Question

Half of the space between the plates of parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates. If initially the capacitance is $C$, then the new capacitance will be

• A. $\frac{2KC}{K+1}$
  
• B. $\frac{(K+1)C}{2}$
  
• C. $\frac{KC}{K+1}$
  
• D. $KC$

Answer 1

The correct option is A $\frac{2KC}{K+1}$


If the area of the plates is $A$ and the initial separation between them is $d$ then the capacitance will be$$C=\frac{{\epsilon }_{0}A}{d}....\left(1\right)$$As we know that capacitance of partially filled dielectric slab of thickness $t$ and dielectric constant $K$ is given by$$C^{\prime }=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}....\left(2\right)$$Since, here half of the space is filled with dielectric so, $t=\frac{d}{2}$. From equation $\left(2\right)$,

$\Rightarrow C^{\prime }=\frac{{\epsilon }_{0}A}{d-\frac{d}{2}+\frac{d}{2K}}$

$\Rightarrow C^{\prime }=\frac{{\epsilon }_{0}A}{\frac{d}{2}+\frac{d}{2K}}$

$\Rightarrow C^{\prime }=\frac{2{\epsilon }_{0}A}{d}\frac{K}{K+1}$

From equation $\left(1\right)$,

$\Rightarrow C^{\prime }=\frac{2CK}{K+1}$

Hence, option (a) is the correct answer.

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