Three point charges + q, + q and -2q are placed at the corner of an equilateral triangle of side 'a'. The net dipole moment of the system is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( qa\sqrt{3} \)
The total charge is zero, so the system can be split into two dipoles and their moments added as vectors.
Split the charge \( -2q \) into two charges \( -q \) at the same corner. Each \( -q \) pairs with one \( +q \), giving two dipoles of moment \( qa \) each (charge \( q \), separation \( a \)). Both start at the \( -2q \) corner and point along the two sides of the triangle that meet there, so the angle between them is \( 60^\circ \).
\( \displaystyle p = \sqrt{p_1^{2} + p_2^{2} + 2p_1 p_2\cos 60^\circ} = \sqrt{q^{2}a^{2} + q^{2}a^{2} + q^{2}a^{2}} = \sqrt{3}\,qa \)