Since r > r_{2}:

$\mathit{E}\mathbf{=}\frac{\mathbf{\left(}\mathbf{3}{\displaystyle \raisebox{1ex}{$\mathbf{Q}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\mathbf{\right)}}{\mathbf{4}\mathbf{\pi}{\mathbf{\epsilon}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$

We have:

$\begin{array}{rcl}\mathbf{V}& \mathbf{=}& \mathbf{\int}\frac{\mathbf{3}\mathbf{Q}}{\mathbf{8}\mathbf{\pi}{\mathbf{\epsilon}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\mathbf{\xb7}\mathbf{d}\mathbf{r}\\ & \mathbf{=}& \frac{\mathbf{3}\mathbf{Q}}{\mathbf{8}\mathbf{\pi}{\mathbf{\epsilon}}_{\mathbf{0}}\mathbf{r}}\end{array}$

A hollow spherical conductor, carrying a net charge +Q, has inner radius r_{1 }and outer radius r_{2}=2r_{1} (see the figure (Figure 1)). At the center of the sphere is a point charge +Q/2. Plot V as a function of r from r=0 to r=2r_{2}. Assume V0=3Q/8πϵ0r_{2}.

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