Question

Three point charges lie in the x-y plane as shown in the diagram below: q₁= 66.1 μC and is located at the origin; q₂= 31.7 μC and is located at x = 2.61 m and y = 0; q₃= -19.9 μC and is located at x = 1.61 m and y = 2.61 m. A force diagram showing the forces acting on q₁ and a coordinate system are given in the diagram.
Calculate the y-component of the net electrostatic force acting on q₁ due to charges q₂ and q₃.

g(m) 3 31 d2 F. 21+

Calculate the x-component of the net electrostatic force acting on q₁ due to charges q₂ and q₃.

Calculate the y-component of the net electrostatic force acting on q₁ due to charges q₂ and q₃.

Calculate the magnitude of the net electrostatic force acting on q₁ due to charges q₂and q₃.

Calculate the direction of the net electrostatic force acting on q₁ due to charges q₂and q₃. Enter the angle as a positive value measured counterclockwise from the positive x-axis.

Answer 1

Angle made by q3 is

$\theta =tan^{-1}\left ( \frac{2.61}{1.61} \right )=58.33^{o}$

a)

X-Component of net force acting on q1 due to q2 and q3 is

$F_{x}=F_{31}Cos58.33-F_{21}=\frac{K\left | Q_{1} \right ||Q_{3}|}{r_{31}^{2}}Cos58.33-\frac{K\left | Q_{1} \right ||Q_{2}|}{r_{21}^{2}}$

$F_{x}=\left ( 9\times 10^{9} \right )(66.1\times 10^{-6})\left [ \frac{19.9Cos58.33}{(\sqrt{2.61^{2}+1.61^{2}})^{2}}- \frac{31.7}{2.61^{2}}\right ]\times 10^{-6}$

$F_{x}=-2.10745 N$

b)

Y-Component of net force acting on q1 due to q2 and q3 is

$F_{y}=F_{31}Sin58.33=\frac{K\left | Q_{1} \right ||Q_{3}|}{r_{31}^{2}}Sin58.33$

$F_{y}=\frac{\left ( 9\times 10^{9} \right )(66.1\times 10^{-6})(19.9\times 10^{-6})}{(\sqrt{1.61^{2}+2.61^{2}})}Sin58.33$

$F_{y}=1.0714 N$

c)

Magnitude

$F=\sqrt{(-2.10745)^{2}+1.0714^{2}}=2.364 N$

d)

Direction

$\theta =tan^{-1}\left ( \frac{1.0714}{-2.10745} \right )=180-26.95=153.05^{o}$counterclockwise from +x-axis