Conducting spheres Gauss's law

In summary, there are two conducting spheres, one hollow outer sphere with +60nC of charge and a smaller inner sphere with +20nC of charge. The net charge on the outer sphere is +60nC, but when connected by a wire to the inner sphere, the charge redistributes to +80nC on the outer surface and +20nC on the inner surface. The ambiguity in the question may have caused confusion, but ultimately the net charge on the outer sphere remains the same.
  • #1
bmxicle
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0

Homework Statement



There are two conducting spheres, one hollow outer sphere which originally had +60nC of charge on it, and a smaller inner sphere enclosed by the hollow outer sphere which contains +20nC of charge.


Homework Equations


INT(E.DA) = Qenclosed/epsilon0

The Attempt at a Solution



Well first off obviously the inner sphere has +20nC of charge on it and that can't change. For the outer sphere, if you place a spherical gaussian surface that runs through the middle of the outer sphere the net Qenclosed must be zero as the Electric field is zero everywhere inside a conductor. So by this there must be -20nC on the inside of the shell and 80nC on the outside of the shell.

Now here comes the part I'm confused by. Isn't there still only +60nC of net charge on the outer sphere? The answer I was given says there is +80nC charge on the outer sphere, but i don't understand how that can be the net charge on the outer sphere.
 
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  • #2
bmxicle said:
Now here comes the part I'm confused by. Isn't there still only +60nC of net charge on the outer sphere? The answer I was given says there is +80nC charge on the outer sphere, but i don't understand how that can be the net charge on the outer sphere.
Are you sure they said that the net charge on the outer sphere was +80nc? In any case, you are correct. But I suspect that the answer key meant that the outer surface had a charge of +80nC.
 
  • #3
That must be what they meant, though the question seems to be somewhat ambiguous. I just wanted to make sure I wasn't making a conceptual error though so it doesn't really matter all that much.

Thanks.
 
  • #4
I would think this too. The outer surface would have a +80 nC charge as stated, and the net charge would be 60 nC.

Indeed, conservation of charge would require the outer shell to have a net charge of 60 nC.
 
  • #5
bmxicle said:
That must be what they meant, though the question seems to be somewhat ambiguous.
Just for fun, can you post the question exactly as given?
 
  • #6
"Two pairs of charged, hollow, spherical, conducting shells are connected with wires and switches. The system AB is very far from CD. The radius of each large shell is four times that of the small shell. Before the switches are closed, the small shells have charges of +20nC and the large shells have +60nC. After the switches are closed, rank the net charge on each shell. "

There is also a diagram Where shells A and B with ra>rb are connected by a wire, and C encloses shell d. I only wasn't sure about the part I asked about but that was the entire question.
 
  • #7
bmxicle said:
"Two pairs of charged, hollow, spherical, conducting shells are connected with wires and switches. The system AB is very far from CD. The radius of each large shell is four times that of the small shell. Before the switches are closed, the small shells have charges of +20nC and the large shells have +60nC. After the switches are closed, rank the net charge on each shell. "
The connecting wire makes a difference! (You left that out in your description.) Can I assume that the shells you were talking about were not connected by a wire? (One pair of shells had a wire, the other didn't?)

In any case, you are right. They do seem to ask for the net charge.
 
  • #8
Well in the question i asked, yes they were not connected by a wire, but now that you mention that, that's not what the actual question asked :p and i can now see a very faint partial wire on the diagram.

If they're connected the charge will all go to the outside because it should essentially act as one large conductor with the wire attached so 80nC makes sense.
 
  • #9
bmxicle said:
If they're connected the charge will all go to the outside because it should essentially act as one large conductor with the wire attached so 80nC makes sense.
Yep, there you go.
 

Related to Conducting spheres Gauss's law

1. What is Gauss's law and how does it apply to conducting spheres?

Gauss's law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed charge. For conducting spheres, this law states that the electric field at any point on the surface of the sphere is directly proportional to the charge enclosed within the sphere and inversely proportional to the radius of the sphere.

2. How is the electric field inside and outside of a conducting sphere related?

Inside a conducting sphere, the electric field is zero. This is because charges inside the sphere will redistribute themselves in such a way that the electric field cancels out. Outside the sphere, the electric field follows the inverse square law, meaning it decreases as the distance from the center of the sphere increases.

3. What is the significance of a conducting sphere in Gauss's law?

A conducting sphere is a useful and commonly used example in Gauss's law because it allows for simplification of the calculations. Due to the symmetry of a sphere, the electric field can be easily calculated using the law, making it a useful tool in understanding the behavior of electric fields.

4. How does the charge distribution on the surface of a conducting sphere affect the electric field?

The charge distribution on the surface of a conducting sphere has no effect on the electric field inside the sphere. However, it does affect the electric field outside the sphere. A non-uniform charge distribution on the surface will result in a non-uniform electric field outside the sphere.

5. Can Gauss's law be used to calculate the electric field for non-spherical conducting objects?

Yes, Gauss's law can be applied to any closed surface, regardless of its shape. However, for non-spherical objects, the calculation of the electric field may be more complex and require other mathematical techniques.

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