Question

Three point charges each of value + q are placed on three vertices of a square of side a metre. What is the magnitude of the force on a point charge of value -q coulomb placed at the centre of the square?

Answer 1

The force will be of magnitude (1/2π∈)(q²/a²) towards the vertex A of the square (given in figure).

Three point charges each of value + q are placed on three vertices of a square of side a meter.

• As shown in the figure let the charges +q be kept at the vertices A, B and D.
• The charge -q is kept at the center.
• There will be a force of attraction between the three positive charges at the vertices and the negative charge at the center.
• This force of attraction is given by :

        F = (1/4π∈)(q²/r²)      where r is the distance between the two charges

        As the charge -q is placed at the center of the square with side a, the distance between the two charges will be $\frac{a}{\sqrt{2} }$ .

• As we can see in the figure, the forces due to charge at point B and charge at point D are equal and opposite in nature. Hence they will cancel out each other.
• So the resultant force will be due to charge at point A.
• F = (1/4π∈)[q²/(a/√2)²]

        F = (1/2π∈)[q²/a²]  towards point A