Find the Electric potential from surfaces with uniform charge density

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  • #1
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Homework Statement:

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Relevant Equations:

Gauss Law,
I do not have the solutions to this problem so I'm wondering if my attempt is correct.

My attempt at solution: We have two surfaces which we can calculate the area of. I think we can use gauss law to find the electric field and then integrate the E-field to find the electric potential.

So for the circular surface the area is pi*a^2. We get E*pi*a^2 = (1/e0)*ps.

For the spherical surface the area is (4*pi*a^2) / 2 = 2*pi*a^2. We get E*2*pi*a^2 = (1/e0)*ps

Next we solve for V by integrating E with regards to r with the limits being 0 to a.

Am I on the right track here?
 

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  • #2
etotheipi
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I could be wrong, but I don't think Gauss' law is the right thing to do here. For the hemispherical surface you might consider the potential at the centre of a full spherical shell, and use symmetry considerations. For the flat surface, you might want to consider an integration approach.
 
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  • #3
BvU
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Next we solve for V by integrating E with regards to r with the limits being 0 to a.
Am I on the right track here?
There is no ##r## in ##\vec E## when you write ##E \,\pi a^2 = \displaystyle {\rho_s\over \varepsilon_0}##

And your integration limits are not right: V is not 0 at the origin
 
  • #4
haruspex
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@etotheipi is right wrt the appropriate method.

There are a couple of problems with the method you attempted.
The fields vary according to position, yet you seem to have constant values. E.g. at the origin there is no field from the disc.

By convention, the potential should be taken as zero at infinity, so if you want to find the potential at some other point by integrating the field you need to integrate from infinity to the point.
 
  • #5
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Let me try again:
Using the formula from my textbook:
V = 1/(4*pi*e0) integral ps/R ds , where ps is surface charge density.

This formula above is used when there is a continuous distribution which is the case here?

Can we apply the formula to both surfaces (we know the area)?

Then the integration limits for the voltage be from infinity to a in both cases?
 
  • #6
haruspex
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Let me try again:
Using the formula from my textbook:
V = 1/(4*pi*e0) integral ps/R ds , where ps is surface charge density.

This formula above is used when there is a continuous distribution which is the case here?

Can we apply the formula to both surfaces (we know the area)?

Then the integration limits for the voltage be from infinity to a in both cases?
The "integration from infinity" I mentioned is if you are integrating the field.
The better approach, which are now using, is to integrate the contributions to the potential from all the little surface patches ds. So the integration range is now over a surface.
You will need to deal with each surface separately. The hemisphere is easy (why?).
See if you can write the integral for the disc as well.
 
  • #7
BvU
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V = 1/(4*pi*e0) integral ps/R ds , where ps is surface charge density.
Please use ##\rho_s## to get ##\rho_s##, ##\varepsilon_0## to get ##\varepsilon_0## and ##\pi## to get ##\pi##

So: ##V = 1/(4\pi\varepsilon_0) \int \rho_s/R ds ## to get ##V = 1/(4\pi\varepsilon_0) \int \rho_s/R ds ##

But with https://www.physicsforums.com/help/latexhelp/ you quickly learn to improve to something like

$$V = \frac {1} {4\pi\varepsilon_0}\; \int {\rho_s\over R}\ ds $$ to get $$V = \frac {1} {4\pi\varepsilon_0}\; \int {\rho_s\over R} \ ds$$
 
  • #8
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Thanks BvU, was wondering how to write the equations better.

So for example if we want to find the potential at the origin by integrating the E-field, then the limits should be infinity to 0? But that would give us a division by 0.

The other approach was fine then?:

##V = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho_s}{R} ds ## (1)
I'm not sure why the hemisphere would be considered easier than the circle.
Since we can figure out the areas without integrating is it ok to just use that?
We know that the area is : ## a^2\pi ## for the circle

and ## 4*pi*a^2 / 2 ## for the hemisphere.

I guess we could also integrate to find the areas
Hemisphere:

##\int_0^{2\pi} \int_0^\frac{\pi}{2} a \ d\theta d\phi ##​

circle:

## \int_0^{2\pi} \int_0^a r \ dr d\phi ##
Now that we have the area we can "plug it" into the first equation?
 
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  • #9
haruspex
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I'm not sure why the hemisphere would be considered easier than the circle.
Since we can figure out the areas without integrating is it ok to just use that?
That should become clear if you answer this:
##V = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho_s}{R} ds ## (1)
In words, what is R in that integral? What does it equate to in each of the two surfaces?
 
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  • #10
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For the hemisphere R is the distance from origin to each of the ds element, which will be the same at all points.

For the circle the distance will vary. I suppose that is why we need to integrate in this case.

So with this reasoning the hemisphere case should be easier.
##V = \frac{1}{4\pi \epsilon_0 R} * \rho_s * \frac{4 \pi R^2}{2} ##

For the circle we get:
##V = \frac{1}{4\pi \epsilon_0} * \rho_s * \int_0^a \frac {\pi R^2}{R} \ dr##

And then we use superposition at the last step. Is this reasoning correct?

Also I was wondering was since the surfaces are symmetric Gauss Law would be OK to use, but just impractical calculations?
 
  • #11
haruspex
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For the hemisphere R is the distance from origin to each of the ds element, which will be the same at all points.

For the circle the distance will vary. I suppose that is why we need to integrate in this case.

So with this reasoning the hemisphere case should be easier.
##V = \frac{1}{4\pi \epsilon_0 R} * \rho_s * \frac{4 \pi R^2}{2} ##

For the circle we get:
##V = \frac{1}{4\pi \epsilon_0} * \rho_s * \int_0^a \frac {\pi R^2}{R} \ dr##

And then we use superposition at the last step. Is this reasoning correct?

Also I was wondering was since the surfaces are symmetric Gauss Law would be OK to use, but just impractical calculations?
Yes for the hemisphere, but your integral is wrong for the disc.
You need to consider an area element for which all parts are approximately the same distance from the origin. What does that look like? What is its area? What is its contribution to the potential at the origin?
 
  • #12
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My bad, for the circle I think it should be :
## V = \frac{1}{4\pi \epsilon_0} * \rho_s * \int_0^a \frac{2\pi R}{R} \ dR ##
 
  • #13
haruspex
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My bad, for the circle I think it should be :
## V = \frac{1}{4\pi \epsilon_0} * \rho_s * \int_0^a \frac{2\pi R}{R} \ dR ##
Yes.
 
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