Question

In the adjoining figure, capacitor $\text{(l)}$ and $\text{(2)}$ have a capacitance $C$ each. When the dielectric of dielectric constant $K$ is inserted between the plates of one of the capacitors, the total charge flowing through battery is

• A. $\frac{KCE}{K+1}$ from $\text{B}$ to $\text{C}$
• B. $\frac{KCE}{K+1}$ from $\text{C}$ to $\text{B}$
• C. $\frac{(K-1)CE}{2(K+1)}$ from $\text{B}$ to $\text{C}$
• D. $\frac{(K-1)CE}{2(K+1)}$ from $\text{C}$ to $\text{B}$

Answer 1

The correct option is D $\frac{(K-1)CE}{2(K+1)}$ from $\text{C}$ to $\text{B}$

In the intial condition:


Equivalent capacitance,

$C_{eq}=\frac{\left(C\right)\left(C\right)}{2C}=\frac{C}{2}$

Initial charge on the each capacitors,

$\Rightarrow q_i=C_{eq}E=\frac{CE}{2}$

After the dielectric is inserted between the plates of one capacitor:


Now, $C_{eq}=\frac{\left(C\right)\left(CK\right)}{C+CK}=\frac{C^{2}K}{C(1+K)}$

$\Rightarrow C_{eq}=\frac{CK}{K+1}$

Thus, final charge on each capacitor,

$q_f=C_{eq}E=\frac{CEK}{K+1}$

Charge supplied by battery is,

$q_{\text{supplied}}=q_f-q_i$

$\Rightarrow q_{\text{supplied}}=\frac{CEK}{K+1}-\frac{CE}{2}$

$\Rightarrow q_{\text{supplied}}=\frac{CE(K-1)}{2(K+1)}$

As, $\frac{CE}{2}<\frac{CEK}{K+1}$

Therefore, $q_{\text{supplied}}$ is -ve, which means charge is released by capacitor towards the battery.

$\therefore \frac{CE(K-1)}{2(K+1)}$ charge flows from $\text{C}$ to $\text{B}$.

Hence, option (d) is the correct answer.