Gauss's (hard to understand) Law

  • #1
physics noob
41
0
ok here the problem I am having...
a point charge Q rests at the center of an uncharged thin spherical condcting shell. with is the the electric field E, as a function of r...then it asks for r less than the inner radius of the shell... then it asks for inside the shell, and finally beyond the shell.

now how I am invisioning the figure is almost like a washer, with a thickness, like the rings of saturn...

so for the first part i need to find E as a function of r...so i thinks to myself... E0 (constant) = 1/ (4(k)(pi)) and gauss's law states
flux = EA flux also = q/E0 so EA = q/ Eo after a little algebra
i get E= Qk/ R^2 using SA of a sphere for area... ok so I am pretty sure that's right for the first question...if not please tell me where i went wrong...next part asking for r less than inner radius of shell?
i have no idea what to put...one thing i do understand is that when there is a charge in a shell, the electric field continues as if there was no shell present, by this i mean if Q is positive , negative charges accumulate on inner shell circumference, and positive accumlate on SA of total spherical shell... this is all kind of new to me, and i don't think i have a great grasp on all of this,,,,,,also a coule off topic questions...what is the diff,,,between q (test charge) and Q////// as always thanks a million for any help/ information.
 

Answers and Replies

  • #2
Lyuokdea
149
0
I assume that the second part is asking you what is the electric field less than the outer surface of the shell but more than the inner surface of the shell, i.e. what is the electric charge inside the conducting shell.

Your right, outside of the whole shell, you will not see the shell because it will not insulate the charge, so it will be like the shell didn't exist.

And also inside of the shell, it will be like the shell doesn't exist because the charge doesn't need to move through the shell to get from the charge to the point where you are testing.

Hope this helps,

~Lyuokdea
 
  • #3
physics noob
41
0
yes, it does,,, so i guess i would just describe what the electric field does,,, i think I am getting confused because of the lack of numbers, i guess i just explain what's happening. thanks
 
  • #4
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,176
21
You are asked to calculate E(r), when :

(i) r < R(in)

(ii) R(in) < r < R(out)

(iii) r > R(out)

In the first and third cases, the total enclosed charge is Q. So, you can find the fields from Gauss' Law, and they will have the same form. But this is not so in the second case. The charge enclosed here will be different because of the induced charges on the inner surface of the shell. So, find the new value of the enclosed charge, and again calculate the field from GL.

Is the result you get for (ii) surprising, or expected ?
 
  • #5
physics noob
41
0
so would Q=0 inside the shell since all charges are found on the surface of conductors? thanks everyone for the help
 
  • #6
Doc Al
Mentor
45,450
1,911
physics noob said:
so would Q=0 inside the shell since all charges are found on the surface of conductors? thanks everyone for the help
I think you are talking about the total charge Q within a spherical Gaussian surface that is inside the material of the shell. If so, since the electric field is zero within the conducting shell, the total charge Q contained in that Gaussian sphere is zero.
 
  • #7
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,176
21
Doc Al said:
I think you are talking about the total charge Q within a spherical Gaussian surface that is inside the material of the shell. If so, since the electric field is zero within the conducting shell, the total charge Q contained in that Gaussian sphere is zero.
In this case, though, I think you want to reason the other way round. You are asked to find the field on this Gaussian surface, so you must determine the charge enclosed. First there's +Q at the origin, and then there's -Q distributed on the inside of the shell. So the total enclosed charge is zero, and hence, the field within the material of the shell is zero (as expected, since the field inside the body of a conductor must be zero).
 
  • #8
Doc Al
Mentor
45,450
1,911
Gokul43201 said:
In this case, though, I think you want to reason the other way round. You are asked to find the field on this Gaussian surface, so you must determine the charge enclosed. First there's +Q at the origin, and then there's -Q distributed on the inside of the shell. So the total enclosed charge is zero, and hence, the field within the material of the shell is zero (as expected, since the field inside the body of a conductor must be zero).
OK, but how do you know that there's -Q distributed on the inside of the shell? (You need to start with some knowledge; I usually make use of the fact that the field must be zero within the conductor.)
 

Suggested for: Gauss's (hard to understand) Law

Replies
1
Views
359
  • Last Post
Replies
1
Views
286
  • Last Post
Replies
14
Views
628
Replies
3
Views
773
  • Last Post
Replies
20
Views
690
Replies
15
Views
569
Replies
10
Views
576
Top