Question on charged spheres and shells

In summary, we have a metallic sphere with charge Q1 and electric potential V1, surrounded by a metallic spherical shell with charge Q2 and electric potential V2, with inner and outer radii b and c. The charge is distributed on the surfaces due to them being conductors. To calculate the charge on the sphere, one can integrate the electric field from a to b, then from b to c, and finally from c to infinity. The charge on the inner surface of the shell is -Q1, as the total charge inside a conductor must be 0. To calculate the charge on the outer surface, one can use the formula for electric potential (V2=k^*Q2/c) or use the equation Q inner
  • #1
Markv
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Homework Statement


Suppose we are given a metallic sphere with charge Q1 and electric potential V1 and radius a. Then it is surrounded by a metallic spherical shell with charge Q2 and electric potential V2 with inner and outter radius b and c.

What is the charge and the electric potential everywhere??

Homework Equations


dV = E*dr

V=K*Q/r

The Attempt at a Solution



First all the charge has to be in the surfaces because they are all conductors.

To calculate the charge on the sphere I would integrate the electric field from a to b, then from b to c (where it is 0) and finally from c to infinite.

Now that I have Q1, I know that the charge in the inner surface of the shell must the -Q1 because the total charge inside a conductor must be cero.

Now comes the part where I am stuck.

To calculate the charge on the outter surface of the shell should I just use the formula por the electric portential:

V2=k^*Q2/c

And that's the charge on the outter surface, or instead that's the total charge of the shell and I should do something like this:

Q inner surface + Q outter surface = Total Q

And isolate the outter surface charge from there.

Thank you.
 
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  • #2
Actually, i couldn't quite understand why did question give all q v and d, if you know two (in this case) you can get other one.
Anyway, that is easy, don't confuse yourself with integrals.
lets start from outermost place. there we only deal with the charge density at radius=c. by perfect symmetry and gauss' law, we don't touch other things and potantial there is simply kq'/c=V2
no probs with that right?
now as we go in this shell, there is no potantial change till (including) r=b. and there as you mentioned there is a charge distribution of q''=Q1
then, as we go in by gauss' law we only deal with Q1 and potantial difference between b and a is simply Va-Vb=kQ1/a - kQ1/b and we know Vb=Vc=kQ2/c
rest is ok i think, if you have further problems feel free to ask
 

1. How does the charge on a spherical conductor affect the electric field inside and outside the sphere?

The electric field inside a charged spherical conductor is zero, regardless of the magnitude of the charge. This is because the charges within the conductor repel each other and distribute themselves evenly on the surface, resulting in a cancellation of electric field inside. The electric field outside the sphere, however, is dependent on the charge and follows the inverse square law.

2. What is the difference between a conducting and non-conducting spherical shell?

A conducting spherical shell is a hollow sphere made of a material that allows charges to move freely, such as metal. This means that the charge on the outside of the shell will distribute itself evenly on the surface, and the electric field inside the shell will be zero. A non-conducting spherical shell, on the other hand, is made of a material that does not allow charges to move freely, so the charge on the outside of the shell will not distribute evenly and the electric field inside will not be zero.

3. How does the charge on a conducting spherical shell affect the electric field outside the shell?

In a conducting spherical shell, the charge distributes itself evenly on the surface, resulting in a cancellation of electric field inside the shell. However, the charge on the outside of the shell will still contribute to the electric field outside. The electric field outside a charged conducting shell is the same as the electric field of a point charge located at the center of the shell with the same magnitude of charge.

4. Can a charged conducting shell be shielded from external electric fields?

Yes, a charged conducting shell can be shielded from external electric fields by placing a second conducting shell around it. This is known as the Faraday cage effect, where the external electric field induces an equal and opposite charge on the inner surface of the outer shell, cancelling out the electric field inside the inner shell.

5. How does the charge distribution on a conducting shell change when a point charge is placed inside it?

When a point charge is placed inside a conducting shell, the charge on the shell redistributes itself, with more charge accumulating on the side of the shell closest to the point charge. This is because the point charge induces a separation of charges on the shell, with the side closest to the point charge having an opposite charge and the side farthest having the same charge. This redistribution of charge results in a change in the electric field both inside and outside the shell.

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