Current induced in a charged hollow sphere

[Final_HB]

Homework Statement

A hollow spherical conducting shell is suspended in air by an insulating string, so that the sphere
is electrically isolated.
The total charge on the conductor is -6 μC. If an additional point charge of +2μC is placed in the hollow region inside the shell, what is the total charge induced on the inside surface of the shell?
If the outside surface of the shell has radius 8 cm, and the electric eld immediately outside the shell?

Homework Equations

The only one I can think that could be half way useful would be:
The Gauss equation ($\oint$ D.dA= Ʃq$_{i}$)
or the charge at the shell version: $\frac{Q}{4\pi ε r^2}$

The Attempt at a Solution

I don't actually know how much the inital charge actually changes the problem, but it obiously does... otherwise they wouldn't give it Without the 6 micro coulombs, you can just sub into the second formula above for both I am pretty sure, but apart from that, nothing really.

I don't expect a full answer, some help on where to go next would be much appreciated :)

 

[Redbelly98]

Homework Statement

A hollow spherical conducting shell is suspended in air by an insulating string, so that the sphere
is electrically isolated.
The total charge on the conductor is -6 μC. If an additional point charge of +2μC is placed in the hollow region inside the shell, what is the total charge induced on the inside surface of the shell?
If the outside surface of the shell has radius 8 cm, and the electric eld immediately outside the shell?

Homework Equations

The only one I can think that could be half way useful would be:
The Gauss equation ($\oint$ D.dA= Ʃq$_{i}$)
or the charge at the shell version: $\frac{Q}{4\pi ε r^2}$

The Attempt at a Solution

I don't actually know how much the inital charge actually changes the problem, but it obiously does... otherwise they wouldn't give it Without the 6 micro coulombs, you can just sub into the second formula above for both I am pretty sure, but apart from that, nothing really.

I don't expect a full answer, some help on where to go next would be much appreciated :)

Welcome to Physics Forums!

The key to these Gauss's law problems is to choose an appropriate Gaussian surface.

Since they are asking about the charge on the conductor's inner surface, the Gaussian surface should probably enclose that inner surface.

EDIT ADDED:
p.s., your thread title talks about induced current, but the problem is really asking about induced charge -- not the same thing.

 

[Final_HB]

Thanks. Sorry. you slip up on my part

How does the starting charge affect the question though? Do i just use the sum of -6 and +2 (-4)?

 

[Redbelly98]

It depends. You use whatever charge is enclosed by the Gaussian surface. We can't say whether you should include the -6 and +2 until you describe a surface to use.

 

[Final_HB]

Oh okay sorry

am... well its a sphere , so its a spherical Gaussian surface.

so it has an electric flux of:

$\phi_E$ = E.A=$\frac{-Q}{\epsilon_0}$ where A is surface area of the sphere. (4$\pi r^2$ )

Tiny bit of re arranging give you:

E= $\frac{Q}{\epsilon_0 4\pi r^2}$

 

[Redbelly98]

Yes, that's the idea.

Now you have to make a suitable choice of r. Usually, you choose r such that you have either:

(1) A known value of E, so that you can determine Q
or
(2) A known value of Q, so that you can determine E.

Since the question is about finding a charge, it looks like you should proceed along the lines of item (1).

 

[Final_HB]

Am....
Its inside the sphere, so E=0... I think There is no electric field inside the sphere, but this is touching the outside surface. So would there still be electric field at the inside surface of the sphere?

I'll continue as if that's true, just correct me if I am wrong

ASSUMING I am right:

E=0
$\rightarrow$ 0= $\frac{Q}{\epsilon_0 4\pi r^2}$
So Q is equivalent to :
$\epsilon 4\pi r^2$

Which are all known constants, except for the radius. So, as you said, we need a value for r.
The only mention of a radius in the entire question is 8cm form centre to outer surface, but that won't do. Can i just call it a, where a=8-r or something.
That okay so far? or am i horribly wrong about the electric field value

 

[Redbelly98]

Am....
Its inside the sphere, so E=0... I think

Yes.

There is no electric field inside the sphere, but this is touching the outside surface. So would there still be electric field at the inside surface of the sphere?

We don't need to worry about at the surface here, so let's just use a Guassian surface located within the conductor, between the inner and outer surfaces. That's inside the conductor, and E is definitely zero there.

I'll continue as if that's true, just correct me if I am wrong

ASSUMING I am right:

E=0
$\rightarrow$ 0= $\frac{Q}{\epsilon_0 4\pi r^2}$

Yes, good.

So Q is equivalent to :
$\epsilon 4\pi r^2$

You'll want to rethink this part.

zero × $\epsilon 4\pi r^2$ = ___?

Which are all known constants, except for the radius. So, as you said, we need a value for r.
The only mention of a radius in the entire question is 8cm form centre to outer surface, but that won't do. Can i just call it a, where a=8-r or something.
That okay so far? or am i horribly wrong about the electric field value

You're good up to the arithmetic error I mention above, and getting pretty close to solving it.

If I remember, I'll try to check back here during my lunch break (time zone is Eastern USA)

 

[Final_HB]

Oh I don't believe I just did that!

On a forum for the whole world to see as well

So, fixing my little mistake:
Multiplying the massive blue zero by anything gives me zero.

so:
0x ε₀4$\pi$r² = 0

so... what Q=0 ?

Take your time, no rush I am running on irish time, and this is for an exam in about a month.

Im just happy for the help really :shy:

 

[Redbelly98]

Oh I don't believe I just did that!

On a forum for the whole world to see as well

LOL

so:
0x ε₀4$\pi$r² = 0

so... what Q=0 ?

Yes. And that is key in figuring this out. Any discussion of Gauss's law (in your textbook? class notes?) is very specific about what charge the Q refers to. So you'll want to review that Gauss's law discussion, and think about how to apply that to the Q we are working with here.

 

[Final_HB]

It says (very specifically ) that Q is the sum of all charges enclosed by the surface.
I think this is what you meant, or are you talking about charge distribution?

 

[Redbelly98]

Yes, that is exactly what I meant.

So, from Gauss's law, we now know that the sum of all charges enclosed by the surface is zero.

 

[Final_HB]

So is that the answer for part one?
Induced charge is = 0 ?

 

[Redbelly98]

No.

They are asking for the charge on the inner surface of the conductor.

That's different than the total charge enclosed by the Gaussian sphere.

What are the charges that are enclosed by the Gaussian sphere? (Reread the problem statement, if it will help.)

 

[Final_HB]

Oh okay
The sphere has -6μC (on the sphere) and +2μC (as a point charge in the centre of the sphere)
So it only encloses the +2 charge, since the -6 is already on the sphere, not in it... ya?

 

[Redbelly98]

Definitely the +2 μC, since that is at the center of the Gaussian sphere (and conducting shell too). Then there must be something else, to make it a net total of zero charge within the Gaussian sphere.

...the -6 is already on the sphere, not in it...

If the -6 is on the surface of the Gaussian sphere, that would put it within the conducting shell where the Gaussian surface is. But one of the "rules" about conductors is that it can't have any excess charge located within it -- that charge has to be somewhere on the surface of the conducting shell.

By the way, if you haven't already, you should have a picture of the three things we are talking about (the point charge, the charged conducting shell, and the Gaussian sphere). You might be able to visualize this picture in your head, or it might work better for you to draw an actual sketch of where those three things are in relation to each other. Then maybe it's easier to see how the Gaussian surface fits in between the inside and outside surfaces of the conducting spherical shell.

 

[Redbelly98]

Oh okay
The sphere has -6μC (on the sphere) . . .

Which sphere did you mean? I was talking about the Gaussian sphere, but the -6μC in on the conducting shell (which has two spherical surfaces, the inner and outer surfaces).

 

[Final_HB]

Im sorry... you lost me I am not sure where this is going anymore...

Late, I know... Internet went down and we need a replacement.

 

[Redbelly98]

Okay, we'll have to establish just where you are with this.

Are you following the discussion up to your message #15? If not, just where in this discussion do you first get lost?

Have you drawn for yourself a figure showing:

• The hollow conducting shell, including both inside and outside surfaces?
• The point charge inside the hollow region?
• The spherical Gaussian surface that we came up with in post #s 7 & 8, between the inner and outer surfaces of the conducting shell?

 

[Final_HB]

yes, just at #15. After that... nothing

ya, its not pretty... but its not art class either.

If there is a +2 charge enlcosed... and we know that all charge enlcosed is 0... there has to be a charge cnacelling out the +2, yes?

 

[Redbelly98]

yes, just at #15. After that... nothing

ya, its not pretty... but its not art class either.

If there is a +2 charge enlcosed... and we know that all charge enlcosed is 0... there has to be a charge cnacelling out the +2, yes?

Yes, that's right. Have you drawn the figure I mentioned?

 

[Final_HB]

yes I have

So there is a -2 charge induced, which then adds to the -6 that we have already...

So total charge induced is -8 ??

 

[Redbelly98]

yes I have

So there is a -2 charge induced, which then adds to the -6 that we have already...

Yes, no.
Yes there is a -2 μC induced on the inner surface of the conductor.

But, that -2 μC is part of the total -6 μC on the conductor. The conductor is not physically in contact with anything, so there is no path for any charge to leave or enter the conductor. The total charge on the conductor remains the same, -6 μC. The -2 μC comes from that -6 μC total charge.

(You may be thinking of situations where a conductor is connected to "ground" by a wire, in which case extra charge can get induced on the conductor. The wire to ground provides a path for charge to enter or leave the conductor in that situation.)

 

[Final_HB]

So its just -2 to cancel out the +2.
And since there is no extra charge going in... the charge outside the shell is -4.

So to find the electric field just outside the surface. we take that charge of -4 and sub it into the formula for an electric field. (ie. kQ/r²)

EDIT: just noticed the question got messed up when I copied and pasted.

It simply just asks for the electric field outside the shell if the radius is 8cm

 

[Redbelly98]

You got it

 

[Final_HB]

Brilliant Thank you so much seriously.