Question

The space between the plates of a parallel plate capacitor is filled with a dielectric constant varies with distance as per the relation, $K\left(X\right)=1+3X$. If the plate area is $50{\text{cm}}^2$ and separation between the plates is $2\text{mm}$, then the capacitance of this capacitor will be $[\ln \left(1.006\right)=0.0005]$

• A. $265\text{pF}$
• B. $500\text{pF}$
• C. $100\text{pF}$
• D. $500\text{pF}$
  

Answer 1

The correct option is A $265\text{pF}$

We know that for a variation of dielectric medium in between the capacitor plates , the capacitance is given by $$C=\frac{{\epsilon }_{0}Ab}{\ln (1+\frac{bd}{K_0})}$$Substituting the data given in the question we get,
$C=\frac{1}{4\pi }\times \frac{1}{9\times {10}^9}\times \frac{50\times {10}^{-4}\times 3}{\ln (1+\left(\frac{3\times 2\times {10}^{-3}}{1}\right))}$

$\Rightarrow C=265.26\text{pF}\approx 265\text{pF}$

Hence, option (a) is the correct answer.