Electric potential is given by
$V = 6 x - 8 x y^{2} - 8 y + 6 y z - 4 z^{2}$
Then electron force acting on 2C point charge placed on origin will be

2 N

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6 N

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8 N

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20 N

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Solution

The correct option is D 20 N
$E_{x} = - \frac{d V}{d x} = - (6 - 8 y^{2}), E_{y} = - \frac{d V}{d y} = - (- 16 x y - 8 + 6 z$
$E_{z} = - \frac{d V}{d z} = - (6 y - 8 z)$
At origin x = y = z = 0 so, $E_{x} = - 6, E_{y} = 8$ and $E_{z} = 0$
$\Rightarrow E = \sqrt{E_{x}^{2} + E_{y}^{2}} = 10 N / C$ .
Hence force $F = Q E = 2 \times 10 = 20 N$

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Electric potential is given by V=6x−8xy2−8y+6yz−4z2 Then electron force acting on 2C point charge placed on origin will be

The correct option is D 20 NEx=−dVdx=−(6−8y2),Ey=−dVdy=−(−16xy−8+6z Ez=−dVdz=−(6y−8z) At origin x = y = z = 0 so, Ex=−6, Ey=8 and Ez=0 ⇒E=√E2x+E2y=10 N/C. Hence force F=QE=2×10=20N

Electric potential is given by
$V = 6 x - 8 x y^{2} - 8 y + 6 y z - 4 z^{2}$
Then electron force acting on 2C point charge placed on origin will be