JEE Main202622 January 2026Morning ShiftPhysicsElectrostaticsActual
Electric field in a region is given by E =A x i +B y j , where A=10 ~V / m ² and B=5 ~V / m ² . If the electric potential at a point (10,20) is 500 V, then the electric potential at origin is _ _ _ _ V.
Options
- A0
- B2000
- C500
- D1000
Correct answer
B. 2000
Step-by-step solution
The electric field is related to potential by E = - V . Given: E = Ax i + By j with A = 10 V/m² and B = 5 V/m² This means: V x = -Ax and V y = -By Integrating: V(x,y) = - Ax^2 2 - By^2 2 + C = -5x^2 - 5y^2 2 + C Using the boundary condition at (10, 20) where V = 500 V: 500 = -5(100) - 5(400) 2 + C 500 = -500 - 1000 + C C = 2000 V At the origin (0, 0) : V(0,0) = -5(0) - 5(0) 2 + 2000 = 2000 V