JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2019
- Hard
- MCQ
A parallel plate capacitor is made of two square plates of side \(a\), separated by a distance \(d\,(d < < a)\). The lower triangular portion is filled with a dielectric of dielectric constant \(K\), as shown in the figure. Capacitance of this capacitor is
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Answer & Solution
Correct Answer
(A) \(\frac{{K{\varepsilon _0}{a^2}}}{{d\left( {K - 1} \right)}}\,\ln \,K\)
Step-by-step Solution
Detailed explanation
Let's consider a strip of thickness \(dx\) at a distance of \(x\) from the left end as shown in the figure.
\(\frac{y}{x}=\frac{d}{a}\)
\(\Rightarrow \quad y=\left(\frac{d}{a}\right) x\)
\(C_{1}=\frac{\varepsilon_{0} a d x}{(d-y)} \quad ; \quad C_{2}=\frac{k \varepsilon_{0} a d x}{y}\)
\(C_{e q}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{k \varepsilon_{0} a d x}{k d+(1-k) y}\)
Now integrating it from \(0\) to \(a\)
\(\int_0^a {\frac{{{\text{k}}{\varepsilon _0}{\text{adx}}}}{{{\text{kd}} + (1 - {\text{k}})\frac{{\text{d}}}{{\text{a}}}{\text{x}}}}} = \frac{{{\text{k}}{\varepsilon _0}{{\text{a}}^2}{\text{lnk}}}}{{{\text{d}}({\text{k}} - 1)}}\)
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