Question

two points charge q1=+25nC and q2=-75nC are separated by a distance of 3.0cm. Find the Magnitude and direction of electric force that exerts on q2 and electric force that q2 exerts on q1

Answer 1

We can find the magnitude of the electric force exerted on $q_1$ by $q_2$ from the Coulomb's law:

$$F=\dfrac{kq_1q_2}{r^2},$$ $$F=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot25\cdot10^{-9}\ C\cdot(-75\cdot10^{-9}\ C)}{(0.03\ m)^2}=-0.01875\ N.$$

The magnitude of the electric force exerted on $q_1$ by $q_2$ (and $q_2$ by $q_1$) equals 0.01875 N:

$$F_{12}=F_{21}=0.01875\ N.$$

Since the two point charges are opposite in sign they will attract each other, therefore, the direction of the electric force is attractive (also, the sign minus indicates that the direction is attractive).