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 coulomb situated at point (1, 1, 1) is

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$6 \sqrt{5} N$

30 N

24 N

$4 \sqrt{35} N$

answer is D.

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Detailed Solution

$Here, V (x, y, z) = 6 x - 8 x y - 8 y + 6 y z$

$The x, y and z components of electric field \vec{E} are$

$E_{x} = - \frac{\partial V}{\partial x} = - \frac{\partial}{\partial x} (6 x - 8 x y - 8 y + 6 y z)$

$= - (6 - 8 y) = - 6 + 8 y$

$E_{y} = - \frac{\partial V}{\partial y} = - \frac{\partial}{\partial y} (6 x - 8 x y - 8 y + 6 y z)$

$= - (- 8 x - 8 + 6 z) = 8 x + 8 - 6 z$

$E_{z} = - \frac{\partial V}{\partial z} = - \frac{\partial}{\partial z} (6 x - 8 x y - 8 y + 6 y z) = - 6 y$

$∴ \vec{E} = E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k}$

$= (- 6 + 8 y) \hat{i} + (8 x + 8 - 6 z) \hat{j} - 6 y \hat{k}$

At point (1, 1, 1)

$\vec{E} = (- 6 + 8) \hat{i} + (8 + 8 - 6) \hat{j} - 6 \hat{k}$

$= 2 \hat{i} + 10 \hat{j} - 6 \hat{k}$

$The magnitude of electric field \vec{E} is$

$| \vec{E} | = \sqrt{E_{x}^{2} + E_{y}^{2} + E_{z}^{2}} = \sqrt{(2)^{2} + (10)^{2} + (- 6)^{2}}$

$= \sqrt{140} = 2 \sqrt{35} {NC}^{- 1}$

Electric force experience by the charge

$F = q E = 2 C \times 2 \sqrt{35} {NC}^{- 1} = 4 \sqrt{35} N$

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