Physics - Elecrostatics Question with Solution | TestHub

Electric potential $V=3x^2$

Electric potential and electric field are related as $E=-\frac{dV}{dx}=-6x$.

At the point $\left(1, 0, 3\right) \mathrm{m}$, 

$\Rightarrow E=-6\times 1=-6 \mathrm{V} {\mathrm{m}}^{-1}$

Thus, electric field is $6 \mathrm{V} {\mathrm{m}}^{-1}$ along negative $x‐$ axis.