In a region,the potential is represented by V | vedclass

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

Vedclass · Accepted Answer

$4\sqrt{35} \ N$

Vedclass · Answer

(D) The electric field $\vec{E}$ is given by the negative gradient of the potential $V$: $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} ight)$.Given $V(x, y, z) = 6x - 8xy - 8y + 6yz$.Calculating partial derivatives:$E_x = -\frac{\partial V}{\partial x} = -(6 - 8y) = -6 + 8y$.$E_y = -\frac{\partial V}{\partial y} = -(-8x - 8 + 6z) = 8x + 8 - 6z$.$E_z = -\frac{\partial V}{\partial z} = -(6y) = -6y$.At point $(1, 1, 1)$:$E_x = -6 + 8(1) = 2 \ V/m$.$E_y = 8(1) + 8 - 6(1) = 10 \ V/m$.$E_z = -6(1) = -6 \ V/m$.Thus,$\vec{E} = 2\hat{i} + 10\hat{j} - 6\hat{k} \ V/m$.The magnitude of the electric field is $|\vec{E}| = \sqrt{2^2 + 10^2 + (-6)^2} = \sqrt{4 + 100 + 36} = \sqrt{140} = 2\sqrt{35} \ V/m$.The electric force $\vec{F} = q\vec{E}$. Given $q = 2 \ C$,the magnitude of the force is $F = q|\vec{E}| = 2 	imes 2\sqrt{35} = 4\sqrt{35} \ N$.

In a region,the potential is represented by $V(x, y, z) = 6x - 8xy - 8y + 6yz$,where $V$ is in volts and $x, y, z$ are in metres. The electric force experienced by a charge of $2 \ C$ situated at point $(1, 1, 1)$ is

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