Question

An air capacitor, a capacitor with a dielectric medium of dielectric constant $K=3$, and a capacitor with a conducting slab (thickness one half the separation introducing between the plates of parallel plate air capacitor in both cases) have capacitances $C_1,C_2$ and $C_3$ respectively then :

• A. $C_1>C_2>C_3$
  
• B. $C_2>C_3>C_1$
  
• C. $C_3>C_2>C_1$
  
• D. $C_3>C_1>C_2$
  

Answer 1

The correct option is B $C_2>C_3>C_1$


If the area of the plates is $A$ and the initial separation between them is $d$ with air medium then the capacitance will be $$C_1=\frac{{\epsilon }_{0}A}{d}$$The capacitance with dielectric medium of dielectric constant $K=3$,

$C_2=\frac{K{\epsilon }_{0}A}{d}=\frac{3{\epsilon }_{0}A}{d}$

As we know that capacitance of partially filled dielectric slab of thickness $t$ and dielectric constant $K$ is given by $$C_3=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}$$As , $t=\frac{d}{2}$ and for conductors $K=\infty$

$C_3=\frac{{\epsilon }_{0}A}{d-\frac{d}{2}+\frac{d}{2\times \infty }}$

$\Rightarrow C_3=\frac{2{\epsilon }_{0}A}{d}$

So, clearly we can conclude that, $C_2>C_3>C_1$

Hence, option (b) is the correct answer.

Key Concept: The capacitance of partially filled dielectric slab of thickness $t$ and dielectric constant $K$ is given by$$C^{\prime }=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}$$