Question

The charges and coordinates of two charged particles held fixed in the x-y plane are: q1 = +3.0x10^-6 C; x = 3.5, y = 0.5 cm, and q2 = - 4.0x10-6 C; x = - 2.0 cm, y = 1.5 cm. (a) Find the magnitude and direction of the force on q2. (b) where could you locate the third charge q3 = + 4.0x10^-6 C such the total force on q2 is zero? 

Answer 1

a)

$q_1=+3 \times 10^{-6} \;(3.5, 0.5) \\ q_2=-4 \times 10^{-6} \;(-2,1.5) \\ Force \;F = \frac{k|q_1||q_2|}{d^2} \\ F= \frac{k \times 3 10^{-6} \times 4 \times 10^{-6}}{(\sqrt{(3.5 +2)^2+(0.5-1.5)^2})^2} \\ = \frac{9 \times 10^9 \times 12 \times 10^{-12}}{(\sqrt{5.5^2+1^2})^2} \\ = \frac{9 \times 12 \times 10^{-3}}{(\sqrt{31.25})^2} \\ = \frac{108 \times 10^{-3}}{5.6^2} \\ = 0.344 \times 10^{-2} \;N$

Its direction is toward the charge q₂.

b) Given

$q_3=-4\times 10^{-6}$

Force on $q_2$ is zero

$Net \;force = \frac{kq_1q_2}{5.6^2}+ \frac{kq_2q_3}{r^2}=0 \\ \frac{q_1q_2}{5.6^2}+ \frac{q_2q_3}{r^2}=0 \\ \frac{q_1}{5.6^2} + \frac{q_3}{r^2} = 0 \\ \frac{3 \times 10^{-6}}{5.6^2}+ \frac{(-4) \times 10^{-6}}{r^2}=0 \\ \frac{3}{5.6^2}- \frac{4}{r^2}=0 \\ r^2=\frac{4}{3} \times 5.6^2 \\ r = 6.119 \;cm$