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

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Question

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

Vedclass · Accepted Answer

$8$ along negative $X$-axis

Vedclass · Answer

(A) The electric potential is given by $V(x, y, z) = 4x^2 \, V$.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 $\vec{E}$:$\vec{E} = -(8x \hat{i} + 0 \hat{j} + 0 \hat{k}) = -8x \hat{i} \, V/m$.At the point $(1 \, m, 0, 2 \, m)$,we substitute $x = 1 \, m$:$\vec{E} = -8(1) \hat{i} = -8 \hat{i} \, V/m$.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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