The electric potential V at any point (x, y, | vedclass

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Question

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

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$8$ along negative $x$-axis

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

The electric potential $V$ at any point $(x, y, z)$ in space is given by the equation $V = 4x^2 \, V$,where $x, y$,and $z$ are all in meters. The electric field at the point $(1 \, m, 0, 2 \, m)$ in $V/m$ is:

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