NEET UG 2016 — Physics Electromagnetism

Here the capacitance, $C=\frac{A k \epsilon_0}{d}$C=dAkϵ0​​ $C_{K_1}, C_{K_2}, C_{K_3}$CK1​​,CK2​​,CK3​​ are in parallel combination and they are connected in series with $\mathrm{C}_{\mathrm{K}_4}$CK4​​ Here, $\mathrm{C}_{\mathrm{K}_1}=\frac{(\mathrm{A} / 3) \mathrm{K}_1 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{~K}_1}{3} \mathrm{C}$CK1​​= d/2(A/3)K1​ϵ0​​=32 K1​​C, $\mathrm{C}_{\mathrm{k}_2}=\frac{(\mathrm{A} / 3) \mathrm{k}_2 \mathrm{\epsilon}_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_2}{3} \mathrm{C}$Ck2​​= d/2(A/3)k2​ϵ0​​=32k2​​C $C_{K_3}=\frac{(\mathrm{A} / 3) \mathrm{k}_3 \epsilon_0}{\mathrm{~d} / 2} \frac{2 \mathrm{k}_3 \mathrm{C}}{3} \mathrm{C}$CK3​​= d/2(A/3)k3​ϵ0​​32k3​C​C $\mathrm{C}_{\mathrm{K}_4}=\frac{(\mathrm{A}) \mathrm{k}_4 \epsilon_0}{\mathrm{~d} / 2}=2 \mathrm{k}_4 \mathrm{C}$CK4​​= d/2(A)k4​ϵ0​​=2k4​C Now the equivalent capacitance for the combination of four capacitors is $\begin{aligned} & \frac{1}{C_{c q}}=\frac{1}{\left(C_{k_1}+C_{k_2}+C_{k_3}\right)}+\frac{1}{C_{k_4}} \\ & C_{e q}=k C \end{aligned}$​Ccq​1​=(Ck1​​+Ck2​​+Ck3​​)1​+Ck4​​1​Ceq​=kC​ $\frac{1}{k C}=\frac{3}{2 C}\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{1}{2 k_4 C}$kC1​=2C3​[k1​+k2​+k3​1​]+2k4​C1​ or $\frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2\left(k_4\right)}$k1​=2(k1​+k2​+k3​)3​+2(k4​)1​ or $k=\frac{2\left(k_1+k_2+k_3\right)}{3}+2\left(k_4\right)$k=32(k1​+k2​+k3​)​+2(k4​)