Question

Three charges each of q are arranged at three corners of an equilateral triangle of side a. The electric field at the mid point of one side

• A. $\frac{1}{3\pi\epsilon_0}\frac{q}{a^2}$
• B. $\frac{1}{4\pi\epsilon_0}\frac{q}{a^2}$
• C. $\frac{1}{4\pi\epsilon_0}\frac{q}{a^2}\sqrt{3}$
• D. Zero

Answer 1

Answer: (A)

$\frac{1}{3\pi\epsilon_0}\frac{q}{a^2}$

Answer 2

Let the equilateral triangle be ABC with side length 'a'. Let charges +q be placed at each vertex A, B, and C. We want to find the electric field at M, the midpoint of side BC.

1. Electric field at M due to charge at B:
   The distance BM is $a/2$. The electric field $\vec{E}_B$ at M due to the charge at B is directed along the line segment BM, away from B. The magnitude is $E_B = \frac{1}{4\pi\epsilon_0}\frac{q}{(a/2)^2} = \frac{1}{4\pi\epsilon_0}\frac{4q}{a^2}$. The direction is along BC, pointing from B to C.

2. Electric field at M due to charge at C:
   The distance CM is $a/2$. The electric field $\vec{E}_C$ at M due to the charge at C is directed along the line segment CM, away from C. The magnitude is $E_C = \frac{1}{4\pi\epsilon_0}\frac{q}{(a/2)^2} = \frac{1}{4\pi\epsilon_0}\frac{4q}{a^2}$. The direction is along CB, pointing from C to B.

3. Vector sum of electric fields due to B and C:
   The vectors $\vec{E}_B$ and $\vec{E}_C$ have the same magnitude ($E_B = E_C$) but are in opposite directions along the line BC. Therefore, their vector sum is zero: $\vec{E}_B + \vec{E}_C = \vec{0}$.

4. Electric field at M due to charge at A:
   The distance AM is the altitude of the equilateral triangle from vertex A to the midpoint M of the opposite side BC. The length of the altitude is $h = \frac{\sqrt{3}}{2}a$.
   The electric field $\vec{E}_A$ at M due to the charge at A is directed along the line segment AM, away from A. The magnitude is $E_A = \frac{1}{4\pi\epsilon_0}\frac{q}{(AM)^2} = \frac{1}{4\pi\epsilon_0}\frac{q}{(\frac{\sqrt{3}}{2}a)^2} = \frac{1}{4\pi\epsilon_0}\frac{q}{\frac{3}{4}a^2} = \frac{1}{4\pi\epsilon_0}\frac{4q}{3a^2}$. The direction is along AM, pointing from A towards M, which is perpendicular to BC.

5. Total electric field at M:
   The total electric field at M is the vector sum of the fields due to the three charges: $\vec{E}_M = \vec{E}_A + \vec{E}_B + \vec{E}_C$.
   Since $\vec{E}_B + \vec{E}_C = \vec{0}$, the total electric field is $\vec{E}_M = \vec{E}_A$.
   The magnitude of the total electric field at M is $E_M = E_A = \frac{1}{4\pi\epsilon_0}\frac{4q}{3a^2}$.

Comparing this magnitude with the given options: $\frac{1}{4\pi\epsilon_0}\frac{4q}{3a^2} = \frac{4}{3} \cdot \frac{1}{4\pi\epsilon_0} \frac{q}{a^2} = \frac{1}{3\pi\epsilon_0}\frac{q}{a^2}$.