Question

Identical dielectric slabs are inserted into two identical capacitors $\text{A}$ and $\text{B}$. These capacitors and a battery are connected as shown in the figure. Now the slab of capacitor $\text{B}$ is pulled out slowly with battery remaining connected.


Consider the following statements:-

$1.)$ During the process positive charge flows from $a$ to $b$.

$2.)$ Final charge on capacitors $\text{B}$ will be less than that on capacitor $\text{A}$.

$3.)$ During the process, a work is done by the external force which completely appears as heat in the circuit.

$4.)$ During the process internal energy of battery increases.

The correct statement(s) is/are:

• A. $2$ & $3$ only
• B. only $1$
• C. $1$ & $4$ only
• D. $2,3$ & $4$ only

Answer 1

The correct option is C $1$ & $4$ only

When dielectric is taken out of capacitor $\text{B}$, the charge on capacitor $\text{B}$ will decrease $(\because C_f<C_i;$ after dielectric removed).
Thus charge from the positive plate of $\text{B}$ will move towards the battery.

$\Rightarrow$ As charge on $\text{B}$ decreases, charge on capacitor $\text{A}$ will also reduce proportionally because they are connected in series.


Therefore, the charge flows from $a\to b$.

During the process, work is done by external force on battery. Thus, the battery gets charged and a portion of work done by external gets dissipated in the form of heat.

$\therefore$ Internal energy of battery increases due to charging.

Correct statements $\to \left(1\right)\&\left(4\right)$ only.

Why this question? Tip: The internal energy of a battery is on account of ''electro-chemical forces operating inside it.''