Answer to Question #177779 in Field Theory for Charles

Question

Three charges are placed at the following locations on the xaxis:+2.0μC at x=0, -3.0μC at x=40cm, -5.0μC at x=120cm. Find (a) on the -3.0μC charge,(b) on the -5.0μC charge.

Answer 1

(a) Let's find the net electric force on the -3.0μC charge:

$$F_{net}=F_{12}-F_{32},$$ $$F_{net}=k(\dfrac{|q_1q_2|}{(r_{12})^2}-\dfrac{|q_3q_2|}{(r_{13}-r{_{12}})^2}),$$

$F_{net}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(\dfrac{|2\cdot10^{-6}\ C\cdot(-3\cdot10^{-6}\ C)|}{(0.4\ m)^2}-\dfrac{|(-5\cdot10^{-6}\ C)\cdot(-3\cdot10^{-6}\ C)|}{(1.2\ m-0.4\ m)^2})=0.13\ N.$

(b) Let's find the net electric force on the -5.0μC charge:

$$F_{net}=F_{13}+F_{23},$$ $$F_{net}=k(\dfrac{|q_1q_3|}{(r_{13})^2}+\dfrac{|q_2q_3|}{(r_{13}-r_{12})^2}),$$

$F_{net}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(\dfrac{|2\cdot10^{-6}\ C\cdot(-5\cdot10^{-6}\ C)|}{(1.2\ m)^2}+\dfrac{|(-3\cdot10^{-6}\ C)\cdot(-5\cdot10^{-6}\ C)|}{(1.2\ m-0.4\ m)^2})=3.04\cdot10^{-11}\ N.$