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Spherical shells

  1. Mar 4, 2010 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations



    3. 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: Mar 4, 2010
  2. jcsd
  3. Mar 4, 2010 #2
    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]
     
  4. Mar 4, 2010 #3
    Ya right? This is exactly what I said.

    And this will be the answer for r=0 and even the potential difference right?
     
  5. Mar 8, 2010 #4
    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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