The electric potential in some region is expr | vedclass

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(D) The electric field $\vec{E}$ is given by the negative gradient of the potential: $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat

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$20$

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(D) The electric field $\vec{E}$ is given by the negative gradient of the potential: $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} ight)$.Calculating the partial derivatives:$\frac{\partial V}{\partial x} = 6 - 8y^2$$\frac{\partial V}{\partial y} = -16xy - 8 + 6z$$\frac{\partial V}{\partial z} = 6y - 8z$At the origin $(0, 0, 0)$:$\frac{\partial V}{\partial x} = 6 - 0 = 6$$\frac{\partial V}{\partial y} = 0 - 8 + 0 = -8$$\frac{\partial V}{\partial z} = 0 - 0 = 0$Thus,$\vec{E} = -(6 \hat{i} - 8 \hat{j} + 0 \hat{k}) = -6 \hat{i} + 8 \hat{j} 	ext{ V/m}$.The magnitude of the electric field is $|\vec{E}| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 	ext{ N/C}$.The electric force is $F = qE = 2 	ext{ C} 	imes 10 	ext{ N/C} = 20 	ext{ N}$.

The electric potential in some region is expressed by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \text{ volt}$. The magnitude of the electric force acting on a charge of $2 \text{ C}$ situated at the origin will be $-$ (in $\text{ N}$)

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