Given below are two statements:
Statement I: Charges \(q\), \(q\) and \(-q\) are placed at the corners of an equilateral triangle. The vector sum of the net forces on the three charges is zero.
Statement II: In this arrangement, the net force on the charge \(-q\) has the same magnitude as the net force on each charge \(q\).
In the light of the above statements, choose the most appropriate answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Statement I is correct but Statement II is incorrect
Each pair of charges is at the same distance \( l \), so each pair exerts forces of the same size \( F=q^2/4\pi\varepsilon_0l^2 \). The net force on a charge is the vector sum of the two forces on it.
Statement I Charges \( q \), \( q \) and \( -q \) are placed at the corners of an equilateral triangle. The vector sum of the net forces on the three charges is zero. — True. Coulomb’s law agrees with Newton’s third law. The force of one charge on another is equal and opposite to the force of the second on the first. So all the forces add to zero in pairs, and the sum of the three net forces is zero.
Statement II In this arrangement, the net force on the charge \( -q \) has the same magnitude as the net force on each charge \( q \). — False. On each charge \( q \), one force is a repulsion and the other an attraction. These two forces of size \( F \) are at \( 120^\circ \) to each other, so their resultant is \( F \). On \( -q \), both forces are attractions at \( 60^\circ \) to each other. Their resultant is larger:
\( \begin{aligned} F_3 &= 2F\cos30^\circ \\ &= \sqrt{3}\,F \end{aligned} \)
NCERT (2026-27):
Example 1.6 – “… The total force F2 on charge q at B is thus F2 = F r̂2, where r̂2 is a unit vector along AC.”
“Similarly the total force on charge –q at C is F3 = √3 F n̂, where …”
Example 1.6 – “It is interesting to see that the sum of the forces on the three charges is zero, i.e., F1 + F2 + F3 = 0”
“The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. …”