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JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2013
- Hard
- MCQ
A parallel plate capacitor having a separation between the plates \(d\) , plate area \(A\) and material with dielectric constant \(K\) has capacitance \(C_0\). Now one-third of the material is replaced by another material with dielectric constant \(2K\), so that effectively there are two capacitors one with area \(\frac{1}{3}\,A\) , dielectric constant \(2K\) and another with area \(\frac{2}{3}\,A\) and dielectric constant \(K\). If the capacitance of this new capacitor is \(C\) then \(\frac{C}{{{C_0}}}\) is
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Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(C_{0}=\frac{k \epsilon_{0} A}{d}\)
\(C=\frac{k \epsilon_{0} 2}{3 d}+\frac{2 k \epsilon_{0} A}{3 d}=\frac{4}{3} \frac{k \epsilon_{0} A}{d}\)
\(\therefore\) \(\frac{{\text{c}}}{{{{\text{C}}_0}}} = \frac{{\frac{{4k{\varepsilon _0}A}}{{3d}}}}{{\frac{{k{\varepsilon _0}A}}{d}}} = \frac{4}{3}\)
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