Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20265 April 2026Evening ShiftPhysicsElectrostaticsActual

A particle of charge q and mass m is projected from origin with an initial velocity v = ( v₀ 2 x + v₀ 2 y ) . There exists a uniform magnetic field B = B₀ z and a space varying electric field E = E₀ e^ - x x within the region 0 x L . After travelling a distance such that x -coordinate has changed from x=0 to x=L , the change in the kinetic energy is _______.

Options

  1. AqE₀ [1 - e^ - L ]
  2. B( v₀ q B₀ 2 )[2 - e^ -2 L ]
  3. CqE₀ [1 + e^ - L ]
  4. Dq ( E₀ + v₀ B₀ )[1 - e^ - L/2 ]

Correct answer

A. qE₀ [1 - e^ - L ]

Step-by-step solution

According to the work-energy theorem, the change in kinetic energy of a particle is equal to the total work done by all the forces acting on it. The forces acting on the particle are the electric force and the magnetic force. The magnetic force is given by F _B = q( v B ) . Since the magnetic force is always perpendicular to the velocity vector of the particle, the work done by the magnetic field is zero. The change in kinetic energy is equal to the work done by the electric field alone. The electric force is F _E

Practice Electrostatics on Quantrex Academy →

More from Electrostatics

Two charges Q₁ = q and Q₂ = mq are placed at the points P₁(a, b) and P₂(ma, mb) , respectively, in the XY plane, where a, b 0 and m 0, 1 . If V₁ is the potential at a point in the 2026Consider an electric dipole comprising two charges +q and -q each with mass m , separated by a fixed distance d and initially at rest with its dipole moment pointing along i . A un 2026Two point charges q₁=3 , C and q₂=-4 , C are placed at points (2 i +3 j +3 k ) and ( i + j + k ) respectively. Force on charge q₂ is ________ N. ( Take 1 4 ₀ = 9 10^9 SI Units ) 2026The electric potential as a function of x, y is given by V = 5(x^2 - y^2) V. The electric field at a point (2, 3) m is __________ V/m. 2026A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb. If the magnitude of the electric field at centre of the half ring is 100 V/m, then the value 2026A three coulomb charge moves from the point (0, -2, -5) to the point (5, 1, 2) in an electric field expressed as E = 2x i + 3y^2 j + 4 k N/C. The work done in moving the charge is 2026A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field E ₁=E₀ x . If another electric field E ₂=2E₀( y + z ) is introduced to the s 2026Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: In electrostatics, a conductor does not store any net charge insid 2026 Full Electrostatics list All JEE Main PYQs