Question

Q 3/60 Three charges each of 1 uC are fixed on vertices of an
equilateral triangle of side 1 cm. The force on a unit test
charge kept at centroid of the triangle will be
Zero
27 kN
o 9 KN
4N​

Answer 1

Given :

The charge of each particle = 1μC

Length of side of a triangle = 1cm

To Find :

The force on charged particle present at centroid of the triangle

Solution :

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

          r = 1/√3

• The force between two particles is  F = kQq/r²

          F₁ =9×10⁹×10⁻⁶×1 / (1/√3)²

          F₁ = 900 / (1/3)

          F₁ = 2700 N

• 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  =  √(F₁² + F₂² + 2F₁F₂ Cos 120°)

                                           = √(2F₁² - F₁²)  

                                           = F₁ = 2700 N

• Net force = 2700 - 2700 N = 0N

The net force on a unit charge present on the centroid of triangle os zero