Answer
$\begin{aligned}
\Delta V & =7.49 \times 10^5 \mathrm{~V}
\end{aligned}$
Work Step by Step
The potential difference is
$
\begin{aligned}
\Delta V & =\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}-\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}=\left(8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{C}^2\right)\left(\frac{15 \times 10^{-6} \mathrm{C}}{0.060 \mathrm{~m}}-\frac{5.0 \times 10^{-6} \mathrm{C}}{0.030 \mathrm{~m}}\right) \\
& =7.49 \times 10^5 \mathrm{~V} .
\end{aligned}
$
