Calculating Electric Potential in a Uniformly Charged Spherical Shell

[sm09]

A non conducting spherical shell is uniformly charged. The electrostatic potential at the centre is 200v and the electrostatic potential at the distance r=50cm from the centre is 40v. Find the radius of sphere a.

I am not sure where go with this.

I know that the potential difference is equal to the work done in moving from one point to another.

Any help or direction would be appreciated,

Thanks

 

[Infinitum]

Do you know the shell theorem?

If so, what does it tell you about the potential inside a uniformly charged shell? And what is the potential outside the shell?

Frame equations based on answers to the above questions, and you should be able to solve for radius.

 

[sm09]

But the question states that the potential inside the shell is 200v

 

[Infinitum]

But the question states that the potential inside the shell is 200v

Yes it does. How can you express the potential inside the shell mathematically? Say the shell has a charge Q, radius R.

 

[sm09]

Q/r^2

 

[Infinitum]

Q/r^2

Noo!

You need to add in the constant of proportionality k=1/4\pi \epsilon before you can use that. And even then, kQ/r² is the electric field outside the shell. You need the potential inside the shell...

 

[sm09]

Is the potential energy constant through the shell?

 

[Infinitum]

Is the potential energy constant through the shell?

Yes it is. Or equivalently, the electric potential is constant throughout the shell. That's indirectly what the shell theorem states...

 

[sm09]

So obviously r has to be less that 50cm.

Can we say Q/r^2= 200v and Q/(r^2 - 50^2) = 40v?

Then we can find Q and substitute it into one of the equations so find r?

 

[Infinitum]

So obviously r has to be less that 50cm.

Yes!

Can we say Q/r^2= 200v and Q/(r^2 - 50^2) = 40v?

You didn't pay attention to this post..

You need to add in the constant of proportionality k=1/4πϵ before you can use that. And even then, kQ/r2 is the electric field outside the shell. You need the potential inside the shell...

 

[sm09]

Yes my mistake. Thanks for your help. Was slightly confused at first but now makes sense.

Very much appreciated

 

[Infinitum]

Yes my mistake. Thanks for your help. Was slightly confused at first but now makes sense.

Very much appreciated

Glad you figured it out.