How Does Charge Distribute on a Conducting Shell Surrounding a Charged Sphere?

[semc]

Suppose a solid insulating sphere of radius a carries a net positive charge uniformly distributed throughout its volume. A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere. Determine the induced charge per unit area on the outer surface of the conducting shell.

I think the inner surface of the conducting shell has an induced charge of -Q but how do we know the charge on the outer surface since the net charge is not given?

 

[kuruman]

You are absolutely correct. The net charge on the outer surface must be given otherwise the problem is ambiguous. The best you can do is to assume that the net charge is zero. This is an educated guess, because if the net charge were not zero, the problem would have specified its value.

 

[collinsmark]

I think the inner surface of the conducting shell has an induced charge of -Q

Sounds right to me so far...

but how do we know the charge on the outer surface since the net charge is not given?

I'm guessing you are supposed to express your answer in terms of Q. 'Kind of like you did above.

Gauss' law will give you the answer to this problem (Gauss' law by the way, is how I assume you came up with the induced charge on the inner surface of the conducting shell -- Gauss' law is what predicts that, starting with the fact that the electrostatic field within a conductor is zero).

[Edit: Hello Kuruman! Looks like you beat me to the punch. But realize this problem can be solved even if the conducting shell has its own net charge, because the problem is not asking for total charge per unit area (on the outer surface); but rather just induced charge per unit area, which ends up being independent of any net charge on conducting shell itself. ]

 

[kuruman]

... But realize this problem can be solved even if the conducting shell has its own net charge, because the problem is not asking for total charge per unit area (on the outer surface); but rather just induced charge per unit area, which ends up being independent of any net charge on conducting shell itself. ]

Point well taken. I read the problem rather hastily, the price of beating you to the punch, collinsmark.

 

[semc]

Yes I got the -Q on the inner surface using Gauss's law. So are we considering the whole sphere with the shell as a point charge when taking r>c? So the charge in this gaussian surface is Q? So that makes the induced charge on the outer layer +Q?

 

[collinsmark]

Yes I got the -Q on the inner surface using Gauss's law. So are we considering the whole sphere with the shell as a point charge when taking r>c?

As long as spherically symmetric, charge distribution is maintained (as it is in this problem), then yes. Right!

So the charge in this gaussian surface is Q? So that makes the induced charge on the outer layer +Q?

Yes! the induced charge (i.e. charge that was induced merely because there is charge on the inner solid sphere) is +Q. Make sure to phrase your answer in induced charge per unit area as your final answer though.

 

[semc]

Got it now thanks guys

 

[collinsmark]

Sorry hope you don't mind if I digress a bit I have another question on inductor in circuit. Is the current passing through the inductor solely the backward current? Does it mean that after a long time no current passes through the inductor suppose a battery is connected throughout the whole time?

Normally you should start a new thread if you have a new, unrelated question.

But if I may give it a shot anyway, I'm pretty sure the answer is no (I could be more specific if you showed a circuit diagram, but perhaps you should do that with a new thread). DC current passes (flows) through an inductor quite easily, as though the inductor was just a wire. But inductors impede quickly changing currents (flowing through them), by changing the voltage across the inductor's terminals, $v = Ldi/dt$. Capacitors are sort of the opposite. Capacitors are governed by $i = Cdv/dt$.

 

[Superstring]

The most obvious thing that comes to mind is that the outer shell becomes polarized.