Determining electric field using gauss's law--different distributions

Click For Summary
Gauss's law can be applied effectively when there is symmetry in the charge distribution. The discussion highlights that distributions A and D lack the necessary symmetry for using Gauss's law, while B and C are more suitable. The key to applying Gauss's law is to choose a surface where the electric flux remains constant. Participants suggest that distribution A could be analyzed as a line of charge with an imaginary cylinder, but the asymmetry complicates its application. Overall, understanding the symmetry of charge distributions is crucial for determining the electric field using Gauss's law.
Helenah
Messages
3
Reaction score
0
Homework Statement
We are supposed to determine which (if any) of the distributions will need to/could use Gauss's law to determine the electric field
Relevant Equations
flux = Q/electric constant = EA
Screen Shot 2021-03-18 at 11.41.03 PM.png

These are the 4 distributions shown, and I have to determine which two distributions (or none at all) can use Gauss's law to determine the electric field.

So electric flux = EA = Q/electric constant.

Since all of them have charges, I could do something like Q/(A*electric constant) to get the electric field—that's where I'm confused, because I think I could actually use Gauss's law on all four of them. The only other way I can think of is that Gauss's law is applied to surfaces so that probably excludes lines (A) and disks (D), which do not really have surface area.
Am I right in thinking so? Is there a better / correct way to get to the answer?
 
Last edited:
Physics news on Phys.org
You perhaps need to think about how you would practically use Gauss's law to determine the electric field.

Hint: think about symmetry.
 
@PeroK
Hmm. Distribution A could potentially be viewed as line of charge + imaginary cylinder to get some electric field, but the charges aren't symmetric and hence will not be able to use gauss's law (?).
I'm not sure where else it's not symmetric though.
 
Helenah said:
@PeroK
Hmm. Distribution A could potentially be viewed as line of charge + imaginary cylinder to get some electric field, but the charges aren't symmetric and hence will not be able to use gauss's law (?).
I'm not sure where else it's not symmetric though.
The trick to use Gauss's law effectively is to have a surface across which the electric flux must be constant.

Your intuition is right that A and D are problematic and B and C are better. Can you explain more clearly how you do things in cases B and C?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
9
Views
733
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K