Electric potential at any point is $V = - 5 x + 3 y + \sqrt{15} z$ , then the magnitude of the electric field is

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$3 \sqrt{2}$

$4 \sqrt{2}$

$5 \sqrt{2}$

answer is D.

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Detailed Solution

$E_{x} = - \frac{dV}{dx} = - (- 5) = 5; E_{y} = - \frac{dV}{dy} = - 3$

$\begin{array}{l} and E_{z} = - \frac{dV}{dz} = - \sqrt{15} \\ E_{net} = \sqrt{E_{x}^{2} + E_{y}^{2} + E_{z}^{2}} = \sqrt{(5)^{2} + (- 3)^{2} + (- \sqrt{15})^{2}} = 7 \end{array}$

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