Question

Three point charges are placed in close proximity to one another. The charges and distances between them are shown. What is the resultant electrostatic force on q2 as a result of the other two charges?....(q1=+6mC, q2=-8mC and q3=+9mC.....q1 to q2 is 60mm and q2 to q3 is 85mm

Answer 1

The resultant electrostatic force on charge $q_2$ is the vector sum of two contributions: the attractive force due to charge $q_1$ (toward to $q_1$) and the attractive force due to charge $q_3$ (toward to $q_3$). Then, the resultant electrostatic force is the sum of these two forces, taken algebraically:

$$F_{res}=F_1+F_2,$$ $$F_{res}=k(\dfrac{|q_2q_3|}{r_{23}^2}-\dfrac{|q_1q_2|}{r_{12}^2}),$$

$F_{res}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(\dfrac{|(-8\cdot10^{-6}\ C)\cdot9\cdot10^{-6}\ C|}{(0.085\ m)^2}-\dfrac{|6\cdot10^{-6}\ C\cdot(-8\cdot10^{-6}\ C)|}{(0.06\ m)^2}),$

$$F_{res}=-30.3\ N.$$

The sign minus means that the resultant electrostatic force on charge $q_2$ directed to the left (toward charge $q_1$).

Answer:

$F_{res}=30.3\ N$, toward charge $q_1$.