Problem 24 - A Complete Resource Book in Physics for JEE Main

Question

\(A B C\) is an equilateral triangle. Charges \(+q\) are placed at each corner. The electric field intensity at centroid \(O\) will be (A) \(\frac{1}{2 \pi \varepsilon_{0}} \times \frac{q}{r^{2}}\) (B) \(\frac{1}{2 \pi \varepsilon_{0}} \times \frac{3 q}{r^{2}}\) (C) \(\frac{1}{2 \pi \varepsilon_{0}} \times \frac{\sqrt{3} q}{r^{2}}\) (D) Zero

Step-by-Step Solution

Answer

The net electric field intensity at the centroid O of the equilateral triangle is zero, as the electric fields due to charges at vertices A, B, and C cancel each other out (Option D).

Step 1: Understand the properties of equilateral triangle and centroid

An equilateral triangle is a triangle in which all three sides are equal in length, and all three angles measure 60 degrees. The centroid O is the point where the medians (the line segment that connects each vertex to the midpoint of the opposite side) of the triangle intersect. In an equilateral triangle, the centroid, circumcenter, and the incenter coincide, resulting in all medians having equal lengths.

Step 2: Calculate distance and angles

Let's call the vertices A, B, and C. Since O is the centroid of the equilateral triangle, the distance from O to each vertex A, B, and C is equal. We'll call this distance r. Given that each angle in an equilateral triangle is 60 degrees, the angle between two consecutive electric field vectors, for example, E_A (from charge at A) and E_B (from charge at B), will be 120 degrees (180° - 60°).

Step 3: Calculate the electric field of individual charges at the centroid

Let's focus on finding the electric field due to the charge +q at vertex A. Using Coulomb's law, the intensity of the electric field E_A at the centroid O can be represented as: \[E_A = \frac{kq}{r^2}\] where k is the Coulomb constant, \(k = \frac{1}{4 \pi \varepsilon_0 }\), and r is the distance from A to O. Similar expressions can be written for E_B and E_C.

Step 4: Find the net electric field at the centroid O

To find the total electric field at O, we'll focus on adding the electric field vectors \(\vec{E_A}\), \(\vec{E_B}\), and \(\vec{E_C}\). Since the angle between any two consecutive electric field vectors is 120 degrees, we can use the vector addition formula: \[\vec{E_{total}} = \vec{E_A} + \vec{E_B} + \vec{E_C}\] Notice that the three vectors form a closed triangle with internal angles of 120 degrees each. This means that the sum of the three vectors is zero. Therefore, the net electric field at the centroid O is: \[\vec{E_{total}} = 0\] Hence, the correct answer is: (D) Zero