Question

What is the strength of the electric field 3.0 cm from a small glass bead that has been charged to + 8.0 nc ?

Answer 1

Thinking the small glass bead as a single point charge, the electric field generated by it is given by
$E(r) = k_e \frac{Q}{r^2}$
where
$k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant
$Q=8.0 nC=8.0 \cdot 10^{-9} C$ is the charge of the bead
$r=3.0 cm=0.03 m$ is the distance at which we calculate the field.

Using these data, we find:
$E=(8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{(8.0 \cdot 10^{-9} C}{0.03 m)^2}=8.0 \cdot 10^4 N/C$