Answer
$|\Delta V|=q / 8 \pi \varepsilon_0 R$
Work Step by Step
The potential difference is
$$
\Delta V=V_s-V_c=\frac{2 q}{8 \pi \varepsilon_0 R}-\frac{3 q}{8 \pi \varepsilon_0 R}=-\frac{q}{8 \pi \varepsilon_0 R},
$$
or $|\Delta V|=q / 8 \pi \varepsilon_0 R$.
