Question

3charges each of 1ucoloum are fixed of vertices of an equilateral triangle of side1cm. The force on a unit test charge kept at centroid of the triangle will be
0
27kN
9kN
4N

Answer 1

It is given that three charges, each of 1μC are fixed of vertices of an equilateral triangle of side 1 cm.

we have to find the force on a unit test charge kept at centroid of the triangle.

distance between centroid and each vertex, r = 1/√3 cm

so, F = kQq/r²

here Q = 10^-6 C and q = 1 C

so, F₁ = (9 × 10^9 × 10^-6 × 1)/(1/√3)²

= (900)/(1/3) = 2700 N

similarly, F₂ = F₃ = 2700 N

angle between F₁ and F₂ is 120° so, F = √{F₁² + F₂² + 2F₁F₂cos120°}

= 2700N directed opposite direction of F₃ as shown in figure.

so, Fnet = 2700N - 2700N = 0

therefore net force on a unit test charge kept at centroid of the triangle would be zero.

Answer 2

Answer:

distance between centroid and each vertex, r = 1/√3 cm

so, F = kQq/r²

here Q = 10^-6 C and q = 1 C

so, F₁ = (9 × 10^9 × 10^-6 × 1)/(1/√3)²

= (900)/(1/3) = 2700 N

similarly, F₂ = F₃ = 2700 N

angle between F₁ and F₂ is 120° so, F = √{F₁² + F₂² + 2F₁F₂cos120°}

= 2700N directed opposite direction of F₃ as shown in figure.

so, Fnet = 2700N - 2700N = 0