Answer :
Given :
Charge on glass bead, Q = 8 nC .
To Find :
The magnitude of the electric field 2.0 m from the center of the bead.
Solution :
Electric field at position r is given by :
[tex]E = \dfrac{kQ}{r^2}\\\\E = \dfrac{9\times 10^9\times 8\times 10^{-9}}{2^2}\\\\E = 18\ N/C[/tex]
Therefore, the magnitude of the electric field 2.0 m from the center of the bead is 18 N/C .