In a certain region of space,the potential is | vedclass

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(B) Given,$V = k[2x^2 - y^2 + z^2]$.The electric field is given by the negative gradient of the potential: $\vec{E} = -
abla V = -\left(\frac{\partia

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$2k\sqrt{6}$

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(B) Given,$V = k[2x^2 - y^2 + z^2]$.The electric field 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} = k(4x)$$\frac{\partial V}{\partial y} = k(-2y)$$\frac{\partial V}{\partial z} = k(2z)$Thus,$\vec{E} = -k(4x \hat{i} - 2y \hat{j} + 2z \hat{k})$.At the point $(1, 1, 1)$,the electric field is:$\vec{E}_{(1,1,1)} = -k(4(1) \hat{i} - 2(1) \hat{j} + 2(1) \hat{k}) = -k(4 \hat{i} - 2 \hat{j} + 2 \hat{k})$.The magnitude of the electric field is:$|\vec{E}| = \sqrt{(-4k)^2 + (2k)^2 + (-2k)^2} = \sqrt{16k^2 + 4k^2 + 4k^2} = \sqrt{24k^2} = 2k\sqrt{6}$.

In a certain region of space,the potential is given by: $V = k[2x^2 - y^2 + z^2]$. The electric field at the point $(1, 1, 1)$ has magnitude =

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