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  4. If electric potential at any point is V=-6 x+2 y+9 z, where V, x, y and z are in SI units, then the magnitude of the electric field (in SI unit) is

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Q. If electric potential at any point is $V=-6 x+2 y+9 z$, where $V, x, y$ and $z$ are in SI units, then the magnitude of the electric field (in SI unit) is

Electrostatic Potential and Capacitance

Solution:

$E_{x}=-\frac{d V}{d x}=-(-6)=6 ;$
$E_{y}=-\frac{d V}{d y}=-2$
and $E_{z}=-\frac{d V}{d z}=-9$
$E_{\text {net }}=\sqrt{E_{x}^{2}+E_{y}^{2}+E_{z}^{2}}$
$=\sqrt{36+4+81}=\sqrt{121}=11$