Question

Three points charges are placed at the corners of an equilateral triangle of side L as shown in the figure.

• The potential at the centroid of the triangle is zero.
• The electric field at the centroid of the triangle is zero.
• The dipole moment of the system is $\sqrt{2}qL$.
• The dipole moment of the system is $\sqrt{3}qL$.

• A. The potential at the centroid of the triangle is zero.
• B. The electric field at the centroid of the triangle is zero.
• C. The dipole moment of the system is $\sqrt{2}qL$.
• D. The dipole moment of the system is $\sqrt{3}qL$.

Answer 1

Answer: (A), (D)

The potential at the centroid of the triangle is zero. The dipole moment of the system is $\sqrt{3}qL$.

Answer 2

Let the vertices of the equilateral triangle be A, B, and C. Assume charges +q are at A and B, and -2q is at C.

Statement 1: Potential at the centroid is zero.

The distance from each vertex to the centroid is $r = \frac{L}{\sqrt{3}}$. The potential at the centroid O is:

$V_O = \frac{1}{4\pi\epsilon_0} \left( \frac{q_A}{r_A} + \frac{q_B}{r_B} + \frac{q_C}{r_C} \right)$

$V_O = \frac{1}{4\pi\epsilon_0} \left( \frac{+q}{L/\sqrt{3}} + \frac{+q}{L/\sqrt{3}} + \frac{-2q}{L/\sqrt{3}} \right) = 0$

Statement 2: Electric field at the centroid is zero.

The electric field at the centroid is the vector sum of the electric fields due to each charge.

$E_A = \frac{1}{4\pi\epsilon_0} \frac{3q}{L^2}$, directed away from A.

$E_B = \frac{1}{4\pi\epsilon_0} \frac{3q}{L^2}$, directed away from B.

$E_C = \frac{1}{4\pi\epsilon_0} \frac{6q}{L^2}$, directed towards C.

The resultant electric field is non-zero.

Statement 3 & 4: Dipole moment of the system.

The total charge of the system is 0. Place the origin at vertex C.

$\vec{P} = q_A \vec{r}_A + q_B \vec{r}_B + q_C \vec{r}_C$

$\vec{P} = q (-L/2, L\sqrt{3}/2) + q (L/2, L\sqrt{3}/2) + (-2q) (0,0)$

$\vec{P} = (0, qL\sqrt{3})$

The magnitude of the dipole moment is $|\vec{P}| = \sqrt{3} qL$.

Therefore, statements 1 and 4 are correct.