Question

if three identical charge are kept on vertex of equilateral triangle then find the electric field at its centoid.

Answer 1

Let q be the charge and L be the length of the side of triangle. Then the distance to the center is $L\cdot \sin60^\circ \cdot\dfrac23 = L\cdot \dfrac{\sqrt{3}}{2} \cdot\dfrac23 = \dfrac{L}{\sqrt{3}}$ . We know that according to the Coulomb's law the force between two charges is

$\vec{F}_{12} = k_e \cdot\dfrac{q\cdot q}{r_{12}^3}\cdot\vec{r}_{12}$ and $\vec{E}_{12} = k_e \cdot\dfrac{q}{r_{12}^3}\cdot\vec{r}_{12}.$

We should find the vector sum of vectors E in the centroid:

$\vec{E} = k_e \cdot\dfrac{q}{r_{1}^3}\cdot\vec{r}_{1} + k_e \cdot\dfrac{q}{r_{2}^3}\cdot\vec{r}_{2} + k_e \cdot\dfrac{q}{r_{3}^3}\cdot\vec{r}_{3} = \dfrac{k_eq}{(L/\sqrt3)^2}\cdot \left( \vec{r}_{1} +\vec{r}_{2} +\vec{r}_{3} \right),$

where vectors $\vec{r}_1, \vec{r}_2, \vec{r}_3$ are directed from the vertices of triangle to the centroid. Due to the symmetry of triangle, this sum will be equal to 0, so the electric field is 0.