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 y^2+ k 3 z^2 x
- BE = i (2 x y+ z ^3 )+ j x ^2+ k 3 xz z ^2
- CE = i 2 x y+ j (x^2+y^2 )+ k (3 x z-y^2 )
- DE = i z+ j x y z+ k z^2
Correct answer
B. E = i (2 x y+ z ^3 )+ j x ^2+ k 3 xz z ^2
Step-by-step solution
aligned V & =-x^2 y-x z^3+4 E & =-V=- ( i x + j y + k z ) & .= (2 x y+z^3 ) i +x^2 j +3 x z^2+4 ) aligned