Why my school method doesnt work

[transgalactic]

the question is :
http://i39.tinypic.com/ll89s.gif

the solution is here:
http://i39.tinypic.com/bej341.gif

my question is about part b when calculating the potential between $$r_1$$ and $$r_2$$:
r1<r<r2
we take the potential from the outer sphere (-150)
which is constant inside of it,and we sum with the potential from the inner sphere
which changes by r.

so its -150+( 9*10^9 *10* 10^-9)/r

they get a different expression in another way

my question is,where is my mistake in my way of solving?

 

[HallsofIvy]

The mistake is that you haven't used "with V= 0 at r= $\infty$".

Any potential is "relative" to some base value. If you take V= 0 at $\infty$
you will should find that the constant potential inside $r_2$ is 0.

 

[transgalactic]

i did take it that way.
and the potential in r1<r<r2
is not zero
the answer in the book doesn't say that too??

 

[transgalactic]

i can't understand how i didnt use that the potential in infinity is 0
what part of my solution says other wise?

 

[transgalactic]

and its not legat lo put the integration variable in the interval

??

 

[transgalactic]

http://i44.tinypic.com/20qz3ev.gif

that exactly like the book says.
you didnt say in what place i have an error

 

[Mark44]

and its not legat lo put the integration variable in the interval

??

Sure it's legal. It just isn't done very much. When you say that something isn't legal, you'll be taken more seriously if you cite a reason for your assertion.

I think what you're having a problem with is this integral:
$$\int_{r_1}^r \frac{89.9}{r^2}dr$$
To evaluate this definite integral, find the antiderivative of the integrand (which will be a function of r), and then evaluate this antiderivative at r (which involves exactly zero work), and then subtract its value at r₁, which is 30 cm. What they're doing is finding the potential at a distance r, where r₁ < r < r₂.

 

[Mark44]

so its -150+( 9*10^9 *10* 10^-9)/r

simplify!

 

[transgalactic]

-150+ 90/r

and they have

-450+ 90/r

??

 

[Mark44]

When you integrated, did you put in r₁ (which is 30 cm)? I can see from the answer you posted how they got their answer. You didn't show how you got your result, so I have to guess at what you might have done wrong.

 

[transgalactic]

but i am using a totaly different method

 

[HallsofIvy]

And we don't know what method you are using because you have refused to show what you have done! How can we possibly say what, if anything, you are doing wrong when you don't show what you have done?