Triangle Charges Force Analysis

Let's consider an equilateral triangle with side length \(a\). Three point charges, each of magnitude \(q\), are placed at the corners of the triangle. Let these corners be A, B, and C. Another point charge, \(-Q\), is placed at the centroid G of the triangle.

We are given that the net force on each of the charges \(q\) at the corners vanishes. Let's analyze the forces acting on the charge \(q\) placed at corner A.

There are three forces acting on the charge at A:

1. The force due to the charge \(q\) at corner B, let's call it \(\vec{F}_{AB}\). This force is repulsive and acts along the line AB. Its magnitude is given by Coulomb's law: \[ F_{AB} = \frac{1}{4\pi\epsilon_0} \frac{|q \cdot q|}{a^2} = \frac{k q^2}{a^2} \] where \(k = \frac{1}{4\pi\epsilon_0}\).
2. The force due to the charge \(q\) at corner C, let's call it \(\vec{F}_{AC}\). This force is repulsive and acts along the line AC. Its magnitude is: \[ F_{AC} = \frac{k q^2}{a^2} \] Since it's an equilateral triangle, \(F_{AB} = F_{AC}\). The angle between these two force vectors, \(\vec{F}_{AB}\) and \(\vec{F}_{AC}\), is \(60^\circ\).
3. The force due to the charge \(-Q\) at the centroid G, let's call it \(\vec{F}_{AG}\). This force is attractive and acts along the line AG, directed from A towards G. The distance from a corner to the centroid in an equilateral triangle with side \(a\) is \(R = \frac{a}{\sqrt{3}}\). The magnitude of this force is: \[ F_{AG} = \frac{k |q \cdot (-Q)|}{R^2} = \frac{k q Q}{(a/\sqrt{3})^2} = \frac{k q Q}{a^2/3} = \frac{3k q Q}{a^2} \]

The resultant force on the charge \(q\) at A is the vector sum of these three forces: \(\vec{F}_{net, A} = \vec{F}_{AB} + \vec{F}_{AC} + \vec{F}_{AG}\).

First, let's find the resultant of \(\vec{F}_{AB}\) and \(\vec{F}_{AC}\). Since \(|\vec{F}_{AB}| = |\vec{F}_{AC}| = \frac{k q^2}{a^2}\) and the angle between them is \(60^\circ\), the magnitude of their resultant, \(\vec{F}_{BC\_resultant}\), is:

The direction of this resultant force \(\vec{F}_{BC\_resultant}\) is along the angle bisector of the \(60^\circ\) angle between \(\vec{F}_{AB}\) and \(\vec{F}_{AC}\). This bisector is the median of the triangle from A to the midpoint of BC. The centroid G lies on this median.

The force \(\vec{F}_{AG}\) due to the charge \(-Q\) at the centroid acts along the line AG. The direction of \(\vec{F}_{BC\_resultant}\) is radially outwards along the median from A, while the direction of \(\vec{F}_{AG}\) is radially inwards along the line AG (towards G).

For the net force on charge \(q\) at A to be zero, the vector sum \(\vec{F}_{BC\_resultant} + \vec{F}_{AG}\) must be zero. Since these two forces act along the same line but in opposite directions, their magnitudes must be equal:

Assuming \(q \neq 0\), we can cancel \(k q / a^2\) from both sides:

We need to find the ratio \(Q/q\). Rearranging the equation:

We can simplify this ratio by rationalizing the denominator or by recognizing that \(3 = \sqrt{3} \times \sqrt{3}\):

This is the required ratio for the force on each corner charge to vanish. The same logic applies to the charges at corners B and C due to the symmetry of the equilateral triangle arrangement.

The resultant of \(F_{AB}\) and \(F_{AC}\) acts along the median from A, radially outwards. For the net force to be zero, \(F_{AG}\) must balance this resultant force.

Final ratio found is \(\frac{Q}{q} = \frac{1}{\sqrt{3}}\).