Electric potential is given by V = 6x - 8x{y^2} - 8y + 6yz - 4{z^2} Then electric force acting on 2,

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2. Electric Potential and Capacitance

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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......$N$

$2$

$6$

$8$

$20$

Solution

(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 $F = QE = 2 \times 10 = 20\,N$

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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......$N$

Solution

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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......$N$