Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main202621 January 2026Evening ShiftPhysicsElectrostaticsActual

Consider two identical metallic spheres of radius R each having charge Q and mass m . Their centers have an initial separation of 4 R . Both the spheres are given an initial speed of u towards each other. The minimum value of u , so that they can just touch each other is : (Take k= 1 4 ₀ and assume k Q²>G m² where G is the Gravitational constant)

Options

  1. Ak Q² 2 m R (1- G m² k Q² )
  2. Bk Q² 4 m R (1- G m² k Q² )
  3. Ck Q² 2 m R (1- G m² 2 k Q² )
  4. Dk Q² 4 m R (1+ G m² k Q² )

Correct answer

B. k Q² 4 m R (1- G m² k Q² )

Step-by-step solution

Using energy conservation, the system's kinetic energy must overcome the increase in electrostatic potential energy as the spheres approach. Initial state: separation 4R, speed u each. Final state: touching (separation 2R), speed zero. Change in potential energy: U = k Q^2 2R - k Q^2 4R = k Q^2 4R . Initial kinetic energy: KE_i = mu^2 . Energy equation: mu^2 = k Q^2 4R (1 - Gm^2 kQ^2 ) . Therefore, u = kQ^2 4mR (1 - Gm^2 kQ^2 ) .

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 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 ele 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 2026 Full Electrostatics list All JEE Main PYQs