JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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
- A \(\frac{{K{\varepsilon _0}{a^2}}}{{d\left( {K - 1} \right)}}\,\ln \,K\)
- B \(\frac{{K{\varepsilon _0}{a^2}}}{{2d\left( {K + 1} \right)}}\)
- C \(\frac{{K{\varepsilon _0}{a^2}}}{d}\,\ln \,K\)
- D \(\frac{1}{2}\frac{{K{\varepsilon _0}{a^2}}}{d}\)
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. \(\frac{y}{x}=\frac{d}{a}\) \(\Rightarrow \quad y=\left(\frac{d}{a}\right) x\)…
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