• Support PF! Buy your school textbooks, materials and every day products Here!

Gauss' Law and charge

  • #1

Homework Statement


Charge is uniformly distributed along the x axis with density ß. Use Gauss' Law to find the electric field it produces, and use this to calculate the work done on a charge Q that moves along the y axis from y = a to y = b.

Homework Equations


Gauss' Law: e0 * electric flux = qencl
where e0 is epsilon0

The Attempt at a Solution


I thought dq = Bdx
q = B integral (x)
= (Bx2)/2
Then if we say EA = qencl/eo
where E is the electric field and A is the area
The areas wind up canceling and I got
E = B/e0
But I'm not sure that's right at all. Particularly since I can't see how I use this electric field to calculate the work done on a charge Q that moves along the x axis. Wouldn't I have to know if the charges were positive or negative? Because presumably opposite charges would cause the charge Q to speed up as it approaches the x axis while same would cause it to slow down... or am I on the wrong track completely?
 

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
2,894
1,232
I thought dq = Bdx
That sounds reasonable to me. :approve:
q = B integral (x)
Wait, where did the x come from? And what happened to your dx? :wink:
 
  • #3
Oh, I see. So I've found the charge, then Q = Bx. But then how do I go about finding the electric field? I can find the flux by dividing by e0 of course. But the flux = integral (flux * dA) which I'm not sure applies to this problem. Or I can say flux = E*A, which I've seen my book randomly do but I don't know if it applies in this case. Is there a general way of trying to find the electric field? Like once you know the flux, shouldn't you theoretically be able to calculate the electric field somehow?
 
  • #4
collinsmark
Homework Helper
Gold Member
2,894
1,232
Oh, I see. So I've found the charge, then Q = Bx.
That's better. :approve:
But then how do I go about finding the electric field? I can find the flux by dividing by e0 of course. But the flux = integral (flux * dA) which I'm not sure applies to this problem. Or I can say flux = E*A, which I've seen my book randomly do but I don't know if it applies in this case. Is there a general way of trying to find the electric field? Like once you know the flux, shouldn't you theoretically be able to calculate the electric field somehow?
Look for symmetry. You need to mentally find some sort of surface, called a Gaussian surface, such that the E·dA is constant over any given section of that surface (and it's okay to break up the closed integral into sections if you have to). Look for the situations where the direction dA is parallel to E for all dA of that section of the closed surface (dA points in a direction normal to the surface). If you can find such a Gaussian surface, then

[tex] \int_S \vec E \cdot \vec {dA} = \int_S EdA = E \int_S dA [/tex]

over that section of the surface. (Remember the above situation does not apply for everything -- you need to have a Gaussian surface such that E·dA is constant.) Note how in this situation, the dot product is merely the multiplication of the the magnitudes, and now E becomes a simple, constant scalar and can be pulled out from under the integral.

By the way, there are only a few situations that you can use this method to find E. You can use it for problems like this one where there is cylindrical symmetry, and the charged line or charged cylinder is infinitely long. You can also use for problems with spherical symmetry. And finally, you can use it for infinitely large planes. But that's about it.

Don't get me wrong, Gauss' law is always true in all situations (classical electrodynamics, that is). And if you happened to already know E for all space, you can always use Gauss' law to find the charge enclosed within any closed surface. But the above situations are the only ones that you will use it to find E.
 
Last edited:

Related Threads on Gauss' Law and charge

  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
11
Views
4K
  • Last Post
Replies
1
Views
727
  • Last Post
Replies
1
Views
13K
  • Last Post
Replies
2
Views
880
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
15K
Replies
5
Views
3K
Top