3D Forces on a Charge at the Centroid
Interactive 3D Simulation
This simulation demonstrates the forces acting on a charge (Q) placed at the centroid of an equilateral triangle with three identical charges (q) at its vertices.
Example
Question:
Consider three charges \( q_1, q_2, q_3 \), each equal to \( q \), at the vertices of an equilateral triangle of side \( l \). What is the force on a charge \( Q \) (with the same sign as \( q \)) placed at the centroid of the triangle, as shown in Fig. 1.6?
Solution:
In the given equilateral triangle \( ABC \) of sides of length \( l \), if we draw a perpendicular \( AD \) to the side \( BC \), \( AD = AC \cos 30^{\circ} = (\sqrt{3}/2)l \) and the distance \( AO \) of the centroid \( O \) from \( A \) is \((2/3)AD = (1/\sqrt{3})l \). By symmetry, \( AO = BO = CO \).
Thus,
Force \( \mathbf{F}_1 \) on \( Q \) due to charge \( q \) at \( A \) is
\[
\frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2}
\]
along \( AO \).
Force \( \mathbf{F}_2 \) on \( Q \) due to charge \( q \) at \( B \) is
\[
\frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2}
\]
along \( BO \).
Force \( \mathbf{F}_3 \) on \( Q \) due to charge \( q \) at \( C \) is
\[
\frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2}
\]
along \( CO \).
By symmetry and parallelogram law, the total force on \( Q \) is
\[
\frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2} (\mathbf{\hat{r}} - \mathbf{\hat{r}})
= 0,
\]
where \( \mathbf{\hat{r}} \) is the unit vector along \( OA \).
So, the three forces sum to zero by symmetry.
Results
Explanation
The example shows three identical charges (q) at the vertices of an equilateral triangle and a charge (Q) at the centroid.
Key Concepts:
Coulomb's Law: The force between two point charges is given by: F = k·q₁·q₂/r² where k = 1/(4πε₀) ≈ 9×10⁹ N·m²/C²
Centroid Distance: In an equilateral triangle with side length l, the distance from any vertex to the centroid is: r = l/√3
Symmetry Principle: Due to the symmetrical arrangement:
- Each vertex charge exerts a force of equal magnitude on the central charge
- The force vectors are at 120° angles to each other
- The vector sum of all three forces is zero



