Question

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of the particles have a negative charge: q1 = -8.2 nC and q2 = -16.4 nC . The remaining particle has a positive charge, q3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Answer 1

Answer:

$1.44\times 10^{-3}$ N

Explanation:

We are given that three charged particle are placed at each corner  of equilateral triangle.

$q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC$

$q_1=-8.2\times 10^{-9} C$

$q_2=-16.4\times 10^{-9} C$

$q_3=8.0\times 10^{-9} C$

Side of equilateral triangle =3.3 cm=$\frac{3.3}{100}=0.033m$

We know that each angle of equilateral angle=$60^{\circ}$

Net force=F =$\sum\frac{kQq }{d^2}$

Where k=$9\times 10^9 Nm^2/C^2$

If we bisect the angle at $q_3$ then we have 30 degrees from there to either charge.

Direction of vertical force  due to charge $q_1$ and $q_2$

Therefore, force will be added

Vertical  force=$9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})$

Vertical net force=$9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}$

Vertical  force =$9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}$

Vertical  force=$-1.41\times 10^{-3}N$ (towards $q_1}$

Horizontal component are opposite in direction then will b subtracted

Horizontal force=$9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}$

Horizontal force=$0.27\times 10^-3}$ N(towards $q_2$

Net electric force acting on particle 3 due to particle =$\sqrt{F^2_x+F^2_y}$

Net force=$\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}$

Net force=$1.44\times 10^{-3}$ N