Electric Field via partial derivative

AI Thread Summary
The electric potential is defined as V(x,y,z) = 1000x - 2000y - 1500z Volts, leading to the electric field components Ex = -1000, Ey = 2000, and Ez = 1500. The electric field is constant, indicating parallel field lines, which suggests a uniform charge distribution. A capacitor is proposed as a possible source of this uniform field, with calculations indicating a surface charge density of σ = 4.76 x 10^-8 C/m². The discussion emphasizes the importance of understanding the relationship between electric potential and electric field in determining charge distributions.
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Homework Statement


The electric potential in a certain region of space is given by: V(x,y,z) = 1000x-2000y-1500z(Volts). a.)Find the electric field corresponding to the given electric potential. Draw some electric field lines. b.) What charge distribution can create this electric field? Give all possible numericical information about the charge distribution that you can find from the given data.



Homework Equations


V=dv dv=-E*dl
dv(x,y,z) = ∂v/∂x *dx + ∂v/∂y *dy + ∂v/∂z *dz

E*dl = Exdx+Eydy+Ezdz

Ex = - ∂v/∂x Ey = - ∂v/∂y Ez = - ∂v/∂z


The Attempt at a Solution


V(x,y,z) = 1000x-2000y-1500z
dv = 1000-2000-1500
Ex = -1000 Ey = 2000 Ez = 1500

I have a gut feeling that I am missing some rather important steps here. I also have no idea where to begin in terms of drawing this, any help is highly appreciated.
 
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welcome to pf!

hi fornax! welcome to pf! :smile:
fornax said:
The electric potential in a certain region of space is given by: V(x,y,z) = 1000x-2000y-1500z(Volts). a.)Find the electric field corresponding to the given electric potential. Draw some electric field lines. b.) What charge distribution can create this electric field? Give all possible numericical information about the charge distribution that you can find from the given data.

V(x,y,z) = 1000x-2000y-1500z
dv = 1000-2000-1500
Ex = -1000 Ey = 2000 Ez = 1500

yes, that looks ok :smile:

(sometimes, the first part of an exam question is deliberately easy! :wink:)

now select a few random points, and draw arrows to show the magnitude and direction of the force (the field) …

what's the pattern?​
 
I haven't worked with these much, so just put in any value in place of the Ex Ey, Ez? If I had to make an educated guess right now, I would say all of the vectors are pointingtoward the left, up and out of the screen, or page. As far as the charge distribution, it would have to be an infinite sheet, correct? Especialy since the field remains constant...
 
fornax said:
I haven't worked with these much, so just put in any value in place of the Ex Ey, Ez? If I had to make an educated guess right now, I would say all of the vectors are pointingtoward the left, up and out of the screen, or page. As far as the charge distribution, it would have to be an infinite sheet, correct? Especialy since the field remains constant...

yes :smile:

to put it simply, the E field is constant, so the lines are parallel

it doesn't have to be an infinite sheet

(they are quite difficult to buy! :biggrin:)

the question says "a certain region", so you only need something small …

can you think of something you could buy that does have a uniform field in a small region? :wink:
 
that's pretty easy, a capacitor. Now to find "all possible numerical information", I just backtrack, with each individual component?

ie. Ex = σ/ 2ε0

-1000 = σ/ 2ε0

σ = -5.7x10^13
 
fornax said:
…, with each individual component?

uhh? how many capacitors are you using?? :confused:

i] which direction will the capacitor have to face?

ii] what is the magnitude of the electric field?

iii] so what is the surface charge density? :smile:
 
Alright

It would have to be facing in the direction of the E field, so that it's face is perpindicular to the E field.

[ii] llVll = √(x2 + y2 + z2
llVll = √-10002 + 20002 + 15002
llVll = √7250000
llVll = 2693

so then

[iii]

E = σ/ 2ε0

2693 = σ/ 2ε0

σ = 4.76 x 10-8 C/m2

that seems too low :/
 
shouldn't it be E = σ/ε0 ?
 
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