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## 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.

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