Answer to Question #284324 in Electric Circuits for Mr Engineer

Question

Point charge q1 = -6.00 * 10-6 C is on the x-axis at x = -0.200 m. Point charge q2 is

on the x-axis at x = +0.400 m. Point charge q3 = +3.00 * 10-6 C is at the origin. What

is q2 (magnitude and sign) (a) if the net force on q3 is 6.00 N in the +x direction; (b)

if the net force on q3 is 6.00 N in the -x direction?

Answer 1

Part (a)

$F=\frac{k(q_1q_2)}{r^2}$

$F_{net}=F_{13}+F_{23}$

$6=8.988×10^9[\frac{(-6×10^{-6})(3×10^{-6})}{(-0.2)^2}+\frac{(q_2)(3×10^{-6})}{(0.4)^2}]$

$q_2=+5.96×10^{-5}\>C$

Part (b)

$-6=8.988×10^9[\frac{(-6×10^{-6})(3×10^{-6})}{(-0.2)^2}+\frac{(q_2)(3×10^{-6})}{(0.4)^2}]$

$q_2=-1.16×10^{-5}\>C$