Answer
(a) (i) $V$ = 180 V, (ii) $V$ = -270 V, (iii) $V$ = -450 V
(b) $\Delta V$ = 720 V
Work Step by Step
(a) $R$ is the radius of the sphere and $r$ is the distance where the potential is measured.
When ($rR$), the potential is
\begin{equation}
V = \dfrac{1}{4 \pi \epsilon_{\circ}} \dfrac{q}{r}
\end{equation}
(i) $r$ = 0, the potenatil $V$ is
\begin{align*}
V &= \dfrac{1}{4 \pi \epsilon_{\circ}} \left[\dfrac{q_{1}}{R_{1}} + \dfrac{q_{2}}{R_{2}} \right] \\
&= (9.0 \times 10^{9} \mathrm{~N\cdot m^{2}/C^{2}}) \left[\dfrac{ 6.00 \times 10^{-9} \,\text{C}}{0.03 \,\text{m}} + \dfrac{- 9.00 \times 10^{-9} \,\text{C}}{0.05 \,\text{m}} \right] \\
&=\boxed{180 \,\text{V}}
\end{align*}
(ii) $r$ = 4.0 cm the potenatil $V$ is
\begin{align*}
V&= \dfrac{1}{4 \pi \epsilon_{\circ}} \left[\dfrac{q_{1}}{r} + \dfrac{q_{2}}{R_{2}} \right] \\
&= (9.0 \times 10^{9} \mathrm{~N\cdot m^{2}/C^{2}}) \left[\dfrac{ 6.00 \times 10^{-9} \,\text{C}}{0.04 \,\text{m}} + \dfrac{- 9.00 \times 10^{-9} \,\text{C}}{0.05 \,\text{m}} \right] \\
&=\boxed{-270 \,\text{V}}
\end{align*}
(iii) $r$ = 6.0 cm the potenatil $V$ is
\begin{align*}
V &= \dfrac{1}{4 \pi \epsilon_{\circ}} \left[\dfrac{q_{1}}{r} + \dfrac{q_{2}}{r} \right] \\
&= (9.0 \times 10^{9} \mathrm{~N\cdot m^{2}/C^{2}}) \left[\dfrac{ 6.00 \times 10^{-9} \,\text{C}}{0.06 \,\text{m}} + \dfrac{- 9.00 \times 10^{-9} \,\text{C}}{0.06 \,\text{m}} \right] \\
&=\boxed{-450\,\text{V}}
\end{align*}
(b) $V_1$ is obtained at $r$ = 0 and $V_2$ is obtained at $r = R$ = 5 cm and could be caclauted by
\begin{align*}
V_{2} &= \dfrac{1}{4 \pi \epsilon_{\circ}} \left[\dfrac{q_{1}}{r} + \dfrac{q_{2}}{R_{2}} \right] \\
&= (9.0 \times 10^{9} \mathrm{~N\cdot m^{2}/C^{2}}) \left[\dfrac{ 6.00 \times 10^{-9} \,\text{C}}{0.05 \,\text{m}} + \dfrac{- 9.00 \times 10^{-9} \,\text{C}}{0.05 \,\text{m}} \right] \\
&= -540\,\text{V}
\end{align*}
Hence, the potential difference $\Delta V$ will be
$$\Delta V = V_{1} - V_{2} = 180 \,\text{V} - (\text{-}540\,\text{V}) = \boxed{720 \,\text{V}} $$
