A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as

ε(x) = ε? + kx, for 0 < x ≤ d/2
ε(x) = ε? + k(d-x), for d/2 ≤ x ≤ d

Please find the answer below

Now, using junction analysis
We can say, q? + q? + q? = 0
2 (x - 6) + 4 (x - 6) + 5 (x) = 0
x = 36/11, q? = 36 (5)/11 = 180/11
q? = 16.36µC

Capacitance of element

Capacitance of element, $C^{\mathrm{\text{'}}}=\frac{K(1+\alpha x){\epsilon }_{0}A}{dx}$ ${}^{}\frac{{}_{}}{}$

$\sum \frac{1}{C^{\mathrm{\text{'}}}}={\int }_0^d?\frac{dx}{K{\epsilon }_0\mathrm{A}(1+\alpha \mathrm{x})}$

$\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{K}{\epsilon }_0\mathrm{A}\alpha }\mathrm{l}\mathrm{n}?(1+\alpha \mathrm{d})$

Given: $\alpha \mathrm{d}?1$ $$

$\frac{1}{C}=\frac{1}{K{\epsilon }_{0}A\alpha }\left(\alpha d-\frac{{\alpha }^2{d}^2}{2}\right);\mathrm{}\frac{1}{C}=\frac{d}{K{\epsilon }_{0}A}\left(1-\frac{\alpha d}{2}\right)$

$C=\frac{K{\epsilon }_{0}A}{d}\left(1+\frac{\alpha d}{2}\right)$

When positive charge moves from low potential to high potential region, work is done against electric field and potential energy increases.