Q.54 A 2.0-mm-diameter glass bead is ... [FREE SOLUTION]

Chapter 25: Q.54 (page 712)

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

Expert verified

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

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

螖 V = k q (\frac{1}{r} - \frac{1}{r + d})

Step 3. Finding the charge

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

$螖 V = k q \frac{r + d - r}{r (r + d)} q = \frac{r (r + d) 螖 V}{k d}$