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QJEEM19Ph0ef4: JEE Main 2019 Physics - Class 12 JEEM19Ph0ef4 Solution: A parallel plate capacitor with square plates is filled with…

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

Multiple Choice Options

  • A K=K1K2K1+K2+K3K4K3+K4K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}
  • B K=(K1+K2)(K3+K4)2(K1+K2+K3+K4)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)}}
  • C K=(K1+K2)(K3+K4)K1+K2+K3+K4K = {{\left( {{K_1} + {K_2}} \right)\left( {{K_3} + {K_4}} \right)} \over {{K_1} + {K_2} + {K_3} + {K_4}}}
  • D K=(K1+K4)(K2+K3)2(K1+K2+K3+K4)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)}}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th January Evening Slot Physics - Capacitor Question 124 English Explanation
Here, C1=K1ε0.L.L2d2=K1ε0L2d{C_1} = {{{K_1}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_1}{\varepsilon _0}{L^2}} \over d}

C2K2ε0.L.L2d2=K2ε0L2d{C_2}{{{K_2}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_2}{\varepsilon _0}{L^2}} \over d}

C3=K3ε0L2d{C_3} = {{{K_3}{\varepsilon _0}{L^2}} \over d}

C4=K4ε0L2d{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 == C1C2C1+C2+C3C4C3+C4{{{C_1}{C_2}} \over {{C_1} + {C_2}}} + {{{C_3}{C_4}} \over {{C_3} + {C_4}}}

=C12+C34= {C_{12}} + {C_{34}}

C12  =  K1ε0L2d×K2ε0L2dK1ε0L2d+K2ε0L2d{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}}}

=  K1K2K1+K2×ε0L2d= \,\,{{{K_1}{K_2}} \over {{K_1} + {K_2}}} \times {{{\varepsilon _0}{L^2}} \over d}

Similarly,

C34=  K3K4K3+K4×ε0L2d{C_{34}} = \,\,{{{K_3}{K_4}} \over {{K_3} + {K_4}}} \times {{{\varepsilon _0}{L^2}} \over d}

∴\therefore   Ceq == [K1K2K1+K2+K3K4K3+K4]ε0L2d\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    Kε0L2d{{K{\varepsilon _0}{L^2}} \over d}    =[K1K2K1+K2+K3K4K3+K4]ε0L2d= \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=K1K2K1+K2+K3K4K3+K4K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}

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