If the electric potential at any point $$(x, y, z)$$ m in space is given by $$V = 3x^2$$ volt. The electric field at the point $$(1, 0, 3)$$ m will be:

Given: $$V = 3x^2$$ volt. Find the electric field at $$(1, 0, 3)$$.

$$\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$$

$$\frac{\partial V}{\partial x} = 6x, \quad \frac{\partial V}{\partial y} = 0, \quad \frac{\partial V}{\partial z} = 0$$

$$\vec{E} = -6x\hat{i}$$

$$\vec{E} = -6(1)\hat{i} = -6\hat{i} \text{ Vm}^{-1}$$

The magnitude is 6 Vm$$^{-1}$$, directed along the negative $$x$$-axis.

Hence, the correct answer is Option C: 6 Vm$$^{-1}$$, directed along negative $$x$$-axis.