Potential difference:

$\overline{){\mathbf{V}}{\mathbf{=}}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi}{\mathbf{\epsilon}}_{\mathbf{0}}\mathbf{r}}}$

Radius, r = 1.50/2 = 0.75 × 10^{- 3} m

The potential difference between the two points:

$\begin{array}{rcl}{\mathbf{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{V}}_{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{q}}{\mathbf{4}{\mathbf{\pi \epsilon}}_{\mathbf{0}}{\mathbf{r}}_{\mathbf{1}}}\mathbf{-}\frac{\mathbf{q}}{\mathbf{4}{\mathbf{\pi \epsilon}}_{\mathbf{0}}{\mathbf{r}}_{\mathbf{2}}}\\ & \mathbf{=}& \frac{\mathbf{q}}{\mathbf{4}{\mathbf{\pi \epsilon}}_{\mathbf{0}}}\mathbf{(}\frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{1}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{2}}}\mathbf{)}\end{array}$

A 1.50 mm-diameter glass bead is positively charged. The potential difference between a point 1.50 mm from the bead and a point 3.60 mm from the bead is 460V.

Note: The given distances are measured from the surface of the bead.

What is the charge on the bead?

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