Suppose three equal charges, each equal to +q, are placed at the vertices of an equilateral triangle of side 1, then the force exerted on a charge Q (with the same sign as q) placed at the centroid of the triangle is:

Question

Question

This question was previously asked in

AAI ATC Junior Executive 27 July 2022 Shift 3 Official Paper

Answer 1

Concept:

Coulomb's Law:

According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges
And inversely proportional to the square of the distance between them.
It acts along the line joining the two charges considered to be point charges.

Coulomb's Formula

F =

In vector form, , where ϵ₀ = permittivity in the air, r = magnitude of the distance between two charge

Calculation:

Here, l = 2 unit

From the above diagram, AD is perpendicular to the BC

The distance AO of the centroid O from A

The force on Q at O due to q₁ = q at A

along AO

The force on Q at O due to q₂ = q at B

along BO

The force on Q at O due to q₃ = q at C

along CO

The angle between F₂ and F₃ is 120º.

The resultant of F2 and F₃ is along AO.

The force on charge Q is,

Hence, the force at the centroid is zero.