# Problem: 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?

###### FREE Expert Solution

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:

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

###### Problem Details

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?