Answer:
[tex]1.44\times 10^{-3}[/tex] N
Explanation:
We are given that three charged particle are placed at each corner of equilateral triangle.
[tex]q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC[/tex]
[tex]q_1=-8.2\times 10^{-9} C[/tex]
[tex]q_2=-16.4\times 10^{-9} C[/tex]
[tex]q_3=8.0\times 10^{-9} C[/tex]
Side of equilateral triangle =3.3 cm=[tex]\frac{3.3}{100}=0.033m[/tex]
We know that each angle of equilateral angle=[tex]60^{\circ}[/tex]
Net force=F =[tex]\sum\frac{kQq }{d^2}[/tex]
Where k=[tex]9\times 10^9 Nm^2/C^2[/tex]
If we bisect the angle at [tex]q_3[/tex] then we have 30 degrees from there to either charge.
Direction of vertical force due to charge [tex]q_1[/tex] and [tex]q_2[/tex]
Therefore, force will be added
Vertical force=[tex]9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})[/tex]
Vertical net force=[tex]9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}[/tex]
Vertical force =[tex]9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}[/tex]
Vertical force=[tex]-1.41\times 10^{-3}N[/tex] (towards [tex] q_1}[/tex]
Horizontal component are opposite in direction then will b subtracted
Horizontal force=[tex]9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}[/tex]
Horizontal force=[tex]0.27\times 10^-3}[/tex] N(towards [tex]q_2[/tex]
Net electric force acting on particle 3 due to particle =[tex]\sqrt{F^2_x+F^2_y}[/tex]
Net force=[tex]\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}[/tex]
Net force=[tex]1.44\times 10^{-3}[/tex] N
