Answer to Question #230974 in Physics for Gee

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 $q_1$.