Two capacitances in series
$\therefore \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
$1$ capacitor in parallel
$\therefore $ $C_{\mathrm{eq}}=C_{3}+C=4+2=6\, \mu \mathrm{F}$


Charge density potential Electric intensity
