The electric potential at a point (x, y, z) is given by V=-x2 y-x z3+4The electric field vecE at that point is

Question

The electric potential at a point (x, y, z) is given by V=-x2 y-x z3+4The electric field vecE at that point is (A)  vecE= hati(2 x y+z3)+ hatj x2+ hatk 3 x z2 (B)  vecE= hati 2 x y+ hatj(x2+y2)+ hatk(3 x z-y2) (C)  vecE= hati z3+ hatj x y z+ hatk z2 (D)  vecE= hati(2 x y-z3)+ hatj x y2+ hatk 3 z2 x

Tardigrade Admin · Accepted Answer

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 = \frac{- d V}{d r}

E = - \frac{\partial V}{\partial r}

= [- \frac{\partial V}{\partial x} \hat{i} - \frac{\partial V}{\partial y} \hat{j} - \frac{\partial V}{\partial x} \hat{k}]

= \hat{i} (2 x y + z^{3}) + \hat{j} x^{2} + \hat{k} 3 x z^{2}

Q. 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

$E = \hat{i} (2 x y + z^{3}) + \hat{j} x^{2} + \hat{k} 3 x z^{2}$

$E = \hat{i} 2 x y + \hat{j} (x^{2} + y^{2}) + \hat{k} (3 x z - y^{2})$

$E = \hat{i} z^{3} + \hat{j} x yz + \hat{k} z^{2}$

$E = \hat{i} (2 x y - z^{3}) + \hat{j} x y^{2} + \hat{k} 3 z^{2} x$

Q. The electric potential at a point (x,y,z) is given by V=−x2y−xz3+4The electric field E at that point is

E=i^(2xy+z3)+j^​x2+k^3xz2

E=i^2xy+j^​(x2+y2)+k^(3xz−y2)

E=i^z3+j^​xyz+k^z2

E=i^(2xy−z3)+j^​xy2+k^3z2x

Q. 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

$E = \hat{i} (2 x y + z^{3}) + \hat{j} x^{2} + \hat{k} 3 x z^{2}$

$E = \hat{i} 2 x y + \hat{j} (x^{2} + y^{2}) + \hat{k} (3 x z - y^{2})$

$E = \hat{i} z^{3} + \hat{j} x yz + \hat{k} z^{2}$

$E = \hat{i} (2 x y - z^{3}) + \hat{j} x y^{2} + \hat{k} 3 z^{2} x$