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QJEEM19Ph14af: JEE Main 2019 Physics - Class 12 JEEM19Ph14af Solution: Two identical parallel plate capacitors, of capacitance C each, …

Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1, K2 and K3. The first capacitor is filled as shown in fig.I, and the second one is filled as shown in fig II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E1 refers to capacitor (I) and E2 to capacitor (II)): JEE Main 2019 (Online) 12th April Morning Slot Physics - Capacitor Question 108 English

Multiple Choice Options

  • A E1E2=(K1+K2+K3)(K2K3+K3K1+K1K2)K1K2K3{{{E_1}} \over {{E_2}}} = {{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)} \over {{K_1}{K_2}{K_3}}}
  • B E1E2=K1K2K3(K1+K2+K3)(K2K3+K3K1+K1K2){{{E_1}} \over {{E_2}}} = {{{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}
  • C E1E2=(K1+K2+K3)(K2K3+K3K1+K1K2)9K1K2K3{{{E_1}} \over {{E_2}}} = {{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)} \over {9{K_1}{K_2}{K_3}}}
  • D E1E2=9K1K2K3(K1+K2+K3)(K2K3+K3K1+K1K2){{{E_1}} \over {{E_2}}} = {{9{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}

Step-by-Step Verified Solution

1C1=d3Aε0(1K1+1K2+1K3){1 \over {{C_1}}} = {d \over {3A{\varepsilon _0}}}\left( {{1 \over {{K_1}}} + {1 \over {{K_2}}} + {1 \over {{K_3}}}} \right)

C1=3Aε0(K1K2K3)d(K1K2+K2K3+K3K1){C_1} = {{3A{\varepsilon _0}\left( {{K_1}{K_2}{K_3}} \right)} \over {d\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} \right)}}

C2=Aε03d(K1+K2+K3){C_2} = {{A{\varepsilon _0}} \over {3d}}\left( {{K_1} + {K_2} + {K_3}} \right)

E1E2=C1C2=3K1K2K3(K1K2+K2K3+K3K1)×3(K1+K2+K3){{{E_1}} \over {{E_2}}} = {{{C_1}} \over {{C_2}}} = {{3{K_1}{K_2}{K_3}} \over {\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} \right)}} \times {3 \over {\left( {{K_1} + {K_2} + {K_3}} \right)}}

⇒E1E2=9K1K2K3(K1+K2+K3)(K1K2+K2K3+K3K1)\Rightarrow {{{E_1}} \over {{E_2}}} = {{9{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} \right)}}

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