Calculating Charge at the Center of a Spherical Shell has me stumped

[Physicsman69]

Edit: Forgot to type "stumped" at the end of the title

1. Homework Statement
Instead of typing it out, a link to a scanned document of the problem is here: http://imgur.com/Be3jSLp.

Homework Equations

The equations to use are stated in the problem here: http://imgur.com/Be3jSLp

The Attempt at a Solution

For 3.a: Wouldn't you have to draw a gaussian sphere around the outer edge of the sphere, which would yield an enclosed charge of Q_o? Then after that, wouldn't you draw another one inside the conducting shell, resulting in an enclosed charge of zero? Therefore, wouldn't the charge at the center be -Q_o?

For 3.b: Using the equation given, I don't know exactly how to integrate the equation after you substitute in the E value given in the beginning of the problem.

Any advice?

 

[SammyS]

Edit: Forgot to type "stumped" at the end of the title

1. Homework Statement
Instead of typing it out, a link to a scanned document of the problem is here: http://imgur.com/Be3jSLp.

Homework Equations

The equations to use are stated in the problem here: http://imgur.com/Be3jSLp

The Attempt at a Solution

For 3.a: Wouldn't you have to draw a gaussian sphere around the outer edge of the sphere, which would yield an enclosed charge of Q_o? Then after that, wouldn't you draw another one inside the conducting shell, resulting in an enclosed charge of zero? Therefore, wouldn't the charge at the center be -Q_o?

For 3.b: Using the equation given, I don't know exactly how to integrate the equation after you substitute in the E value given in the beginning of the problem.

Any advice?

Here's the image:

Have you tried applying Gauss's Law ?

 

[Buzz Bloom]

Hi Physicsman69:

I suggest you keep the following hints in mind:
1. The total field is radially symmetric.
2. The field out side of the sphere at radius r is exactly the same as if the total charge inside

Q₀ + q₀​

was at the center.

Hope this helps.

Regards,
Buzz

 

[Physicsman69]

Have you tried applying Gauss's Law ?

Yes, you can draw a gaussian surface outside of the sphere which would yield $$E4\pi r^2 = Q_0/\epsilon$$

And, then a gaussian surface drawn inside the conducting shell would yield $$E4\pi r^2 = 0/\epsilon$$

Wouldn't this mean the charge in the center equals $$-Q_0$$ To offset the positive charge on the surface of the shell?

I suggest you keep the following hints in mind:
1. The total field is radially symmetric.
2. The field out side of the sphere at radius r is exactly the same as if the total charge inside
Q0 + q0
was at the center.

Being that it is radially symmetric, the integral for Gauss' law would yield: $$E4\pi r^2$$ That is what you are implying right?

And for #2 are you saying that the unknown charge at the center, q0, would have to be opposite and equal of Q0?

 

[Physicsman69]

Wait, for:

Hi Physicsman69:

I suggest you keep the following hints in mind:
1. The total field is radially symmetric.
2. The field out side of the sphere at radius r is exactly the same as if the total charge inside

Q₀ + q₀​

was at the center.

Hope this helps.

Regards,
Buzz

For #2, this would mean that q0 would be actually be zero, correct? A charge uniformly distributed over the surface, can act like a charge concentrated at the center, therefore Q0 + q0 = Q0, so q0 = 0?

 

[azizlwl]

Just make a spherical net around that spherical shell and shrink it to just a point at the centre carrying unaltered total of charge.

Edit
Maybe it's right to call it Gauss's net?

 

[Physicsman69]

Just make a spherical net around that spherical shell and shrink it to just a point at the centre carrying unaltered total of charge.

A gaussian surface just around the point at the center wouldn't yield anything. The charge at the center is unknown.

 

[azizlwl]

Gauss's Law
dQ=D.dA
For spherical surface,
Q=DA, where D=εE.
Where Q is net charge at the center.

 

[SammyS]

Yes, you can draw a gaussian surface outside of the sphere which would yield $$E4\pi r^2 = Q_0/\epsilon$$

The total charge enclosed is Q₀ + q₀, is it not?

And, then a gaussian surface drawn inside the conducting shell would yield $$E4\pi r^2 = 0/\epsilon$$Wouldn't this mean the charge in the center equals $$-Q_0$$ To offset the positive charge on the surface of the shell?

Being that it is radially symmetric, the integral for Gauss' law would yield: $$E4\pi r^2$$ That is what you are implying right?

And for #2 are you saying that the unknown charge at the center, q0, would have to be opposite and equal of Q0?