Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20266 April 2026Evening ShiftPhysicsElectrostaticsActual

The 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.

Options

  1. A(-20 i + 30 j )
  2. B(20 i - 30 j )
  3. C(20 i + 45 j )
  4. D(-4 i + 6 j )

Correct answer

A. (-20 i + 30 j )

Step-by-step solution

The electric field E is given by the negative gradient of the electric potential V : E = - ( V x i + V y j ) Given V = 5(x^2 - y^2) , we find the partial derivatives: V x = 10x V y = -10y Substituting these into the expression for E : E = -(10x i - 10y j ) = -10x i + 10y j At the point (2, 3) , substitute x = 2 and y = 3 : E = -10(2) i + 10(3) j = -20 i + 30 j V/m Answer: (-20 i + 30 j )

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 ) 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 2026Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: In electrostatics, a conductor does not store any net charge insid 2026 Full Electrostatics list All JEE Main PYQs