Exams › JEE Main › Physics

A small ball of mass m carrying charge +q is attached to a string of length L and whirled in a vertical circle. In addition to gravity, a uniform horizontal electric field E acts on it. At what angular position (measured from a suitable reference) is the tension in the string minimum?

1. theta = arctan(qE/(mg))
2. theta = pi
3. theta = pi - arctan(qE/(mg))
4. theta = pi + arctan(qE/(mg))

Correct answer: theta = pi - arctan(qE/(mg))

Solution

The constant forces are gravity mg (downward) and electric force qE (horizontal); their resultant defines an effective gravity g_eff pointing at angle arctan(qE/mg) from the vertical. Tension is least at the point of the circle diametrically opposite to the direction of g_eff (the effective 'top'). Measuring the angle from the downward vertical, this top position is at theta = pi - arctan(qE/mg).

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 →