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Spherical shells

  • #1
138
0

Homework Statement



A steel sphere of radius 2.5 cm has a positive charge of 400 nC and it is surrounded by a plastic sphere of radius 10 cm that has the same properties as free space. This plastic sphere is coated with a thin copper metal and has a charge of -400 nC.

What is the value for the voltage at r=0 and the potential difference between the steel sphere and the copper shell?

Homework Equations





The Attempt at a Solution



I am getting like 2 answers. I got one of them from my professor, I got the other one using a method which was described on my ebook's site.



This is the way that I did:


-[tex]\int[/tex] [tex]\frac{ K Qinside}{r^2}[/tex] dr

where the upper limit of the integral is a and lower limit is b. And in this case a=2.5cm and b = 10 cm.

I arrive at that expression for V at r=0 with this

V = -[tex]\int[/tex] [tex]\frac{ K Qinside}{r^2}[/tex] dr -[tex]\int[/tex] [tex]\frac{ K Qinside}{r^2}[/tex] dr -[tex]\int[/tex] [tex]\frac{ K Qinside}{r^2}[/tex] dr

And here the limit of the 1st integral is upper limit is b and lower is [tex]\infty[/tex], 2nd integral: upper is a and lower is b, 3rd: upper is 0 and lower is a

Now the integral from [tex]\infty[/tex] to b becomes 0 because the charge (400-400) will add up to 0.

Integal from a to 0 is 0 because voltage inside a conductor is always 0.

The remaining stuff is from a to b and this is how I arrived.
The answer for that is 107880


If you want to understand it further clearly
download this file: http://www.sendspace.com/file/3cx5bj


What my professor did is:

-[tex]\int[/tex] [tex]\frac{ K Qinside}{r^2}[/tex] dr

Without any limits and solved it with r = 2.5 cm which gives answer of 144000V


All this is for r=0 the first part of the question.

And I don't know what will the 2nd part be.
 
Last edited:

Answers and Replies

  • #2
1,860
0
You're doing
[tex]V(r)=\int E \cdot dl[/tex]
right?

You have to have limits, I don't know how your professor would do it any other way. You of course want to look at the region between a and b, as you said.

Did you get this?
[tex]V=\frac{1}{4\pi \epsilon_0}(\frac{q}{b}-\frac{q}{a})[/tex]
 
  • #3
138
0
Ya right? This is exactly what I said.

And this will be the answer for r=0 and even the potential difference right?
 
  • #4
1,860
0
So let's see here:

V=9*10^9*(400*10^-9/(10*10^-2)-400*10^-9/(2.5*10^-2))=-108000V

I think you're right. I'm really curious to see if we missed something as the problem is very straightforward.
 

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