Electric potential is given by
$V = 6x - 8x{y^2} - 8y + 6yz - 4{z^2}$
Then electric force acting on $2\,C$ point charge placed on origin will be......$N$
Medium
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(d)${E_x} = - \frac{{dV}}{{dx}} = - (6 - 8{y^2}),$${E_y} = - \frac{{dV}}{{dy}} = - ( - \,16xy - 8 + 6z)$
${E_z} = - \frac{{dV}}{{dz}} = - \,(6y - 8z)$
At origin $x = y = z = 0$ so,${E_x} = - \,6,\,{E_y} = 8$ and ${E_z} = 0$
$==>$ $E = \sqrt {E_x^2 + E_y^2} = 10\,N/C$.
Hence force $F = QE = 2 \times 10 = 20\,N$
${E_z} = - \frac{{dV}}{{dz}} = - \,(6y - 8z)$
At origin $x = y = z = 0$ so,${E_x} = - \,6,\,{E_y} = 8$ and ${E_z} = 0$
$==>$ $E = \sqrt {E_x^2 + E_y^2} = 10\,N/C$.
Hence force $F = QE = 2 \times 10 = 20\,N$
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