JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2019
- Hard
- MCQ
A parallel palate capacitor with square plates is filled with four dielectrics of dielectric constants \(K_1, K_2, K_3, K_4\) arranged as shown in the figure. The effective dielectric constant \(K\) will be
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Answer & Solution
Correct Answer
(A) \(K = \frac{{({K_1} + {K_3})({K_2} + {K_4})}}{{{K_1} + {K_2} + {K_3} + {K_4}}}\)
Step-by-step Solution
Detailed explanation
\(C_{1}=\frac{\varepsilon_{0} K_{1} \frac{L^{2}}{2}}{\frac{d}{2}}+\frac{\varepsilon_{0} K_{3} \frac{L^{2}}{2}}{\left(\frac{d}{2}\right)}=\frac{\varepsilon_{0} L^{2}}{d}\left(K_{1}+K_{3}\right)\)
\(C_{2}=\frac{\varepsilon_{0} K_{2} \frac{L^{2}}{2}}{\frac{d}{2}}+\frac{\varepsilon_{0} K_{4} \frac{L^{2}}{2}}{\frac{d}{2}}=\frac{\varepsilon_{0} L^{2}}{d}\left(K_{2}+K_{4}\right)\)
\(\therefore \quad \frac{1}{c}=\frac{1}{c_{1}}+\frac{1}{c_{2}}\)
\(\Rightarrow \quad \frac{\mathrm{d}}{\varepsilon_{0} \mathrm{KL}^{2}}=\frac{\mathrm{d}}{\varepsilon_{0} \mathrm{L}^{2}\left(\mathrm{K}_{1}+\mathrm{K}_{3}\right)}+\frac{\mathrm{d}}{\varepsilon_{0} \mathrm{L}^{2}\left(\mathrm{K}_{2}+\mathrm{K}_{4}\right)}\)
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