The electric potential at a point (x, y, z) i | vedclass

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(D) The electric potential is given by $V = -x^2y - xz^3 + 4$.The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} =

Vedclass · Accepted Answer

$\vec{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}$

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(D) The electric potential is given by $V = -x^2y - xz^3 + 4$.The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\vec{
abla} V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} ight)$.Calculating the partial derivatives:$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(-x^2y - xz^3 + 4) = -2xy - z^3$$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(-x^2y - xz^3 + 4) = -x^2$$\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(-x^2y - xz^3 + 4) = -3xz^2$Substituting these into the expression for $\vec{E}$:$\vec{E} = -[(-2xy - z^3)\hat{i} + (-x^2)\hat{j} + (-3xz^2)\hat{k}]$$\vec{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}$.

The electric potential at a point $(x, y, z)$ is given by $V = -x^2y - xz^3 + 4$. The electric field at that point is:

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