The electric potential V at any point (x, y, z) is given by V = 3x² where x is in meters and V in volts. The electric field at the point (1 m, 0, 2m) is:

$\begin{array}{l}\mathrm{V}\left(\mathrm{x}\right)=3{\mathrm{x}}^2\\ {\left(\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{x}}\right)}_{\mathrm{y},\mathrm{z}}=\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{x}}\left(3{\mathrm{x}}^2\right)=6\mathrm{x},{\left(\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{y}}\right)}_{\mathrm{x},\mathrm{z}}=\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{y}}\left(3{\mathrm{x}}^2\right)=0,{\left(\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{z}}\right)}_{\mathrm{x},\mathrm{y}}=\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{z}}\left(3{\mathrm{x}}^2\right)=0\\ \text{ Now }\overrightarrow{\mathrm{E}}=\left(\frac{-\mathrm{\delta V}}{\mathrm{\delta x}}\right)\overrightarrow{\mathrm{i}}+\left(\frac{\mathrm{\delta V}}{\mathrm{\delta y}}\right)\overrightarrow{\mathrm{j}}+\left(\frac{\mathrm{\delta V}}{\mathrm{\delta z}}\right)\overrightarrow{\mathrm{k}}\\ \overrightarrow{\mathrm{E}}=(-6\mathrm{x})\overrightarrow{\mathrm{i}}\end{array}$
Now for given point, x=1m
So $\overrightarrow{\mathrm{E}}=(-6)\overrightarrow{\mathrm{i}}=6(-\overrightarrow{\mathrm{i}})$
So electric field is 6 Vm along x axis.