Find the potentail inside the sphere

[leonne]

Homework Statement

the potential vo(0) is specified on the surface of a hollow sphere find the pot inside .

Homework Equations

The Attempt at a Solution

This is a example problem in the book but don't get this one part
The general solution is V=$$\sum$$Alr^lpl(cos@)
The only boundary we are told that at the surface its vo(0)
so
V=$$\sum$$Alr^lpl(cos@)=Vo(0)
now this the part i am lost. To find the constant I know from the other section to find this, I do ex the gen sol is say sin(k) than to find constant it would be sin(k)sin(k^') than i would integrate it than i would plug in some val for k and k^' and see what i get

but here they go to $$\int$$-1 to 1 Pl(x)Pl^'(x)dx= $$\int$$ 0 to pie pl(cos@)p^'(cos@)sin@d@ where did they get this from?
thxs

 

[mark726]

for any help
Thank you for your question. To find the potential inside the hollow sphere, we need to use the general solution V=\sumAlrlpl(cos@), as you have correctly stated. However, to determine the constant, we need to use the boundary condition given in the problem, which is V=Vo(0) at the surface of the sphere. This means that at the surface, the potential is equal to the specified value Vo(0).

To find the constant, we can use the method of separation of variables. This involves assuming that the potential can be written as a product of two functions, one depending only on the radial coordinate (r) and the other depending only on the angular coordinate (@). So, we can write V(r,@)=R(r)P(@). Substituting this into the general solution, we get V(r,@)=\sumAlrlpl(cos@)=Vo(0).

Now, we can use the boundary condition at the surface to determine the value of the constant A. Since the potential is constant at the surface, we can write V(r,@)=A. This means that at the surface, the potential is equal to A. So, we can equate this to Vo(0) and solve for A. This will give us the value of the constant A.

To find the constant A, we can use the method you have mentioned, which is to integrate the general solution over the range of the variables. In this case, we need to integrate over the range of the angular coordinate (@). This is where the integral \int-1 to 1 Pl(x)Pl'(x)dx= \int 0 to pie pl(cos@)p'(cos@)sin@d@ comes in. This integral is a standard result in mathematical tables and can be easily evaluated. Once we have the value of A, we can substitute it back into the general solution to get the potential inside the hollow sphere.

I hope this helps to clarify the solution. Let me know if you have any further questions. Keep up the good work in your studies!