Question

A 2.0-mm-diameter glass bead is positively charged. The potential difference between a point 2.0 mm from the bead and a point 4.0 mm from the bead is 500 V. What is the charge on the bead?

Short answer

The charge on the bead is$q=1.39·{10}^{-13}\mathrm{C}$.

Step-by-step solution

Step 1. Given information

A 2.0-mm-diameter glass bead is positively charged. The potential difference between a point 2.0 mm from the bead and a point 4.0 mm from the bead is 500 V .

Step 2.Simplify

We'll use the same old "trick" that you have hopefully mastered by now: consider the potential on the surface of the sphere to be the potential created by a point charge at a distance equal to the radius of the sphere. This allows us to write down our given potential difference as

$\Delta V=kq\left(\frac{1}{r}-\frac{1}{r+d}\right)$.

Step 3. Finding the charge

Where $\mathit{r}$is the radius of the bead, d is the distance from the bead in which the potential falls by $\Delta V$and q is the charge we're looking for. We can find the latter as

$$\begin{gathered} \Delta V=kq\frac{r+d-r}{r(r+d)} \\ q=\frac{r(r+d)\Delta V}{kd} \end{gathered}$$

Numerically, in our case, we'll have

$q=\frac{0.001(0.001+0.004)}{9·{10}^9·0.004}=1.39·{10}^{-13}\mathrm{C}$