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Calculating Electric Field Strength using Electric Potential?

  • #1
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Homework Statement


How do I calculate the Field Strength if I know the Potential?


Homework Equations


E = -[itex]\nabla[/itex]V
V = Electric Potential
E = Electric Field Strength

The Attempt at a Solution


There's a problem in which I calculated the Potential, it's constant however. So according to the equation that means that the Field Strength is 0?
In the solution they calculated them individually, and the Field Strength wasn't equal to 0.
In the problem, the "body" is a ball and I calculated the potential with respect to origo (centrum of the ball) and so in my answer all I got was in terms of the largest radius (constant) and the volume charge distribution. My potential is correct too, according to the solution, and is a constant.
I think they got that the Field Strength is equal to the potential (in magnitude) in the direction of the negative z-axis.
So how do I get the Field Strength by using the Potential I calculated?
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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For a uniform E field (as between parallel plates), E = V/d where V is the potential difference between plates and d the distance between them.

It isn't clear to me what problem you are working on, but if it involves a sphere, likely the potential varies with distance r from the center. In that case you can find the E if you know how V(r) varies. E = dV/dr.
 
  • #3
128
0
For a uniform E field (as between parallel plates), E = V/d where V is the potential difference between plates and d the distance between them.

It isn't clear to me what problem you are working on, but if it involves a sphere, likely the potential varies with distance r from the center. In that case you can find the E if you know how V(r) varies. E = dV/dr.
The problem I was working on involves a ball (not a sphere since then there's only charge on the surface?) with a uniform charge distribution. So my volume charge density is the total charge divided by the volume of the sphere. The thing is that in my answer (which is correct) I get that the potential is constant. So I'm not sure how I can calculate the field strength by taking the derivative of a constant. In the solution they calculated the field strength separately and didn't get that it was equal 0.
As I said, they got that the field strength only has a z-component, which is equal to the potential, in the direction of the negative z-axis.
 
  • #4
ehild
Homework Helper
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The potential of a charged ball is not constant. Perhaps you get the value of the potential at a given distance from the centre of the ball. Show the text of the problem and your work.

ehild
 
  • #5
128
0
The potential of a charged ball is not constant. Perhaps you get the value of the potential at a given distance from the centre of the ball. Show the text of the problem and your work.

ehild
Sorry, my mistake. It's a half ball. Well what I did was I changed dq into ρdxdydz, ρ is the volume charge distribution (uniform). Then switched to spherical coordinates. dxdydz = r2sin(θ)drdθdψ and then I let r vary from 0 to a (the maximum radius), θ from 0 to π/2 and ψ from 0 to 2π. The 1/r from the formula for calculating the potential turns r2 into r. So what I got was that V = (ρ/4πε0)*(a2/2)*2π. ρ = Q * 1/(2πa3/3).
 
  • #6
ehild
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I still do not know what was the question. I guess you had to find the potential outside the half ball, on the axis? Do not mix the position of a point you need the potential at with the radius of the ball. The "r" in the formula for the electric field refers to the distance from the point charge.

You can consider the charged ball as a lot of tiny charges dq. Their contribution to the potential at a distance d is k dq/r. Integrate this expression for the half ball using the charge density and the volume element in spherical coordinates.

ehild
 
Last edited:
  • #7
128
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I still do not know what was the question. I guess you had to find the potential outside the half ball, on the axis? Do not mix the position of a point you need the potential at with the radius of the ball. The "r" in the formula for the electric field refers to the distance from the point charge.

You can consider the charged ball as a lot of tiny charges dq. Their contribution to the potential at a distance d is k dq/r. Integrate this expression for the half ball using the charge density and the volume element in spherical coordinates.

ehild
The question is to find the potential at origo, I did pretty much what you said also.
 
  • #8
ehild
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So you got the potential at the origin, but you need to know the potential function to get the field strength, by taking the negative gradient at the point of question. Where do you need the field strength?

ehild
 
  • #9
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0
So you got the potential at the origin, but you need to know the potential function to get the field strength, by taking the negative gradient at the point of question. Where do you need the field strength?

ehild
Is potential not a function?
I need to find the field strength at origo as well.

Edit: Oh right, the potential depends on the point and the potential I calculated is for origo.
 
Last edited:
  • #10
ehild
Homework Helper
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At the origin, the electric field has only z component, because of symmetry.
Ez=-dU/dz. Calculate the potential at a point z on the axis, then take the derivative with respect to z.

ehild
 

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