Question

Ex. Given an equilateral triangle with side $l$. Find electric field strength at the centroid. The linear charge density is as shown in figure.

• A. $\frac{3\lambda}{\pi\epsilon_0 l}$
• B. Zero
• C. $\frac{\sqrt{2}\lambda}{\pi\epsilon_0 l}$
• D. $\frac{\lambda}{\pi\epsilon_0 l}$

Answer 1

Answer: (B)

Zero

Answer 2

The electric field at the centroid of an equilateral triangle with uniformly distributed charge on its sides can be calculated by considering the contribution from each side. Let the side length of the equilateral triangle be $l$ and the linear charge density on each side be $\lambda$. The centroid is equidistant from each side. The distance from the centroid to each side is the inradius of the equilateral triangle, which is $r = \frac{l}{2\sqrt{3}}$.

Consider one side of the triangle. The centroid is located on the perpendicular bisector of this side. The electric field at a point P at a perpendicular distance $d$ from the midpoint of a uniformly charged rod of length $l$ with linear charge density $\lambda$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{d} \sin\theta$, where $2\theta$ is the angle subtended by the rod at the point P. In our case, the point is the centroid, the rod is a side of length $l$, and the perpendicular distance is $d = r = \frac{l}{2\sqrt{3}}$. The angle subtended by each half of the side at the centroid is $60^\circ$, so $2\theta = 120^\circ$ and $\theta = 60^\circ$.

The magnitude of the electric field at the centroid due to one side is $E_1 = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{r} \sin 60^\circ = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{l/(2\sqrt{3})} \frac{\sqrt{3}}{2} = \frac{1}{4\pi\epsilon_0} \frac{4\sqrt{3}\lambda}{l} \frac{\sqrt{3}}{2} = \frac{1}{4\pi\epsilon_0} \frac{6\lambda}{l} = \frac{3\lambda}{2\pi\epsilon_0 l}$.

The direction of the electric field due to a uniformly charged line segment at a point on its perpendicular bisector is along the perpendicular bisector, pointing away from the line if the charge is positive.

Let the three sides be $S_1$, $S_2$, and $S_3$. The electric fields at the centroid due to these sides are $\vec{E}_1$, $\vec{E}_2$, and $\vec{E}_3$. The magnitude of each field is $E_1$. The direction of each field is perpendicular to the corresponding side and points away from the side. The perpendiculars from the centroid to the sides are along the lines joining the centroid to the midpoints of the sides. These lines make angles of $120^\circ$ with each other.

Let's place the centroid at the origin. Let the direction of $\vec{E}_1$ be along the positive y-axis. Then its direction vector is $(0, 1)$. The directions of $\vec{E}_2$ and $\vec{E}_3$ are at angles of $120^\circ$ and $240^\circ$ with the direction of $\vec{E}_1$. So, the direction vector of $\vec{E}_2$ makes an angle of $90^\circ + 120^\circ = 210^\circ$ with the positive x-axis, or an angle of $120^\circ$ with the positive y-axis. Its components are $E_1 \cos(210^\circ) = -E_1 \cos 30^\circ = -E_1 \frac{\sqrt{3}}{2}$ and $E_1 \sin(210^\circ) = -E_1 \sin 30^\circ = -E_1 \frac{1}{2}$. The direction vector of $\vec{E}_3$ makes an angle of $90^\circ + 240^\circ = 330^\circ$ with the positive x-axis, or an angle of $240^\circ$ with the positive y-axis. Its components are $E_1 \cos(330^\circ) = E_1 \cos 30^\circ = E_1 \frac{\sqrt{3}}{2}$ and $E_1 \sin(330^\circ) = -E_1 \sin 30^\circ = -E_1 \frac{1}{2}$.

The total electric field is the vector sum: $\vec{E} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3$ $\vec{E} = (0, E_1) + (-\frac{\sqrt{3}}{2}E_1, -\frac{1}{2}E_1) + (\frac{\sqrt{3}}{2}E_1, -\frac{1}{2}E_1)$ $\vec{E} = (0 - \frac{\sqrt{3}}{2}E_1 + \frac{\sqrt{3}}{2}E_1, E_1 - \frac{1}{2}E_1 - \frac{1}{2}E_1)$ $\vec{E} = (0, E_1 - E_1) = (0, 0)$

Thus, the total electric field strength at the centroid is zero. This result is expected due to the symmetry of the equilateral triangle and the uniform charge distribution on its sides. The electric fields due to the three sides cancel each other out vectorially at the centroid.