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

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(A) The electric potential is given by $V(x, y, z) = 4x^2 	ext{ volt}$.The electric field $\vec{E}$ is related to the potential $V$ by the relation $

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$8$ along negative $X$-axis

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(A) The electric potential is given by $V(x, y, z) = 4x^2 	ext{ volt}$.The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -
abla V = -\left( \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} ight)$.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 expression for $\vec{E}$:$\vec{E} = -(8x \hat{i} + 0 \hat{j} + 0 \hat{k}) = -8x \hat{i} 	ext{ V/m}$.At the point $(1 	ext{ m}, 0, 2 	ext{ m})$,the $x$-coordinate is $1 	ext{ m}$.Therefore,$\vec{E} = -8(1) \hat{i} = -8 \hat{i} 	ext{ V/m}$.The negative sign indicates that the electric field is directed along the negative $X$-axis with a magnitude of $8 	ext{ V/m}$.

The electric potential $V$ at any point $(x, y, z)$ (all in metres) in space is given by $V = 4x^2 \text{ volt}$. The electric field at the point $(1 \text{ m}, 0, 2 \text{ m})$ in $\text{volt/metre}$ is

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