QJEEAdv15Ph2455: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph2455 Solution: A parallel plate capacitor having plates of area S and plate…

We can think of this configuration to be made up of three parts:

(a) Capacitor of plate area s/2, plate separation d, and filled with a dielectric of relative permittivity $\in$ ₁ = 2 (lower half). The capacitance of this part is ${C_a} = { \in _1}{ \in _0}{{s/2} \over d} = { \in _0}{s \over d} = {C_1}$ .

(b) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$ ₁ = 2 (left upper half). The capacitance of this part is ${C_b} = { \in _1}{ \in _0}{{s/2} \over {d/2}} = 2{ \in _0}{s \over d} = 2{C_1}$ .

(c) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$ ₂ = 4 (right upper half). The capacitance of this part is ${C_c} = { \in _2}{ \in _0}{{s/2} \over {d/2}} = 4{ \in _0}{s \over d} = 4{C_1}$ .

The capacitors C_b and C_c are connected in series with equivalent capacitance

${C_{bc}} = {{{C_b}{C_c}} \over {{C_b} + {C_c}}} = {{(2{C_1})(4{C_1})} \over {(2{C_1}) + (4{C_1})}} = {4 \over 3}{C_1}$ .

The capacitor C_bc is connected in parallel with C_a. The equivalent capacitance is

$\left. {{C_2} = {C_{abc}} = {C_a}} \right\|{C_{bc}} = {C_1} + {4 \over 3}{C_1} = {7 \over 3}{C_1}$ .