Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
99 Percentile Qs Bank for JEE MainPhysicsElectrostatics

The uniform electric field intensity between the two plates of a parallel plate capacitor is 1 103 Vm ⁻¹ acting vertically upwards as shown in the figure. The plates are sufficiently long and have separation 2 ~cm . A particle of negative charge 1 C and mass 2 ~g is projected at an angle 45^ with the electric field from the lower plate with a velocity ' u '. The maximum velocity acquired by the particle, if it is not

Options

  1. A2 ~ms ⁻¹
  2. B1 ~ms ⁻¹
  3. C0.1 ~ms ⁻¹
  4. D0.2 ~ms ⁻¹

Correct answer

D. 0.2 ~ms ⁻¹

Step-by-step solution

We have, aligned & h_ = u^2 ^2 2 ( q E m ) [ Here g q E m ] & u^2= 2 h m q E m ^2 45^ u^2= 2 2 10⁻² 10⁻⁶ 10^3 2 10⁻³ 1 / 2 & u^2=0.04 u=0.2 ~m / s aligned

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 99 Percentile Qs Bank for JEE Main PYQs