Calculating q2 and q3 on a Conducting Shell?

[dragonrider]

Consider an insulating sphere with radius of 7 cm. A charge of 13.8561 µC is uniformly distributed throughout this sphere. It is surrounded by a conducting shell. Denote the charge on the inner surface of the shell to be q₂ and the charge on the outer surface of the shell to be q₃. The total charge q on the shell is of 64.0681 µC.

http://img381.imageshack.us/img381/5829/problem2jy1.th.gif

Find q₂ and q₃? Answer in units of µC

For this problem I am not sure where to start because I only have one radius.
https://www.physicsforums.com/attachments/14744

 

[Doc Al]

You don't even need that radius. See what you can deduce by putting Gaussian surfaces in various places and applying Gauss's law. Hint: Take advantage of the fact that it's a conducting shell.

 

[dragonrider]

I am not sure what you mean by conducting shell. Because isn't E = -q/(4*π*ε₀*r²).

 

[Doc Al]

I am not sure what you mean by conducting shell.

An electrical conductor--like a metal. Hint: What's the field within a conductor?

 

[dragonrider]

An electrical conductor--like a metal. Hint: What's the field within a conductor?

I know that the field within a conductor is zero. But I need the charge of the inner an outer surface of the shell.

 

[Doc Al]

Use Gauss's law!

 

[dragonrider]

Yes if I apply Gauss' Law E = q/(4*π*ε₀*r²) but I don't know both E and r. and if I use E = 0 wouldn't that make q₂ = 0

 

[Doc Al]

since the E = 0 inside does q2 = 0?

What does Gauss's law tell you? (Use a Gaussian surface inside the conducting shell, where E = 0.)

 

[dragonrider]

Yes if I apply Gauss' Law E = q/(4*π*ε₀*r²)

I used E = 0 but then q₂ = 0.

 

[Doc Al]

Yes if I apply Gauss' Law E = q/(4*π*ε₀*r²) but I don't know both E and r. and if I use E = 0 wouldn't that make q₂ = 0

The q in Gauss's law stands for the total charge. (q2 is just the charge on the inside of the shell. You'll use the total charge to determine q2.)

 

[dragonrider]

So how do I apply Gauss Law to find q2 and q3 without knowing either E nor r?

P.S.
I know I am asking very basic stuff but I just learned the concept and I haven't fully understand on how to use Gauss Law properly yet.

 

[Doc Al]

So how do I apply Gauss Law to find q2 and q3 without knowing either E nor r?

But you do know E inside the conducting shell. So what does that tell you about the total charge contained within a Gaussian surface inside that shell? (Hint: q2 is just one of the charges inside that surface. What's the other?)

 

[dragonrider]

So the charge contained within the Gaussian surface = q1

 

[Doc Al]

So the charge contained within the Gaussian surface = q1

Nope. (But that is one of the charges within the Gaussian surface.) Draw a diagram showing where the Gaussian surface goes. The problem only defines three charges: q1, q2, and q3. Which are inside the Gaussian surface? Which are outside?

 

[dragonrider]

Inside are q1 and q2. Outside are q3

 

[Doc Al]

Inside are q1 and q2. Outside are q3

Yes! So what's the total charge inside the surface? What must it equal from Gauss's law?

 

[dragonrider]

So [(q1+q2) /(4*π*ε0*r2)] = [q3/(4*π*ε0*r2)]

Does q2 + q3 = 64.0681 µC

 

[Doc Al]

So [(q1+q2) /(4*π*ε0*r2)] = [q3/(4*π*ε0*r2)]

Does q2 + q3 = 64.0681 µC

Answer these questions:
(1) What's the total charge within the Gaussian surface in terms of q1, q2, and q3?
(2) What does E equal at all points on the Gaussian surface?
(3) What does Gauss's law tell you about the total charge?
(4) Write a simple equation expressing the above.

 

[dragonrider]

q1 = -q2
q3 = 2*q1

 

[Doc Al]

I should have said:

So [(q1+q2) /(4*π*ε0*r2)] = [q3/(4*π*ε0*r2)]

Does q2 + q3 = 64.0681 µC

Good. This doesn't follow from Gauss's law, but from the fact that you were given the total charge of the shell.

q1 = -q2

Yes! This does follow from Gauss's law, since Qtotal = 0 = q1 + q2.

q3 = 2*q1

Where does this come from?

 

[dragonrider]

I got it thank you very much for helping me.