Question
The electric potential at a point $$\left( {x,y,z} \right)$$ is given by $$V = - {x^2}y - x{z^3} + 4.$$ The electric field $${\vec E}$$ at that point is
A.
$$\vec E = \hat i\,2xy + \hat j\left( {{x^2} + {y^2}} \right) + \hat k\left( {3xz - {y^2}} \right)$$
B.
$$\vec E = \hat i\,{z^3} + \hat j\,xyz + \hat k\,{z^2}$$
C.
$$\vec E = \hat i\left( {\,2xy - {z^3}} \right) + \hat j\,x{y^2} + \hat k\,3{z^2}x$$
D.
$$\vec E = \hat i\left( {\,2xy - {z^3}} \right) + \hat j\,{x^2} + \hat k\,3x{z^2}$$
Answer :
$$\vec E = \hat i\left( {\,2xy - {z^3}} \right) + \hat j\,{x^2} + \hat k\,3x{z^2}$$
Solution :
The electric field at a point is equal to negative of potential gradient at that point.
$$\eqalign{
& \vec E = - \frac{{\partial V}}{{\partial r}} = \left[ { - \frac{{\partial V}}{{\partial x}}\hat i - \frac{{\partial V}}{{\partial y}}\hat j - \frac{{\partial V}}{{\partial z}}\hat k} \right] \cr
& = \left[ {\left( {2xy + {z^3}} \right)\hat i + \hat j\,{x^2} + \hat k\,3x{z^2}} \right] \cr} $$