Question

Electric potential in a region is varying according to the relation $V=\frac{3x^2}{2}-\frac{y^2}{4}$ , where $x$ and $y$ are in $\text{meter}$ and $V$ is in $\text{volt}$. Electric field intensity in $\text{N/C}$ at a point $(1\text{m},2\text{m})$ is

• A. $3\hat{i}-\hat{j}$
• B. $-3\hat{i}+\hat{j}$
• C. $6\hat{i}-2\hat{j}$
• D. $-6\hat{i}+2\hat{j}$

Answer 1

The correct option is B $-3\hat{i}+\hat{j}$

Electric potential, $V=\frac{3x^2}{2}-\frac{y^2}{4}$

Relation between electric potential and electric field intensity is given by,

$\overrightarrow{E}=-[\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}]....\left(1\right)$

Now,

$\frac{\partial V}{\partial x}=\frac{\partial (\frac{3x^2}{2}-\frac{y^2}{4})}{\partial x}=3x$

Similarly,

$\frac{\partial V}{\partial y}=-\frac{y}{2}$ and $\frac{\partial V}{\partial z}=0$

From $\left(1\right)$ equation, we have

$\overrightarrow{E}=-3x\hat{i}+\frac{y}{2}\hat{j}$

At point $(1\text{m},2\text{m})$

$\overrightarrow{E}=(-3\times 1)\hat{i}+\frac{2}{2}\hat{j}$
$\therefore \overrightarrow{E}=-3\hat{i}+\hat{j}$

Hence, option (b) is the correct answer.