Question

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)

Short answer

The value of the fourth charge q, when placed at the center of an equilateral triangle with three charges of value Q at its vertices such that all charges remain stationary, is \(q = -\frac{Q}{\sqrt{3}}\).

Step-by-step solution

Analyzing the forces acting on each charge

Let's denote the 3 charges at the vertices of the triangle as A, B, and C, and the charge at the center as D. Now let's analyze the forces acting on them: - Charge at A experiences forces from charges B, C, and D. - Charge at B experiences forces from charges A, C, and D. - Charge at C experiences forces from charges A, B, and D. - Charge at D experiences forces from charges A, B, and C. The objective is to find the value of charge q, at point D, such that all charges remain stationary.

Using Coulomb's Law

Coulomb's Law states that the force F between two point charges is given by: \(F = k \frac{q_1 q_2}{r^2}\), where q1 and q2 are charges, r is the distance between them, and k is the electrostatic constant. First, let's examine the forces acting on charge A: 1. The force between A and D, facing in the AD direction 2. The force between A and B, acting in the AB direction 3. The force between A and C, acting in the AC direction Since we have an equilateral triangle with equal charges, the angle between AB and AC is 60 degrees. The force between A and D (F_AD) should balance the net force from charges B and C (F_BC) such that all charges remain stationary. This balancing can be expressed as: \(F_{AD} = F_{BC}\), or combining the Coulomb's law equations for each force: \(\frac{kQq}{r^2} = \frac{kQ^2}{r^2}(\frac{1}{2}) + \frac{kQ^2}{r^2}\ cos(60°)\) (considering the angle between the vectors), where r is the distance between A and D.

Solving the equation for q

Now, we can simplify the equation and solve for q: \(\frac{kQq}{r^2} = \frac{kQ^2}{r^2}\left(\frac{1}{2} + \frac{1}{2}\right)\) Cancelling the constants and r^2 terms, we get: \(q = Q\) Now let's calculate the distance between charges A and D (r), and the side length of the equilateral triangle (l): \(r = \frac{\sqrt{3}}{2}l\) Now, substituting this back into the equation for Coulomb's Law: \(\frac{2Qq}{l^2} = Q\left(\frac{1}{2} + \frac{1}{2}\right)\) Solving for q: \(q = -\frac{Q}{\sqrt{3}}\) Thus, the correct answer is (B) \(q = -\frac{Q}{\sqrt{3}}\).