Finding the equipotential surfaces of a finite line of charge

In summary: Can I ask which book? Our standard text for the class is Technically Griffiths but our professor sometimes likes using Jackson since he himself learned from Jackson. Also I go to Cal if that helps at all.Richard Becker's old book
  • #1

Homework Statement


Consider a line of charge stretching along the z-axis from -L to +L. Find
the potential everywhere. What are the surfaces of constant potential. (The next question answers the previous question and says its a prolate ellipsoid.

Homework Equations


I will upload an image of my attempted solution. dV=kdq/r

The Attempt at a Solution


I got something of the form ax^2 + by^2 = cz
IMG_2449.jpg
 

Attachments

  • IMG_2449.jpg
    IMG_2449.jpg
    31.5 KB · Views: 456
Physics news on Phys.org
  • #2
Its supposed to be 4z'l = x'^2 +y'^2
 
  • #3
Faizan Samad said:

Homework Statement


Consider a line of charge stretching along the z-axis from -L to +L. Find
the potential everywhere. What are the surfaces of constant potential. (The next question answers the previous question and says its a prolate ellipsoid.

Homework Equations


I will upload an image of my attempted solution. dV=kdq/r

The Attempt at a Solution


I got something of the form ax^2 + by^2 = cz
[ ATTACH=full]230780[/ATTACH]
It's extremely difficult to read that image. I did manage to get some idea of what you did.

I wrote essentially what you have for the potential, V, but I write it as the potential at point (x, y, z) and use z' as the integration variable.

Here's the your integral for the potential with the above (minor) modifications in variable names:

## \displaystyle V(x,\,y,\,z) = \frac{\lambda } { 4\pi \varepsilon_0} \int_{-L}^L \frac{ dz' } { \, \sqrt{ x^2 + y^2 + (z' -z)^2 } \,} ##

After following the next few steps I see a problem.

You use a form of the anti-derivative which is in terms of a logarithm. Nothing wrong with that. But if we call that anti-derivative, ##\ \ln(F(z')) \,,\ ## then plugging in the limits of integration gives ##\ \ln(F(L))-\ln(F(-L)) \ \ \rightarrow \ \ \ln \left( \frac{F(L)} {F(-L)} \right)##.

The problem is that then you pick the seemingly convenient choice of solving ## \ F(L) = F(-L) \ since that will make the argument of the logarithm be a constant, namely 1 . The problem with that is this would make the integral equal to zero. That's not possible. The integrand is greater than zero for all values of z' .

As for the answer of being a "prolate ellipsoid", I suppose they mean a prolate spheroid I'm somewhat skeptical of that. Perhaps the equi-potential surfaces approach this shape at a distance sufficiently far from the line charge.
 
  • #4
SammyS said:
It's extremely difficult to read that image. I did manage to get some idea of what you did.

I wrote essentially what you have for the potential, V, but I write it as the potential at point (x, y, z) and use z' as the integration variable.

Here's the your integral for the potential with the above (minor) modifications in variable names:

## \displaystyle V(x,\,y,\,z) = \frac{\lambda } { 4\pi \varepsilon_0} \int_{-L}^L \frac{ dz' } { \, \sqrt{ x^2 + y^2 + (z' -z)^2 } \,} ##

After following the next few steps I see a problem.

You use a form of the anti-derivative which is in terms of a logarithm. Nothing wrong with that. But if we call that anti-derivative, ##\ \ln(F(z')) \,,\ ## then plugging in the limits of integration gives ##\ \ln(F(L))-\ln(F(-L)) \ \ \rightarrow \ \ \ln \left( \frac{F(L)} {F(-L)} \right)##.

The problem is that then you pick the seemingly convenient choice of solving ## \ F(L) = F(-L) \ since that will make the argument of the logarithm be a constant, namely 1 . The problem with that is this would make the integral equal to zero. That's not possible. The integrand is greater than zero for all values of z' .

As for the answer of being a "prolate ellipsoid", I suppose they mean a prolate spheroid I'm somewhat skeptical of that. Perhaps the equi-potential surfaces approach this shape at a distance sufficiently far from the line charge.
Thanks for responding and for the insight. I was unaware that uploading photos is not the right way to go in the future I will learn latex and use it.
 
  • #5
SammyS said:
As for the answer of being a "prolate ellipsoid", I suppose they mean a prolate spheroid I'm somewhat skeptical of that. Perhaps the equi-potential surfaces approach this shape at a distance sufficiently far from the line charge.
I happen to remember this problem being worked out in book I have. Interestingly, it turns out that the equipotential surfaces are prolate spheroids even when close to the line source.
 
  • Like
Likes SammyS
  • #6
TSny said:
I happen to remember this problem being worked out in book I have. Interestingly, it turns out that the equipotential surfaces are prolate spheroids even when close to the line source.
Can I ask which book? Our standard text for the class is Technically Griffiths but our professor sometimes likes using Jackson since he himself learned from Jackson. Also I go to Cal if that helps at all.
 
  • #8
TSny said:
I happen to remember this problem being worked out in book I have. Interestingly, it turns out that the equipotential surfaces are prolate spheroids even when close to the line source.
That amazes me ! :oldsurprised:
- especially for locations very close to the line charge relative to L and near the center of the line charge. I would have expected equipotential surfaces there to be virtually cylindrical.

Now, to make a little tweak, to that physical intuition I used to claim and was often proud of.

but ... At my age, maybe no need to bother. :wink:
 
  • #9
SammyS said:
That amazes me ! :oldsurprised:
- especially for locations very close to the line charge relative to L and near the center of the line charge. I would have expected equipotential surfaces there to be virtually cylindrical.
If the charge were spread along a conducting line (needle), then the equipotential surfaces would approach the shape of the needle when you get in close. But, then, the charge density along the needle would be very nonuniform (concentrated near the ends).

Now, to make a little tweak, to that physical intuition I used to claim and was often proud of.

but ... At my age, maybe no need to bother. :wink:
I'm with you.

By the way, whenever I dip into Becker's book to read about a specific topic, I am usually very pleased with his treatment. The book has an interesting history. See the first paragragh of the forward to the book just before the table of contents at the link https://www.amazon.com/gp/product/0486642909/?tag=pfamazon01-20

The lineage of the book goes all the way back to a popular German text of Foppl (1894), which apparently had some influence on Einstein.
http://einstein-special-relativity.com/august-foppl/
 
  • #10
TSny said:
If the charge were spread along a conducting line (needle), then the equipotential surfaces would approach the shape of the needle when you get in close. But, then, the charge density along the needle would be very nonuniform (concentrated near the ends).
Yes, I see. It took me a couple of readings to get what you're saying, but yes, that makes sense.
Considering that I thought the equipotential surfaces would be cylindrical in shape very near the line of charge. If we have such a shape for an equipotential surface, then replacing that surface with a conducting surface of the same shape (and appropriate charge) would produce the the same field at points external to that conductor. Knowing that the majority of charge on such a conductor would migrate to the ends, shows that such an equipotential surface is inconsistent with being due to a uniform distribution on a line segment,

Your explanation did appeal to physical intuition . Thanks!
I'm with you.

By the way, whenever I dip into Becker's book to read about a specific topic, I am usually very pleased with his treatment. The book has an interesting history. See the first paragragh of the forward to the book just before the table of contents at the link https://www.amazon.com/gp/product/0486642909/?tag=pfamazon01-20

The lineage of the book goes all the way back to a popular German text of Foppl (1894), which apparently had some influence on Einstein.
http://einstein-special-relativity.com/august-foppl/
Yes. This book looks very interesting. I may order a copy. (Maybe I'm not too old to see things in a different light.) I would say new light, but the book's roots are even older than I am.
 

What is an equipotential surface?

An equipotential surface is a surface in space where all points have the same potential energy. This means that any charge placed on the surface will experience the same electric potential.

How do you find the equipotential surfaces of a finite line of charge?

To find the equipotential surfaces of a finite line of charge, you can use the following equation: V = kλ/ρ, where V is the potential, k is the Coulomb's constant, λ is the linear charge density, and ρ is the distance from the line of charge.

What is the significance of finding equipotential surfaces?

Finding equipotential surfaces is important because it allows us to understand the behavior of electric fields and how charges interact with each other. It also helps us visualize the distribution of charges in a given system.

What factors affect the shape of equipotential surfaces?

The shape of equipotential surfaces is affected by the distribution of charges in a system, the distance from the charges, and the strength of the electric field. In the case of a finite line of charge, the linear charge density also plays a role in determining the shape of the surfaces.

Can equipotential surfaces intersect?

No, equipotential surfaces cannot intersect. This is because if they were to intersect, it would mean that two different points on the same surface have different potentials, which goes against the definition of an equipotential surface.

Suggested for: Finding the equipotential surfaces of a finite line of charge

Replies
4
Views
599
Replies
13
Views
2K
Replies
3
Views
785
Replies
33
Views
2K
Replies
2
Views
688
Replies
19
Views
584
Replies
3
Views
535
Back
Top