Question

three charges each of 1 microcoulomb are fixed on vertices of an equilateral triangle of side 1 cm.the force on a unit test charge kept on the centroid of the triangle

Answer 1

Given :

The charge of each charged particle = 1μC

Length of side of  triangle = 1cm

To Find :

The force on  particle present at centroid of the triangle

Solution :

• The distance between centroid and the each  of equilateral triangle is  a/√3

         r = 1/√3

• The force between two particles is    $F=\frac{kQq}{r^{2} }$

            $F_{1} =\frac{9\times10^{9}\times10^{-6}}{\frac{1}{\sqrt{3} ^{2}}}$

            $F_{1} = \frac{900}{\frac{1}{3} }$

            $F_{1} =2700N$

• Similarly F₂ = 2700N , F₃ = 2700 N

• As it is equilateral triangle the angle between F₁ and F₂ is 120°

• The resultant of F₁ and F₂  =  $\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos\theta }$

                                                     = $\sqrt{2F_{1}^{2}-F_{1}^{2}}$

                                                     = $F_{1}=2700N$

• Net force = 2700 - 2700 N = 0N

The net force on a unit charge present at the centroid of triangle is zero.