Question

What is the strength of the electric field 2.4 cm from a small plastic bead that has been charged to -9.0 nc ?

Answer 1

We can assimilate the small bead as a single point charge, with charge Q equal to
$Q=-9.0 nC=-9.0 \cdot 10^{-9}C$
We want to find the electric field at a distance $d=2.4 cm= 0.024 m$. The electric field of a single point charge is given by
$E=k_e \frac{Q}{d^2}$
where $k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$
So, substituting, we find the electric field at the distance d:
$E=(8.99 \cdot 10^9 N m^2 C^{-2}) \frac{-9.0\cdot 10^{-9}C}{(0.024m)^2}=-1.40\cdot 10^5 N/C$
And the negative sign means the field is directed towards the charge, since the charge is negative.