Calculate the charge in a hollow spherical conductor

  • Thread starter Renaldo
  • Start date
  • #1
58
0

Homework Statement




Consider a hollow spherical conductor with total charge +5e. The outer and inner radii are a and b, respectively.

(a) Calculate the charge on the sphere's inner and outer surfaces if a charge of -3e is placed at the center of the sphere. (Use the following as necessary: e.)


Homework Equations



Gauss's Law?

The Attempt at a Solution



I honestly don't have much of a clue. I feel like the problem should be pretty simple, but I'm not grasping all the aspects of Gauss's Law.

I hypothesized that the charge was 0 on the inner surface, and surmised that the charge on the outer surface should be equal to the total charge. This was wrong. I then hypothesized that the inner charge would be -3e and that the outer charge would need to be added to this to result in 5e, so 8e. Also not correct.
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,190
6,290
I then hypothesized that the inner charge would be -3e
Close, but you made a small mistake. Think that through again.
 
  • #3
58
0
The inner surface would be +3e because the positive charge would be attracted by the negative field, accumulating on the outside of the sphere. The charge on the outer surface would be +2e because the field is weaker there, and we know the total charge is +5e. Is this correct?
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,190
6,290
The inner surface would be +3e because the positive charge would be attracted by the negative field, accumulating on the outside of the sphere. The charge on the outer surface would be +2e because the field is weaker there, and we know the total charge is +5e. Is this correct?
Sounds right to me. But it also sounds as though you are basing this on a hunch rather than on known principles. Consider the field in the conductor, i.e. between the inner and outer surfaces. What do you know about that? How would you calculate it from the isolated charge in the cavity, the charge on the inner surface, and the charge on the outer surface?
 
  • #5
58
0
All right, I will give it a shot.

q/εo = ∫Eda
q/εo = E4∏r2

E = kq/r2

Inner radius = b

E1 = kq1/b2

Outer radius = a

E2 = kq2/a2

E1+E2 = kq1/b2 + kq2/a2

E3 = E1+E2
E3 = kq3/r2

kq3/r2 = kq1/b2 + kq2/a2

This leaves me with three variables.
 
  • #6
Dick
Science Advisor
Homework Helper
26,260
619
All right, I will give it a shot.

q/εo = ∫Eda
q/εo = E4∏r2

E = kq/r2

Inner radius = b

E1 = kq1/b2

Outer radius = a

E2 = kq2/a2

E1+E2 = kq1/b2 + kq2/a2

E3 = E1+E2
E3 = kq3/r2

kq3/r2 = kq1/b2 + kq2/a2

This leaves me with three variables.
The electric field inside of a conductor is zero, right? Pick a radius r where a<r<b. So it's inside of the conductor. Then use Gauss' law at radius r. What does that tell you?
 
Last edited:
  • #7
58
0
The electric field inside of a conductor is zero, right? Pick a radius r where a<r<b. So it's inside of the conductor. Then use Gauss' law at radius r. What does that tell you?
Perhaps I am confused as to the nature of hollow spherical conductors. If I understand the problem correctly, a point where a<r<b would be inside the metal of the shell. What is the purpose of using Gauss's Law here? Would not the field at this point be = to 0?
 
  • #8
Dick
Science Advisor
Homework Helper
26,260
619
Perhaps I am confused as to the nature of hollow spherical conductors. If I understand the problem correctly, a point where a<r<b would be inside the metal of the shell. What is the purpose of using Gauss's Law here? Would not the field at this point be = to 0?
Yes, it would. So the enclosed charge must be zero, yes?
 
  • #9
58
0
Yes, it would. So the enclosed charge must be zero, yes?
(a) Calculate the charge on the sphere's inner and outer surfaces if a charge of -3e is placed at the center of the sphere.

So the enclosed charge is not equal to -3e? What does -3e represent?
 
  • #10
Dick
Science Advisor
Homework Helper
26,260
619
(a) Calculate the charge on the sphere's inner and outer surfaces if a charge of -3e is placed at the center of the sphere.

So the enclosed charge is not equal to -3e? What does -3e represent?
The enclosed charge is the sum of -3e plus the charge on the inner surface of the conductor. The Gauss integral is zero. Hence, what's the charge on the inner surface on the conductor? Help me out here.
 
  • #11
58
0
The enclosed charge is the sum of -3e plus the charge on the inner surface of the conductor. The Gauss integral is zero. Hence, what's the charge on the inner surface on the conductor? Help me out here.
Oh, I see. The enclosed charge is equal to the charge on the inner radius, b, plus the charge in the center. It's obviously 3e, since the field is equal to 0.

As for the outer radius, I know that the spherical conductor has total charge, +5e. In order for it to have this total charge, I add the charges on the two surfaces together:

3e + 2e = 5e

It's really quite simple, but I was failing to understand the logic behind it. Thanks for the help.
 

Related Threads on Calculate the charge in a hollow spherical conductor

Replies
5
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
0
Views
3K
Replies
4
Views
13K
Replies
4
Views
197
  • Last Post
Replies
18
Views
9K
  • Last Post
Replies
1
Views
4K
Replies
4
Views
3K
Replies
17
Views
24K
Replies
2
Views
1K
Top