In the circuit shown, charge on the $5\mu \mathrm{F}$ $$ capacitor is:

Please find the answer below

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

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

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

dC? = (ε? + kx)A / dx [For 0 < x < d/2]
1/C? = ∫ dx / (ε? + kx)A) from 0 to d/2
= (1/Ak) [ln (ε? + kx)] from 0 to d/2
= (1/kA) ln (1 + kd/ (2ε? )
C? = kA / ln (1 + kd/ (2ε? )

Similarly dC? = (ε? + k (d-x)A / dx [For d/2 ≤ x ≤ d]
C? = kA / ln (1 + kd/ (2ε? )
Clearly, C? = C? = C
For series combination:
C_eq = C? / (C? + C? ) = C/2 = kA / (2ln (2ε? + kd)/2ε? )

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