Exams › JEE Main › Physics

Three point charges 4q, Q and q lie on a straight line at positions x = 0, x = L/2 and x = L respectively. For the net force on the charge q to be zero, the value of Q must be:

1. -q
2. -2q
3. -q/2
4. 4q

Correct answer: -q

Solution

Force on q from 4q at distance L is k(4q)(q)/L². Force from Q at distance L/2 is kQq/(L/2)² = 4kQq/L². For balance these must cancel: 4q + 4Q = 0, so Q = -q (attractive, opposite sign).

Related JEE Main Physics questions

• What is the SI unit of the permittivity of free space, ε₀?
• What is the SI unit used for electric flux?
• A constant electric field E is directed along the positive x-axis. If a charge of 0.5 C is displaced by 2 m along a line inclined at 60° to the x-axis, and the work done in this process is 10 J, what is the magnitude of the electric field?
• An electric dipole is located at the origin O along the x-axis. A point P lies 20 cm from O, and the line OP makes an angle π/3 with the x-axis. If the electric field at P is inclined at an angle θ to the x-axis, what is θ?
• A uniformly charged semicircular wire has radius a and linear charge density λ. The magnitude of the electric field at the centre of the semicircle is
• A point Q is located on the perpendicular bisector of an electric dipole whose dipole moment is p. If Q is at a distance r from the dipole and r is much greater than the dipole size, the electric field at Q varies as

⚔️ Practice JEE Main Physics free + battle 1v1 →