Question

A 10.0cm long thin glass rod is uniformly charged to +50.0nC. A small bead, charged -5.0nC is 4.0cm from the center of the rod. What is the force (both magnitude and direction) on the bead?

Answer 1

The electric field due to the charged glass rod of length $l$ with charge $q$ at the distance $d$ from the center of rod is given by

$$E=k\frac{q}{d\sqrt{d^2+(l/2)^2}}$$

Thus, the force of interaction between rod and point charge $Q$:

$$F=QE=k\frac{qQ}{d\sqrt{d^2+(l/2)^2}}$$

$$F=9*10^9\frac{50.0*10^{-9}*5.0*10^{-9}}{0.04\sqrt{0.04^2+(0.1/2)^2}}=8.8*10^{-4}\:\rm N$$

The force is attractive.