JEE Advance - Physics (2015 - Paper 2 Offline - No. 13)

Explanation
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 Cb and Cc 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 Cbc is connected in parallel with Ca. The equivalent capacitance is
$$\left. {{C_2} = {C_{abc}} = {C_a}} \right\|{C_{bc}} = {C_1} + {4 \over 3}{C_1} = {7 \over 3}{C_1}$$.
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