Question

In the figure shown, a parallel plate capacitor is connected across a source of emf $E$. The plates 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 $v$. Assuming the connecting wires have no resistance, what is the current in the circuit?

• A. $\frac{{\epsilon }_{0}lEKv}{d}$
  
• B. $\frac{{\epsilon }_{0}lEv(K-1)}{d}$
  
• C. $\frac{2{\epsilon }_{0}lEv(K-1)}{d}$
  
• D. $\frac{{\epsilon }_{0}lEv(K-1)}{2d}$
  

Answer 1

The correct option is B $\frac{{\epsilon }_{0}lEv(K-1)}{d}$



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

The capacitance of the system is:

$C=\frac{{\epsilon }_{0}Kxl}{d}+\frac{{ϵ}_{0}l(l-x)}{d}$

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

Charge on capacitor:

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

As we know, current in the circuit:

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

$\Rightarrow i=\frac{{\epsilon }_{0}l(K-1)}{d}\frac{dx}{dt}\times E$

$\Rightarrow i=\frac{{\epsilon }_{0}l(K-1)}{d}v\times E$

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

Hence, option (b) is the correct answer.

Key concepts: Current in the circuit when dielectric is introduced with constant speed inside a parallel plate capacitor.