Question

Electric potential is given by $V=6x-8x{y}^2-8y+6yz-4Z^2$ Then electric force (in $N$) acting on $2C$ point charge placed on origin will be _________ .

Answer 1

Answer: 20

$E_x=-\frac{dV}{dx}=-(6-8y^2)$,
$E_y=-\frac{dV}{dy}=-(-16xy-8+6z)$
$E_z=-\frac{dV}{dz}=-(6y-8z)$
At origin $x=y=z=0$
so, $E_x=-6,E_y=8$ and $E_z=0$
$\Rightarrow E=\sqrt{E_x^2+E_y^2}=10N/C$
Hence force $F=QE=2\times 10=20N$