Spherical shells

1. Mar 4, 2010

sonutulsiani

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:

-$$\int$$ $$\frac{ K Qinside}{r^2}$$ 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 = -$$\int$$ $$\frac{ K Qinside}{r^2}$$ dr -$$\int$$ $$\frac{ K Qinside}{r^2}$$ dr -$$\int$$ $$\frac{ K Qinside}{r^2}$$ dr

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

Now the integral from $$\infty$$ 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

What my professor did is:

-$$\int$$ $$\frac{ K Qinside}{r^2}$$ 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. Mar 4, 2010

Mindscrape

You're doing
$$V(r)=\int E \cdot dl$$
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?
$$V=\frac{1}{4\pi \epsilon_0}(\frac{q}{b}-\frac{q}{a})$$

3. Mar 4, 2010

sonutulsiani

Ya right? This is exactly what I said.

And this will be the answer for r=0 and even the potential difference right?

4. Mar 8, 2010

Mindscrape

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.