Electric Field via partial derivative

In summary: I miss that?thanks a lot for your help tiny-tim! :cool:In summary, the electric potential in a certain region of space is given by V(x,y,z) = 1000x-2000y-1500z (Volts). To find the corresponding electric field, we can use the equations dv = 1000-2000-1500, Ex = -1000, Ey = 2000, and Ez = 1500. This suggests that the electric field lines are parallel and pointing towards the left, up and out of the screen or page
  • #1
fornax
20
0

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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  • #2
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?​
 
  • #3
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...
 
  • #4
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:
 
  • #5
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
 
  • #6
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:
 
  • #7
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 :/
 
  • #8
shouldn't it be E = σ/ε0 ?
 

Related to Electric Field via partial derivative

1. What is an electric field?

An electric field is a region around an electrically charged object or particle where other charged particles experience a force. It is a fundamental concept in physics that helps us understand the interactions between charged particles.

2. How is electric field calculated?

Electric field can be calculated by taking the partial derivative of the electric potential with respect to distance. In other words, it is the change in electric potential per unit distance.

3. What is the significance of using partial derivatives to calculate electric field?

Partial derivatives allow us to calculate the electric field at a specific point in space, taking into account the contributions from all surrounding charges. This helps us understand the strength and direction of the electric field at that point.

4. How does the electric field change with distance?

The electric field decreases with distance from the source charge, following an inverse square law. This means that as distance increases, the electric field strength decreases by a factor of the square of the distance.

5. What are some applications of understanding electric fields?

Understanding electric fields is crucial in many fields, including electrical engineering, electronics, and telecommunications. It also helps us understand phenomena like lightning, the behavior of particles in particle accelerators, and the functioning of electronic devices such as capacitors and transistors.

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