Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE MainPhysicsElectrostatics

Two infinite parallel vertical plane sheets are separated by a distance d . The left sheet has a uniform positive surface charge density +3 and the right sheet has a uniform positive surface charge density + . A particle of mass m and positive charge q is released from rest very close to the left sheet. The time taken by the particle to reach the right sheet is (neglect gravity):

Options

  1. Amd ₀ q
  2. B2md ₀ q
  3. C4md ₀ 3q
  4. Dmd ₀ 2q

Correct answer

B. 2md ₀ q

Step-by-step solution

The electric field between the two charged sheets is the vector sum of the fields produced by each sheet. The field due to the left sheet (charge density +3 ) is E₁ = 3 2 ₀ directed towards the right. The field due to the right sheet (charge density + ) is E₂ = 2 ₀ directed towards the left. The net electric field in the region between the sheets is: E_ net = E₁ - E₂ = 3 2 ₀ - 2 ₀ = 2 2 ₀ = ₀ (directed towards the right). The force experienced by the positively charged particle is F = qE_ net = q ₀ . The accelerati

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