NEET2009PhysicsElectrostaticsActual
The electric potential at a point (x, y, z) is given by V=-x^2 y-x z^3+4 The electric field E at that point is
Options
- AE = i (2 x y+z^3 )+ j x^2+ k 3 x z^2
- BE = i 2 x y+ j (x^2+y^2 )+ k (3 x z-y^2 )
- CE = i z^3+ j x y z+ k z^2
- DE = i (2 x y-z^3 )+ j x y^2+ k z^2 x
Correct answer
A. E = i (2 x y+z^3 )+ j x^2+ k 3 x z^2
Step-by-step solution
Key Idea Electric field at a point is equal to the negative gradient of the electrostatic potential at that point. Potential gradient relates with electric field according to the following relation E= -d V d r aligned E & =- V r = [- V x i - V y j - V x k ] & = [ i (2 x y+z^3 )+ j x^2+ k 3 x z^2 ] aligned