constant $K=2$ is filled in the half region of both the capacitors as shown in figure.




$(C\to$ capacitance, $q\to$ charge stored, $U\to$ energy stored $)$

Match the following column:

Column I	Column II
$\left(A\right)$ $\frac{C_1}{C_2}$	$\left(p\right)$ $8/9$ $\left(q\right)$ $4/9$ $\left(r\right)$ $9/8$ $\left(s\right)$ None of these
$\left(B\right)$ $\frac{q_1}{q_2}$
$\left(C\right)$ $\frac{U_1}{U_2}$

• A. $A\to s,B\to s,C\to s$
• B. $A\to r,B\to p,C\to s$
• C. $A\to p,B\to r,C\to q$
• D. $A\to r,B\to s,C\to p$

Answer 1

The correct option is D $A\to r,B\to s,C\to p$

Let original capacitance of capacitor without dielectric is

$C_0=\frac{A{ϵ}_0}{d}$

In $C_1$, capacitor is in parallel combination. So equivalent capacitance will be

$C_1=\frac{{ϵ}_0\left(A/2\right)}{d}+\frac{K{ϵ}_0\left(A/2\right)}{d}$

$\Rightarrow C_1=\frac{A{ϵ}_0}{2d}+\frac{A{ϵ}_0}{d}(\because K=2)$

$\Rightarrow C_1=\frac{C_0}{2}+C_0=\frac{3C_0}{2}$

Similarly, for the second capacitor,

$C_2=\frac{A{ϵ}_0}{d-t+\frac{t}{K}}$

$\Rightarrow C_2=\frac{A{ϵ}_0}{d-\frac{d}{2}+\frac{d}{2\times 2}}(\because t=\frac{d}{2};K=2)$

$\Rightarrow C_2=\frac{2{ϵ}_{0}A}{d(1+\frac{1}{2})}=\frac{4}{3}\frac{A{ϵ}_0}{d}$

$\therefore C_2=\frac{4}{3}{C}_0$

Hence, $\frac{C_1}{C_2}=\frac{9}{8}$

Since capacitors are in series,

$\therefore q_1=q_2$

$\Rightarrow \frac{q_1}{q_2}=1$

And we know that energy stored in a capacitor is given by

$U=\frac{q^2}{2C}$

$\therefore \frac{U_1}{U_2}=\frac{C_2}{C_1}$

(because $q$ is same for both)

Thus, $\frac{U_1}{U_2}=\frac{4C_0/3}{3C_0/2}=\frac{8}{9}$

Hence, $A\to r,B\to s,C\to p$