 # What is the magnitude of the electric field at the Jessie Lee 2021-12-20 Answered
What is the magnitude of the electric field at the point $\left(3.00i-200j+4.00K\right)m$ if the electric potential in the region is given by $V=200xy{z}^{2}$, where V is in volts and coordinates x, y, and z are in meters?
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$Ex,=-d\frac{V}{dx}=-d\frac{2xy{z}^{2}}{dx}=-2y{z}^{2}=-\left(2\right)×\left(-2\right)×\left({4}^{2}\right)=64V{m}^{-1}$
$Ey,=-d\frac{V}{dy}=-d\frac{2xy{z}^{2}}{dy}=-2x{z}^{2}=-\left(2\right)×\left(3\right)×\left({4}^{2}\right)=-96V{m}^{-1}$
$Ez,=-d\frac{V}{dz}=-d\frac{2xy{z}^{2}}{dx}=-4xyz=-\left(4\right)×\left(3\right)×\left(-2\right)×\left(4\right)=96V{m}^{-1}$
$|E|=\sqrt{{E}_{\left\{x\right\}}^{2}+{E}_{\left\{y\right\}}^{2}+{E}_{\left\{z\right\}}^{2}}=\sqrt{{64}^{2}+{\left(-96\right)}^{2}+{96}^{2}}=150V{m}^{-1}$
$|E|=150V{m}^{-1}$

We have step-by-step solutions for your answer! porschomcl

Given
$V=2.00xy{z}^{2}$
Electric Potential
Point $P=3\stackrel{^}{i}-2\stackrel{^}{j}+4\stackrel{^}{k}$
From Point P
$x=3m$
$y==2m$
$z=4m$
Electric field is given by
$\stackrel{\to }{E}=-\mathrm{△}v$
$=-\left[\begin{array}{c}\frac{g}{gx}\left(2xy{z}^{2}\right)\stackrel{^}{i}+\frac{g}{gy}\left(2xy{z}^{2}\right)\stackrel{^}{j}+\frac{g}{gz}\left(2xy{z}^{2}\right)\stackrel{^}{k}\end{array}\right]$
$\stackrel{\to }{E}=-2y{z}^{2}\stackrel{^}{i}-2x{z}^{2}\stackrel{^}{j}-4xyz\stackrel{^}{k}$
Put x=3m y=-2m z=4m
$\stackrel{\to }{E}=64\stackrel{^}{i}-96\stackrel{^}{j}+96\stackrel{^}{k}$
Magnitude of Electric field is
$|\stackrel{\to }{E}|=\sqrt{{64}^{2}+{\left(-96\right)}^{2}+{\left(96\right)}^{2}}$

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