Rectifying my logic of Gauss's Law

[The Blind Watchmaker]

Homework Statement

Homework Equations

The Attempt at a Solution

E4Πr² = Q/∈₀
49000⋅4Π4.1² =Q/∈₀
Q = 91.6 μC
Q_shell = Q = 91.6 μC
Q_shell = Q_inner + Q_outer
91.6 = -3.3 + Q_outer
Q_outer = 94.9 μC

Can someone point out the error? I have skipped too many lectures and I am catching up right now. Any help would be appreciated! :)

 

[gneill]

Pay close attention to the actual question posed! Precisely what charge are they going for?

 

[The Blind Watchmaker]

Pay close attention to the actual question posed! Precisely what charge are they going for?

Ah I think I get it. Since the electric field inside a conductor must be 0, there must exist a charge on the inner surface of the larger shell that is equal and opposite to 3.3 μC. Thus, the remaining charge is 91.6 - 3.3 = 88.3 μC. Is this what you are hinting at?

 

[gneill]

Ah I think I get it. Since the electric field inside a conductor must be 0, there must exist a charge on the inner surface of the larger shell that is equal and opposite to 3.3 μC. Thus, the remaining charge is 91.6 - 3.3 = 88.3 μC. Is this what you are hinting at?

Something like that, but take care... Sure, the inner shell's charge is matched and effectively "cancelled" by the charge on the inner surface of the outer shell. So what remains on the outer surface must be what Gauss's Law "sees" as the contained charge...

 

[The Blind Watchmaker]

Something like that, but take care... Sure, the inner shell's charge is matched and effectively "cancelled" by the charge on the inner surface of the outer shell. So what remains on the outer surface must be what Gauss's Law "sees" as the contained charge...

So if the question is asking for the "charge of the larger shell" it would be 91.6 + 3.3 μC, but since this question is only asking for the outer surface, the charge is taken to be 91.6 μC?

 

[gneill]

So if the question is asking for the "charge of the larger shell" it would be 91.6 + 3.3 μC, but since this question is only asking for the outer surface, the charge is taken to be 91.6 μC?

Right!

 

[The Blind Watchmaker]

Right!

It is 3 AM here but thanks!