Looking for classes? Ksquare Career Institute, Bengaluru →
The electric potential at a point (x, y, z) is given by $V = -x^2y - xz^3 + 4$. The electric field $\vec{E}$ at that point is
A
$\vec{E} = \hat{i}(2xy - z^3) + \hat{j}xy^2 + \hat{k}3z^2x$
B
$\vec{E} = \hat{i}(2xy + z^3) + \hat{j}x^2 + \hat{k}3xz^2$
C
$\vec{E} = \hat{i}2xy + \hat{j}(x^2 + y^2) + \hat{k}(3xz - y^2)$
D
$\vec{E} = \hat{i}z^3 + \hat{j}xyz + \hat{k}z^2$
Detailed Solution
$\vec{E} = -\left(\hat{i}\frac{\partial V}{\partial x} + \hat{j}\frac{\partial V}{\partial y} + \hat{k}\frac{\partial V}{\partial z}\right)$
$\frac{\partial V}{\partial x} = -2xy - z^3$, so $E_x = 2xy + z^3$
$\frac{\partial V}{\partial y} = -x^2$, so $E_y = x^2$
$\frac{\partial V}{\partial z} = -3xz^2$, so $E_z = 3xz^2$
$\vec{E} = \hat{i}(2xy + z^3) + \hat{j}x^2 + \hat{k}3xz^2$
$\frac{\partial V}{\partial x} = -2xy - z^3$, so $E_x = 2xy + z^3$
$\frac{\partial V}{\partial y} = -x^2$, so $E_y = x^2$
$\frac{\partial V}{\partial z} = -3xz^2$, so $E_z = 3xz^2$
$\vec{E} = \hat{i}(2xy + z^3) + \hat{j}x^2 + \hat{k}3xz^2$