Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main2009PhysicsElectrostaticsActual

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then the Q / q equals

Options

  1. A-2 2
  2. B-1
  3. C1
  4. D- 1 2

Correct answer

A. -2 2

Step-by-step solution

Three forces F₄₁, F₄₂ and f₄₃ acting on Q are shown Resultant of F₄₁+F₄₃ aligned & = 2 F_ each & = 2 1 4 ₀ Q q d^2 aligned Resultant on Q becomes zero only when ' q ' charges are of negative nature. aligned & F_ 4,2 = 1 4 ₀ Q Q ( 2 d)^2 & 2 d Q d^2 = Q Q 2 d^2 & 2 q= Q Q 2 & q=- Q 2 2 or Q q =-2 2 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 JEE Main PYQs