2. Electric Potential and Capacitance
hard

वर्गाकार प्लेटों वाले एक समान्तर प्लेट संधारित्र को चित्रानुसार चार परावैधुतों, जिनके परावैधुतांक $K _{1}, K _{2}, K _{3}$ तथा $K _{4}$ है, से भर दिया जाता है तो प्रभावी परावैधुतांक $K$ का मान होगा?

A

$K = \frac{{({K_1} + {K_3})({K_2} + {K_4})}}{{{K_1} + {K_2} + {K_3} + {K_4}}}$

B

$K = \frac{{({K_1} + {K_2})({K_3} + {K_4})}}{{2({K_1} + {K_2} + {K_3} + {K_4})}}$

C

$K = \frac{{({K_1} + {K_2})({K_3} + {K_4})}}{{{K_1} + {K_2} + {K_3} + {K_4}}}$

D

$K = \frac{{({K_1} + {K_4})({K_2} + {K_3})}}{{2({K_1} + {K_2} + {K_3} + {K_4})}}$

(JEE MAIN-2019)

Solution

$C_{1}=\frac{\varepsilon_{0} K_{1} \frac{L^{2}}{2}}{\frac{d}{2}}+\frac{\varepsilon_{0} K_{3} \frac{L^{2}}{2}}{\left(\frac{d}{2}\right)}=\frac{\varepsilon_{0} L^{2}}{d}\left(K_{1}+K_{3}\right)$

$C_{2}=\frac{\varepsilon_{0} K_{2} \frac{L^{2}}{2}}{\frac{d}{2}}+\frac{\varepsilon_{0} K_{4} \frac{L^{2}}{2}}{\frac{d}{2}}=\frac{\varepsilon_{0} L^{2}}{d}\left(K_{2}+K_{4}\right)$

$\therefore \quad \frac{1}{c}=\frac{1}{c_{1}}+\frac{1}{c_{2}}$

$\Rightarrow \quad \frac{\mathrm{d}}{\varepsilon_{0} \mathrm{KL}^{2}}=\frac{\mathrm{d}}{\varepsilon_{0} \mathrm{L}^{2}\left(\mathrm{K}_{1}+\mathrm{K}_{3}\right)}+\frac{\mathrm{d}}{\varepsilon_{0} \mathrm{L}^{2}\left(\mathrm{K}_{2}+\mathrm{K}_{4}\right)}$

Standard 12
Physics

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