Question

A parallel plate capacitor is connected across a source of emf $\epsilon$. The plates of the capacitor are square shaped with edge length $l$ and separated by a distance $d$. A dielectric slab of dielectric constant $K$ and thickness $d$ is inserted between the plates with constant speed $V$.


The direction of current in connecting wires (during insertion) and its value respectively are:
[Current $\left(i\right)$ is time rate of charge transfer]

• A. $b$ to $a;i=\frac{{ϵ}_{0}l\epsilon V(2K-1)}{d}$
• B. $a$ to $b;i=\frac{{ϵ}_{0}l\epsilon V(K-1)}{d}$
• C. $b$ to $a;i=\frac{{ϵ}_{0}l\epsilon V(2K-1)}{2d}$
• D. $a$ to $b;i=\frac{2{ϵ}_{0}l\epsilon V(2K-1)}{d}$

Answer 1

The correct option is B $a$ to $b;i=\frac{{ϵ}_{0}l\epsilon V(K-1)}{d}$

Consider that the length $x$ of dielectric is inside the capacitor.

The capacitance of the system is,

$C=\frac{{ϵ}_{0}K\left(xl\right)}{d}+\frac{{ϵ}_0\left[l(l-x)\right]}{d}$

$[\because$ Area of capacitor plate with dielectrics part is $(x\times l)$ and area of capacitor plate without dielectric will be $(l\times (l-x\left)\right)]$

$\Rightarrow C=\frac{{ϵ}_{0}l}{d}[(K-1)x+l]$

Both the capacitors can be visualized to be in parallel combination.

Now the charge $\left(q\right)$ supplied by battery will be,

$q=C\epsilon =\frac{{ϵ}_{0}l}{d}[(K-1)x+l]\epsilon$

Current in conducting wires will be,

$i=\frac{dq}{dt}$

$\Rightarrow i=\frac{d}{dt}[\frac{{ϵ}_{0}l(K-1)\epsilon }{d}].\left(\frac{dx}{dt}\right)+0$

$[\because \frac{dx}{dt}=V]$

$\Rightarrow i=\frac{{ϵ}_{0}l(K-1)\epsilon V}{d}$

As the dielectric is occupying the spaces between plates, hence $C_{net}\uparrow$ increases with time elapse, thus more charge will supply by battery.

$\therefore$ direction of current $\to (a\to b)$, during insertion.

Hence, option $\left(b\right)$ is correct.

Why this question ? Bottom line: A constant current $i$ will flow from "a to b" till the dielectric is completely inserted. Once dielectric is fully inside $C$ will become constant and current in circuit ceases $(i=0)$. Tip: when dielectric is taken out from capacitor, again a constant current $i$ will flow but in opposite direction i.e "b to a".