MCQMediumJEE 2025Electric Potential & Potential Energy

JEE Physics 2025 Question with Solution

A particle of mass mm and charge qq is fastened to one end AA of a massless string having equilibrium length ll, whose other end is fixed at point OO. The whole system is placed on a frictionless horizontal plane and is initially at rest. If a uniform electric field is switched on along the direction as shown in the figure, then the speed of the particle when it crosses the xx-axis is:

Coordinate axes with origin O, electric field directed along positive x-axis, and a string OA of length l initially making 60 degrees with the x-axis, with particle at A.
  • A

    qElm\sqrt{\frac{qEl}{m}}

  • B

    2qElm\sqrt{\frac{2qEl}{m}}

  • C

    qEl4m\sqrt{\frac{qEl}{4m}}

  • D

    qEl2m\frac{qEl}{2m}

Answer

Correct answer:B

Step-by-step solution

Work-Energy Method

Given: A particle of mass mm and charge qq is attached to a massless string of length ll on a frictionless horizontal plane. A uniform electric field EE acts along the positive xx-direction.

Find: The speed of the particle when it crosses the xx-axis.

The solution uses the work-energy principle.

The electric force on the particle is

Felectric=qEF_{\text{electric}} = qE

When the particle crosses the xx-axis, the displacement along the field direction is taken as ll in the provided solution.

Hence, the work done by the electric field is

W=qElW = qEl

This work appears as kinetic energy:

12mv2=qEl\frac{1}{2}mv^2 = qEl

Solving for vv,

v=2qElmv = \sqrt{\frac{2qEl}{m}}

Therefore, the speed of the particle is 2qElm\sqrt{\frac{2qEl}{m}}, so the correct option is B.

Common mistakes

  • Using the tension in the string to calculate work. Tension does no work here because it is always along the string, while the instantaneous displacement is perpendicular to it. Use only the work done by the electric force in the energy equation.

  • Substituting the wrong displacement in the work term. In a uniform electric field, work depends on displacement along the field direction, not on the curved path length. Use displacement along the xx-direction while applying W=qEΔxW = qE\,\Delta x.

  • Confusing force with potential energy change directly without writing the work-energy relation. First compute the work done by the electric field, then equate it to the change in kinetic energy.

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