Question

The electric potential $V$ at any point $(x,y,z)$ in space is given by $V=3x^2$ where $x,y,z$ are all in metre. The electric field at the point $(1m,0,2m)$ is

• A. $6V{m}^{-1}$ along negative $x$-axis
• B. $6V{m}^{-1}$ along positive $x$-axis
• C. $12V{m}^{-1}$ along negative $x$-axis
• D. $12V{m}^{-1}$ along positive $x$-axis
• E. $8V{m}^{-1}$ along negative $x$-axis

Answer 1

The correct option is A $6V{m}^{-1}$ along negative $x$-axis

Electric potential $V=3x^2$
$E=-\frac{dV}{dx}$
$E=-\frac{d}{dx}\left(3x^3\right)$
$E=-6x$
At the point $(1,0,2)$
Electric field $E=6\times 1=-6V/m$
Thus electric field at $(1m,0,2m)$ is $6V{m}^{-1}$ along negative $x-$ axis.