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Electric potential is given by
$\large V = 6x - 8x{y^2} - 8y + 6yz - 4{z^2}$

Then electric force acting on 2 C point charge placed at origin will be

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20 N

8 N

6 N

2 N

answer is D.

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Detailed Solution

\large {E_x} = - \frac{{dV}}{{dx}} = - (6 - 8{y^2}),

\large {E_y} = - \frac{{dV}}{{dy}} = - ( - \,16xy - 8 + 6z)

\large {E_z} = - \frac{{dV}}{{dz}} = - \,(6y - 8z)

At origin x = y = z = 0 so,

and

⇒

\large E = \sqrt {E_x^2 + E_y^2} = 10\,N/C

Hence force

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