Flux of Electric field through sphere

Click For Summary
The discussion centers on understanding electric flux through a sphere, highlighting that it is determined solely by the charges enclosed within the sphere, while external charges do not influence the flux. The initial calculation mistakenly included an external charge, leading to confusion. Gauss's law is introduced as the principle explaining that electric flux is related to the net charge inside a closed surface. An analogy is made comparing electric charges to sources and sinks of water, illustrating how flux behaves with enclosed and external charges. Ultimately, the key takeaway is that only enclosed charges affect the electric flux through a sphere, aligning with Gauss's law.
Tesla In Person
Messages
34
Reaction score
13
Homework Statement
What is the flux of electric field through sphere with charges inside and outside of the sphere?
Relevant Equations
Electric field flux = Q/ e0
My attempt: We have 3 charges inside 2 +ve and 1 -ve so i just added them up. 4 + 5 +(-7) = 2q
Then there is a -5q charge outside the sphere. I did 2q + (-5q)= -3q . The electric field flux formula is Flux= q/ E0 . So i got -3q/E0 which is obviously wrong : ) . After quick googling , I discovered that the electric flux through a sphere is only dependent on charges enclosed by the sphere , charges outside don't affect electric flux through sphere. Is it correct and why it is the case? I am not familiar with Gauss' law so how do i get started. Thank you
 

Attachments

  • physics 5.PNG
    physics 5.PNG
    12.3 KB · Views: 167
Physics news on Phys.org
Tesla In Person said:
After quick googling , I discovered that the electric flux through a sphere is only dependent on charges enclosed by the sphere , charges outside don't affect electric flux through sphere. Is it correct and why it is the case?
That is Gauss’ law.
 
The sources of electric flux are charges. Think of them as sources (positive) and sinks (negative) of water. If a closed surfaces encloses source, water comes out of the surface at every point on it. Likewise, if a sink is enclosed, water goes into the surface at every point on it. If you have a combination of sources and sinks you need to add all the positives and negatives and see what the net is.

Now if a source is outside the surface and there are no sources or sinks enclosed by the surface, what goes into it through the side closer to the source must come out the other side. You put all this together and you have an intuitive understanding of Gauss's law.
 
  • Like
Likes Tesla In Person
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

  • · Replies 9 ·
Replies
9
Views
820
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
809
  • · Replies 22 ·
Replies
22
Views
3K
Replies
9
Views
2K
Replies
23
Views
3K
Replies
26
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K