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

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Question

(A) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V$.Where $
abla = \hat{i} \frac{\partial

Vedclass · Accepted Answer

$8$,along negative $X-$ axis

Vedclass · Answer

(A) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V$.Where $
abla = \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}$.Therefore,$\vec{E} = -\left[ \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} ight]$.Given $V = 4x^2$,we calculate the partial derivatives:$\frac{\partial V}{\partial x} = 8x$,$\frac{\partial V}{\partial y} = 0$,and $\frac{\partial V}{\partial z} = 0$.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 $(1, 0, 2)$,we substitute $x = 1$:$\vec{E} = -8(1) \hat{i} = -8 \hat{i} 	ext{ V/m}$.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 $V = 4x^2$ volt. The electric field at the point $(1, 0, 2)$ in volt/meter,is

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