Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20262 April 2026Evening ShiftPhysicsElectrostaticsActual

Two metal plates (A, B) are kept horizontally with separation of ( 12 ) cm, with plate A on the top. An atomizer jet sprays oil (density 1.5 g/cm ^3 ) droplets of radius 1 mm horizontally. All oil droplets carry a charge 5 nC. The potentials V_A and V_B are required on plates A and B respectively in order to ensure the droplets do not descend. The values of V_A and V_B are _______. (Neglect the air resistance to the

Options

  1. A100 , V and 580 , V
  2. B580 , V and 100 , V
  3. C60 , V and 400 , V
  4. D0 , V and -200 , V

Correct answer

A. 100 , V and 580 , V

Step-by-step solution

For the oil droplets to not descend, the upward electric force must balance the downward gravitational force. qE = mg The mass of the oil droplet is given by: m = 4 3 r^3 Substituting the given values r = 10⁻³ m and = 1500 kg/m ^3 : m = 4 3 (10⁻³)^3 1500 = 2 10⁻⁶ kg The electric field E required is: E = mg q = 2 10⁻⁶ 10 5 10⁻⁹ = 4000 V/m The potential difference V between the plates is: V = E d = 4000 ( 12 10⁻² ) = 480 V Since the charge on the droplet is positive, the electric field must be directed upwards to pro

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