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 (3.00i200j+4.00K)m if the electric potential in the region is given by V=200xyz2, where V is in volts and coordinates x, y, and z are in meters?
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Expert Answer

Charles Benedict
Answered 2021-12-21 Author has 32 answers
Ex,=dVdx=d2xyz2dx=2yz2=(2)×(2)×(42)=64Vm1
Ey,=dVdy=d2xyz2dy=2xz2=(2)×(3)×(42)=96Vm1
Ez,=dVdz=d2xyz2dx=4xyz=(4)×(3)×(2)×(4)=96Vm1
|E|=E{x}2+E{y}2+E{z}2=642+(96)2+962=150Vm1
|E|=150Vm1

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porschomcl
Answered 2021-12-22 Author has 28 answers

Given
V=2.00xyz2
Electric Potential
Point P=3i^2j^+4k^
From Point P
x=3m
y==2m
z=4m
Electric field is given by
E=v
=[ggx(2xyz2)i^+ggy(2xyz2)j^+ggz(2xyz2)k^]
E=2yz2i^2xz2j^4xyzk^
Put x=3m y=-2m z=4m
E=64i^96j^+96k^
Magnitude of Electric field is
|E|=642+(96)2+(96)2
|E|=150.09 n/c

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