a
(a) The electric potential \(V\,(x,y,z) = 4{x^2}\,volt\)
Now \(\overrightarrow E = - \,\left( {\hat i\frac{{\partial V}}{{\partial x}} + \hat j\frac{{\partial V}}{{\partial y}} + \hat k\frac{{\partial V}}{{\partial z}}} \right)\)
Now \(\frac{{\partial V}}{{\partial x}} = 8x,\,\frac{{\partial V}}{{\partial y}} = 0\) and \(\frac{{\partial V}}{{\partial z}} = 0\)
Hence \(\overrightarrow E = - \,8x\hat i\), so at point (\(1\,m, 0, 2\,m\))
\(\overrightarrow E = - \,8\hat i\,\,volt/metre\) or \(8\) along negative \(X-\) axis.