Question

Four equal free positive charges e are at the corners a square of side a. What should be the charge placed at the centre of the square for the system to be in equilibrium? Is the equilibrium stable?

Answer 1

Answer: A negative charge q = e(1 + 2√2)/4; the equilibrium of the system is unstable.

Explanation:

On each charge situated at the vertices of the square there act four forces (Fig.): the two forces F₁ = F₂ = e²/a², due to the charges at vertices 1 and 2; the force F₃ = e²/2a², due to the charge at vertex 3; and the force F₄ = 2qe/a², due to charge q.

For equilibrium, it is necessary that the vector sum of these forces be zero, i.e.

To determine the nature of the equilibrium of the system, it is sufficient to consider a small displacement of one of the charges and estimate the character of the change in the magnitude of the forces due to the other charges. For the sake of simplicity, we shall consider a small displacement s of one of the charges +e along a diagonal in the direction away from the centre of the square (Fig.).

Since the distance from this charge to charge q is the smallest, then the displacement s produces a greater relative change in the distance to charge q than the relative change in the distance to the other charges. Consequently, in the case of the displacement s the force F₄ decreases to a greater extent than the forces F₁, F₂ , F₃. Moreover, the displacement s leads to a decrease in the angle between forces F₁ and F₂ due to charges 1 and 2. This decrease in angle leads to an increase in the resultant of forces F₁ and F₂ . Thus, in the new position of the charge, the force F₄ will definitely be less than the vector sum of forces F₁ + F₂ + F₃.

In the new position there will be a resultant force directed away from the equilibrium position.