The electric potential $V$ at any point $(x, y, z)$, all in metres in space is given by $V=4 x^2$ volt. The…
- 8 along positive $X$-axis
- 16 along negative $X$-axis
- 16 along positive $X$-axis
- 8 along negative $X$-axis
Solution
$\mathrm{E}=-\left[\hat{i} \frac{\partial V}{\partial x}+\hat{j} \frac{\partial V}{\partial y}+\mathrm{k} \frac{\partial V}{\partial z}\right]$
So
$\begin{aligned}
& \mathrm{E}=-\hat{i} \frac{\partial V}{\partial x}=-\hat{i} \frac{\partial}{\partial x}\left(4 x^2\right) \\
&=-8 x \hat{i} \mathrm{Vm}^{-1} \\
& \mathrm{E}_{(1,0,2)}=-8 \hat{i} \mathrm{Vm}^{-1}
\end{aligned}$ ~
Asked in: NEET 2011 (Mains)