If the electric potential at any point (x, y, | vedclass

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(D) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x}

Vedclass · Accepted Answer

$6 \, Vm^{-1}$,directed along negative $x$-axis

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(D) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} ight)$.Given $V = 3x^2$,we calculate the partial derivatives:$\frac{\partial V}{\partial x} = \frac{d}{dx}(3x^2) = 6x$$\frac{\partial V}{\partial y} = 0$$\frac{\partial V}{\partial z} = 0$Thus,$\vec{E} = -(6x) \hat{i} = -6x \hat{i}$.At the point $(1, 0, 3)$,the $x$-coordinate is $1$. Substituting this into the expression for $\vec{E}$:$\vec{E} = -6(1) \hat{i} = -6 \hat{i} \, Vm^{-1}$.The magnitude is $6 \, Vm^{-1}$ and the negative sign indicates it is directed along the negative $x$-axis.

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 ............

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