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

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30 N

$6 \sqrt{5} N$

24 N

$4 \sqrt{11}$

answer is D.

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Detailed Solution

V(x,y,z) = 6x–8xy+6yz
$\vec{E} = - i \frac{\partial v}{\partial x} - j \frac{\partial v}{\partial y} - k \frac{\partial v}{\partial z}$
= –i(6-8y) –j(8x+6z) –k(6y)
$\begin{array}{l} \vec{E} (1, 1, 1) = 2 \hat{i} + 2 \hat{j} - 6 \hat{k} \\ | \vec{E} | = \sqrt{2^{2} + 2^{2} + 6^{2}} = 2 \sqrt{11} \end{array}$

$F = qE = 4 \sqrt{11} N$

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