In a certain region,the electric potential is | vedclass

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Question

(D) The electric field $\overrightarrow{E}$ is related to the potential $V$ by $\overrightarrow{E} = -
abla V = -\left( \frac{\partial V}{\partial x}

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) The electric field $\overrightarrow{E}$ is related to the potential $V$ by $\overrightarrow{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} ight)$.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,the electric field at the origin is $\overrightarrow{E} = -(6 \hat{i} - 8 \hat{j} + 0 \hat{k}) = -6 \hat{i} + 8 \hat{j} \, N/C$.The magnitude of the electric field is $|\overrightarrow{E}| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, N/C$.The force on the charge $q = 2 \, C$ is $F = q|\overrightarrow{E}| = 2 	imes 10 = 20 \, N$.

In a certain region,the electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \, V$. The magnitude of the force on a charge of $2 \, C$ at the origin is ........ $N$.

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