From circuit the current,
$$
I=\frac{E}{R}
$$
The equivalent resistance of above circuit
$$
\begin{array}{l}
=3+2+1 \\
=6 \mathrm{k} \Omega \\
=6 \times 10^{3} \Omega \\
E=3 \text { volt }
\end{array}
$$
$$
\begin{aligned}
I &=\frac{3}{6 \times 10^{3}} \\
&=0.5 \times 10^{-3} \mathrm{A}
\end{aligned}
$$
Porential, $\begin{aligned} V_{A D} &=i R=0.5 \times 10^{-3} \times 3 \times 10^{3} \\ &=1.5 \mathrm{V} \end{aligned}$
Charge,
$$
\begin{array}{l}
\frac{1}{C}=\frac{1}{1}+\frac{1}{2}=\frac{3}{2} \\
C=\frac{2}{3} \\
Q=C \cdot V_{A D} \\
=\frac{2}{3} \times 1 \cdot 5 \\
=1 \mu \mathrm{C}
\end{array}
$$
Applying KVL (Kirchhoffs Voltage Law) from $B$ to $C$
$$
\begin{array}{c}
v_{s}-0 \cdot 5 \times 10^{-3} \times 2 \times 10^{3}+\frac{1}{2}=V_{c} \\
V_{n}-V_{c}=1-\frac{1}{2}=0.5 \mathrm{V}
\end{array}
$$