a) The magnitude of the electric field generated by a point charge Q at distance r from the charge is
[tex]E=k \frac{Q}{r^2} [/tex]
where k is the Coulomb's constant. In this problem, the charge is
[tex]Q=6 \mu C=6 \cdot 10^{-6} C[/tex]
and the distance is
[tex]r=20 cm=0.20 m[/tex]
Therefore the electric field is
[tex]E=k \frac{Q}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{6 \cdot 10^{-6} C}{(0.20 m)^2}=1.36 \cdot 10^6 N/C [/tex]
b) The electric force acting on a charge q is given by
[tex]F=qE[/tex]
where E is the intensity of the electric field at the point where the charge q is located.
We have already calculated the magnitude of the electric field, E, at point x, therefore the force acting on the charge is
[tex]F=qE=(-0.20 \cdot 10^{-6} C)(1.36 \cdot 10^6 N/C)=-0.272 N[/tex]
and the negative sign means the force points in the opposite direction of the electric field, since the charge is negative.
