
$V_{A C}=5 \times \frac{2}{9}$ volt $=\frac{10}{9}$ volt
$V_{D B}=4 \times \frac{2}{9}$ volt $=\frac{8}{9}$ volt
$\frac{U_{5 \mu F}}{U_{4 \mu F}}=\frac{\frac{1}{2} \times 5\left(\frac{8}{9}\right)^{2}}{\frac{1}{2} \times 4 \times\left(\frac{10}{9}\right)^{2}}=0.8$

$\varepsilon(x)=\varepsilon_{0}+k x, \text { for }\left(0\,<\,x \leq \frac{d}{2}\right)$
$\varepsilon(x)=\varepsilon_{0}+k(d-x)$, for $\left(\frac{d}{2} \leq x \leq d\right)$

