Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20236 Apr 2023Evening ShiftPhysicsElectrostaticsActual

A dipole comprises of two charged particles of identical magnitude q and opposite in nature. The mass m of the positive charged particle is half of the mass of the negative charged particle. The two charges are separated by a distance l . If the dipole is placed in a uniform electric field E → ; in such a way that dipole axis makes a very small angle with the electric field, E → . The angular frequency of

Options

  1. A3 q E 2 m l
  2. B8 q E m l
  3. C4 q E m l
  4. D8 q E 3 m l

Correct answer

A. 3 q E 2 m l

Step-by-step solution

The data given is l = distance between the charges q = charge m = mass of the positive charge 2 m = mass of negative charge The moment of inertia of the system is I = m × 2 m m + 2 m l 2 = 2 m l 2 3 The angular frequency can be written as ω = p E I       . . . ( i ) Substituting the value of moment of inertia in the equation (i) we get ω = p E 2 m l 2 3 = 3 p E 2 m l 2 The value of the dipole moment is p = q l Hence, the angular frequency becomes ω = 3 q E 2 m l This question was

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