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(A) Step $1$: The electric field $\vec{E}$ is related to the potential $V$ by the gradient formula: $\vec{E} = -
abla V = -\left( \frac{\partial V}{\

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$8 \ V/m$ in the direction of the $(-x)$ axis.

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(A) Step $1$: The electric field $\vec{E}$ is related to the potential $V$ by the gradient formula: $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} ight)$.Given $V = 4x^2$,we calculate the partial derivatives:$\frac{\partial V}{\partial x} = \frac{d}{dx}(4x^2) = 8x$,$\frac{\partial V}{\partial y} = 0$,and $\frac{\partial V}{\partial z} = 0$.Substituting these into the formula,we get $\vec{E} = -(8x) \hat{i} = -8x \hat{i} \ V/m$.Step $2$: Evaluate the electric field at the point $(1, 0, 2)$.At $x = 1$,$\vec{E} = -8(1) \hat{i} = -8 \hat{i} \ V/m$.The negative sign indicates that the electric field is directed along the negative $x$-axis with a magnitude of $8 \ V/m$.

If the electric potential in a region is given by $V = 4x^2$ volts,then the electric field at the point $(1, 0, 2) \ m$ is:

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