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JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 2)

A parallel plate capacitor with square plates is filled with four dielecytrics of dielectrics constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be :

JEE Main 2019 (Online) 9th January Evening Slot Physics - Capacitor Question 124 English
$$K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}$$
$$K = {{\left( {{K_1} + {K_2}} \right)\left( {{K_3} + {K_4}} \right)} \over {2\left( {{K_1} + {K_2} + {K_3} + {K_4}} \right)}}$$
$$K = {{\left( {{K_1} + {K_2}} \right)\left( {{K_3} + {K_4}} \right)} \over {{K_1} + {K_2} + {K_3} + {K_4}}}$$
$$K = {{\left( {{K_1} + {K_4}} \right)\left( {{K_2} + {K_3}} \right)} \over {2\left( {{K_1} + {K_2} + {K_3} + {K_4}} \right)}}$$

Explanation

JEE Main 2019 (Online) 9th January Evening Slot Physics - Capacitor Question 124 English Explanation
Here, $${C_1} = {{{K_1}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_1}{\varepsilon _0}{L^2}} \over d}$$

$${C_2}{{{K_2}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_2}{\varepsilon _0}{L^2}} \over d}$$

$${C_3} = {{{K_3}{\varepsilon _0}{L^2}} \over d}$$

$${C_4} = {{{K_4}{\varepsilon _0}{L^2}} \over d}$$

Here  C1 C2 are in series and C3, C4 are in series.

Equivalent capacitance point A and B is,

Ceq $$=$$ $${{{C_1}{C_2}} \over {{C_1} + {C_2}}} + {{{C_3}{C_4}} \over {{C_3} + {C_4}}}$$

$$ = {C_{12}} + {C_{34}}$$

$${C_{12}}\,\, = \,\,{{{{{K_1}{\varepsilon _0}{L^2}} \over d} \times {{{K_2}{\varepsilon _0}{L^2}} \over d}} \over {{{{K_1}{\varepsilon _0}{L^2}} \over d} + {{{K_2}{\varepsilon _0}{L^2}} \over d}}}$$

$$ = \,\,{{{K_1}{K_2}} \over {{K_1} + {K_2}}} \times {{{\varepsilon _0}{L^2}} \over d}$$

Similarly,

$${C_{34}} = \,\,{{{K_3}{K_4}} \over {{K_3} + {K_4}}} \times {{{\varepsilon _0}{L^2}} \over d}$$

$$ \therefore $$  Ceq $$=$$ $$\left[ {{{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}} \right]{{{\varepsilon _0}{L^2}} \over d}$$

$$ \Rightarrow $$  $${{K{\varepsilon _0}{L^2}} \over d}$$  $$ = \left[ {{{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}} \right]{{{\varepsilon _0}{L^2}} \over d}$$

$$ \Rightarrow $$  $$K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}$$

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