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

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

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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 relation $\overrightarrow{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} ight)$.Given $V = 4x^2$,we calculate the partial derivatives:$\frac{\partial V}{\partial x} = \frac{d}{dx}(4x^2) = 8x$$\frac{\partial V}{\partial y} = 0$$\frac{\partial V}{\partial z} = 0$Thus,$\overrightarrow{E} = -(8x) \hat{i} = -8x \hat{i}$.At the point $(1 \ m, 0, 2 \ m)$,the value of $x$ is $1 \ m$.Substituting $x = 1$,we get $\overrightarrow{E} = -8(1) \hat{i} = -8 \hat{i} \ V/m$.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 $(x, y, z)$ (all in $m$) in space is given by $V = 4x^2 \ V$. The electric field at the point $(1 \ m, 0, 2 \ m)$ in $V/m$ is:

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