Question

Electric potential existing in space is given by $V=K(x^{2}y+y^{2}z+xyz)$. Find the expression for electric field.

• A. $0$
• B. $-K\{(2xy+yz)\hat{i}+(x^2+2yz+xz)\hat{j}+(y^2+xy)\hat{k}\}$
• C. $-K\{[\frac{x^{3}y}{3}+\frac{x^{2}yz}{2}]\hat{i}+(y^2+xy)\hat{k}\}$
• D. $K\{[2xy+yz]\hat{i}+[x^2+2yz+xy]\hat{j}+(y^2+xy)\hat{k}\}$

Answer 1

The correct option is B $-K\{(2xy+yz)\hat{i}+(x^2+2yz+xz)\hat{j}+(y^2+xy)\hat{k}\}$

Given that,
Electric potential, $V=K(x^{2}y+y^{2}z+xyz)$

The relation between electric potential and electric field $E$ is
$E=-\{\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z}\}V$

$\Rightarrow E=-\hat{i}\frac{\partial V}{\partial x}-\hat{j}\frac{\partial V}{\partial y}-\hat{k}\frac{\partial V}{\partial z}.....\left(i\right)$

Partially differentiating individually we get

$\frac{\partial V}{\partial x}=2xy+yz$

$\frac{\partial V}{\partial y}=x^2+2yz+xz$

$\frac{\partial V}{\partial z}=y^2+xy$

Using this in (1) we obtain
$E=-K\{(2xy+yz)\hat{i}+(x^2+2yz+xz)\hat{j}+(y^2+xy)\hat{k}\}$
Hence, option (b) is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept: If potential is given at a point}\\ (x,y,z)\text{or in free space, then electric field is given by}\\ \overrightarrow{E}=-\hat{i}\frac{\partial V}{\partial x}-\hat{j}\frac{\partial V}{\partial y}-\hat{k}\frac{\partial V}{\partial z}\end{array}$