Finding V in a Simple Differential Equation for Concentric Spherical Conductors

[John O' Meara]

Two concentric spherical conductors radii a and b are at potentials V1 and 0 respectively. The potential V at a distance x from their common centre is given by \frac{d[x^2\frac{dV}{dx}]}{dx}=0 \\. Find V in terms of x, a, b and V1 ( Note V=V1 when x=a and V=0 when x=b).
I just find it difficult to get started on this, I seem not able to separate out the variables dV and x. Thanks for the help.

 

[daveb]

Well, if (as you've written it) the derivative is 0, then x^2V' would be constant. That is separable.

 

[John O' Meara]

Do I write down x^2V' =0, then x^2 \int dV =0

 

[arunbg]

No, you write
x^2\frac{dV}{dx} = k(constant)

 

[Office_Shredder]

EDIT: Bah, I screwed up somewhere in the latex. arunbg wrote the essential part anyway

 

[John O' Meara]

Is that constant k not =0? I mean the d[x^2\frac{dV}{dx}}] = dx \times 0 = 0 \\ In other words where did you get the k. Anyway I have the following V=k\int \frac{1}{x^2}dx \\.

 

[cristo]

You can't separate the original equation like that. You have \frac{d}{dx}\left(x^2V'\right)=0. Integrating both sides wrt x gives \int \frac{d}{dx}\left(x^2V'\right)dx=k \Rightarrow x^2V'=k

Your next part is correct. Now integrate that.

 

[John O' Meara]

Thanks cristo for your help.