Question

Two charges \( -q \) each are placed at the vertices A and B of an equilateral triangle ABC. If M is the mid-point of AB, the net electric field at C will point along

• A. CA
• B. CB
• C. MC
• D. CM

Hint:

Fields of negative charges point toward the charges. The two equal fields at C add along the bisector, toward M.

Answer 1

The Correct Option is D

Step 1: Understand the set-up:
Charges \(-q\) sit at A and B. C is the third vertex, and M is the mid-point of AB. In an equilateral triangle, CM is perpendicular to AB, and it bisects the angle ACB.

Step 2: Direction of each field at C:
The field of a negative charge points toward the charge. So the field at C due to the charge at A points along CA. The field at C due to the charge at B points along CB.
Both have the same magnitude \(E = kq/a^2\) because AC = BC = a.

Step 3: Add the two fields:
The two fields make an angle of \(60^{\circ}\) with each other, the angle of the triangle at C. Equal vectors add along the bisector of the angle between them. The resultant is
\[ E_{net} = 2E\cos 30^{\circ} = \sqrt{3}\,E \]
It points along the bisector of angle ACB, which is the line from C toward M. That direction is CM.

Step 4: Why the other options are wrong:
(A) CA and (B) CB are the directions of only one of the two fields each, so neither can be the resultant. (C) MC points from M to C, away from the charges. That would be right if both charges were positive, since a positive charge pushes the field away from itself.

Final Answer:
The net field at C points along CM. This is option (D). \[ \boxed{\text{(D) CM}} \]