Gauss's law help

  • #1
bmandude156
23
0

Homework Statement


A 8.0 charge and a -4.9 charge are inside an uncharged sphere.
What is the electric flux through the sphere?


Homework Equations


It says in the book that Electric flux = E(4piR^2) although I do not know the radius and yet still dont believe thats the way to go to solving this problem.


The Attempt at a Solution


I attempted to add the charges together because I had no other idea what to do with them. And then tried to use the equation of a sphere for electric flux which is above. I do not know for sure if that is the way to proceed but I did not get the answer correct.
 

Answers and Replies

  • #2
sgd37
213
8
you don't need the radius that's the point of gauss's law. Just substitute coulombs law for the electric field i.e. [tex] \frac{q}{4 \pi \epsilon_0 r^2} [/tex] into your equation for flux
 
  • #3
bmandude156
23
0
I have immediate reservations after seeing your reply for i still see r^2 in your denominator. Isnt that going to be a problem and as for the charges, should I merely take the sum of them to be my q.
 
  • #4
betel
318
0
sgd is talking about the electric field. You have to do something to it to get the electric flux.
 
  • #5
bmandude156
23
0
i believe the equation at the moment is Electic flux = q/4piEnotr^2. by doing something I think ur implying..doing something to get rid of r^2. but i havent a clue how or if thats what ur suggesting i do. sgd said that i didnt have to worry about r anymore but thats not true. coloumbs law had one itself. im confused
 
  • #6
betel
318
0
What are the quantities entering the formula for the electric flux. Your last formula is not correct.
 
  • #7
bmandude156
23
0
well now im more confused than ever. sgd gave me the formula. so r u saying that his formula was wrong.
 
  • #8
kaksmet
83
0
In your first formula, you have the electric field, E, on the right hand side. See if you can use sgd37 formula for the electric field to calculate your flux.
 
  • #9
betel
318
0
Do you know the difference between electric field and electric flux?
 
  • #10
bmandude156
23
0
electric flux=E(4piR^2) and Electic flux = q/4piEnotr^2(was sgd's equation). now, what i think ur saying is take sgd's equation of electic flux and put it into mine. so the result is as follows: E(4piR^2) = q/4piEnotr^2. what i hoped would happen is by doing this, is alregbriaclly get rid of R. but i dont think that could happen. and for that matter...what is E now?!?....im afraid this is diogn more harm than good
 
  • #11
betel
318
0
Perhaps you should first think about the equations and not randomly plugging one into the other that has the same letters in it.
 
  • #12
bmandude156
23
0
you don't need the radius that's the point of gauss's law. Just substitute coulombs law for the electric field i.e. [tex] \frac{q}{4 \pi \epsilon_0 r^2} [/tex] into your equation for flux
. im must be misunderstanding what the equation looks like than
 
  • #13
betel
318
0
I have to repeat my question.

Do you know the difference between electric field and electric flux?

We can continuer argueing forever but if you don't know the difference we will get nowhere.
 
  • #14
bmandude156
23
0
Electric flux is proportional to the number of electric field lines going through a surface. NO arguing here mate. just need help. electric field is like the magnetic field that pertains to charges attraction with eachother
 
  • #15
betel
318
0
Well, that is good.
Now what does Gauss's Law state? You already refer to it in the title, so why don't you do it the easy way?
 
  • #16
bmandude156
23
0
Electric flux = E(4piR^2)..that one? But there was a problem with it remember...r^2?
 
  • #17
betel
318
0
Now what is E in that formula?

Maybe Gauss's law in words is easier here.
 
  • #18
bmandude156
23
0
electric field...sorry it took so long for me to respond...didnt see ur response on tihs page
 
  • #19
betel
318
0
And the formula for the electric field is?
 
  • #20
bmandude156
23
0
E=F/q
 
  • #21
betel
318
0
I'm not sure what your F and q are but this formula seems strange.
 
  • #22
bmandude156
23
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hahah right from the book in my previous chapter. what is urrr formula for the electric field so we can progress?
 
  • #23
betel
318
0
Then tell me what F and q are supposed to mean. I dont't know every definition of every book by heart and without that letters are meaningless.

And btw. I dont' have to solve this question.
 
  • #24
bmandude156
23
0
F is the electric force and q is as usual, the charge. perhaps this isnt the proper equation seeing as ur unfamiliar with it
 
  • #25
betel
318
0
Well it is not exactly the equation that will be of much help to us.

Didn't you derive a formula for the electric field of a point charge? I.e. if you have a charge q at radius r, what is the strength of the electric field?

And I still think you should have a statement of Gauss's law in words. This would kill this question in one go.
 
  • #26
bmandude156
23
0
I will try to reply to ur post ASAP and i appreciate ur post. But ironically, I have a physics class at the moment. brb later
 
  • #27
sgd37
213
8
wow this is one funny thread

so for the sake of my sanity; denoting electric FLUX with the Letter [tex] \Phi [/tex] and the electric FIELD as [tex] E [/tex]

Now [tex] \Phi = E 4 \pi r^2 [/tex] and [tex] E = \frac{q}{4 \pi \epsilon_0 r^2} [/tex]

Substituting for [tex] E [/tex] in [tex] \Phi [/tex] you have

[tex] \Phi = \frac{q 4 \pi r^2}{4 \pi \epsilon_0 r^2} = \frac{q}{ \epsilon_0} [/tex]

This is Gauss's law that the flux through a surface is the charge encompassed by that surface divided by the vacuum permittivity
 
  • #28
bmandude156
23
0
ALRIGHT SO... what about the two charges. should i just sum them
 
  • #29
sgd37
213
8
yeah
 
  • #30
bmandude156
23
0
.....but that didnt work good sir. the answer was incorrect. i got 3.5 X 10^-11.
 
  • #31
sgd37
213
8
I'm pretty sure Gauss is correct. But what units are you dealing with. If the charges are in coulombs then you should have a flux of 3.5 * 10^11 Vm which seems ridiculous. But if they were in terms of the electron charge then the flux would be 5.6 * 10^-8 Vm
 
Last edited:
  • #32
bmandude156
23
0
KN*m2/C is the units...i tried 3.5 x 10^-6 but it failed
 
  • #33
sgd37
213
8
i meant the units for charge. What is the answer supposed to be
 
  • #34
bmandude156
23
0
the unit i mentioned. the KN*m^2/C
 
  • #35
bmandude156
23
0
the charges are in mirco coloumbs
 

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