The electric potential (in volt) in a region is given by $V = 6 x - 8 {xy}^{2} - 8 y + 6 yz - 4 x^{2}$
Then electric force acting on a point charge of 2C placed at the origin will be

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2 N

6 N

8 N

20 N

answer is D.

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Detailed Solution

$\begin{array}{l} \vec{E} = \vec{\nabla} V \\ \Rightarrow \vec{E} = - (\hat{i} \frac{\partial U}{\partial x} + \hat{j} \frac{\partial U}{\partial y} + \hat{k} \frac{\partial U}{\partial z}) \end{array} \begin{array}{l} \frac{\partial U}{\partial x} = 6 - 8 y^{2} - 8 x \\ \frac{\partial U}{\partial y} = - 16 xy - 8 + 6 z \\ \frac{\partial U}{\partial z} = 6 y \\ \frac{At (0, 0, 0)}{\frac{\partial U}{\partial x}} = 6, \frac{\partial U}{\partial y} = - 8, \frac{\partial U}{\partial z} = 0 \end{array} \begin{array}{l} \Rightarrow | \vec{E} |_{(0, 0, 0)} = - 6 \hat{i} + 8 \hat{j} \\ \Rightarrow | \vec{E} | = 10 {NC}^{- 1} \\ \Rightarrow | \vec{F} | = q | \vec{E} | \\ \Rightarrow | \vec{F} | = 20 N \end{array}$