Calculate potential form poisson equation

[Kosta1234]

Homework Statement

Calculate potential form poisson equation

Relevant Equations

$\Delta \psi = -\frac { \rho _ 0}{\varepsilon _0 }$

Hi.

I've the following charge density: $\rho = \rho_0 \frac {r}{R}$
I'm getting a trouble to calculate the potential inside a sphere of radius R located in the center of axis with the given charge density (using poisson equation):

the Laplacian in spherical coordinates is: $\frac {1}{r^2} \frac {\partial}{\partial r} (r^2 \cdot \partial \frac {\psi}{\partial r}) = - \frac {\rho _ 0 r}{R \varepsilon _0 }$
(the components theta and phi are canceled because of the symmetry of the problem.

so outside of the sphere it's pretty easy:
$$ \frac {1}{r^2} \frac {\partial}{\partial r} (r^2 \cdot \partial \frac {\psi}{\partial r}) = $$
is a solution of form $$ \psi = \frac {a}{r} + b $$
and $b = 0$ because the potential is infinity is zero. I can find out the $a$ using $a = kq = k \int dq dV$
so that $\psi = \frac {\pi k \rho _ R^3}{r}$ where $r>R$ .

inside
So I'm getting little bit confused to know the solution form of the equation:
$$ \frac {1}{r^2} \frac {\partial}{\partial r} (r^2 \cdot \partial \frac {\psi}{\partial r}) = - \frac {\rho _ 0 r}{R \varepsilon _0 } $$

I tried to figure out that $\psi$ must be proportional to $r^3$ with this way, and tried to use different Laplacian in some website that I found:
$$\frac {\partial ^ 2 \psi}{\partial r^2} + \frac {2}{r} \cdot \frac {\partial \psi}{\partial r} = - \rho _ 0 \frac {r}{R \varepsilon _ 0 } $$
using those I could substitute $\psi '' = u'$ and $\psi ' = u$ and to solve the equation using second order order differential equation using the example of this website at page 5 ( http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt1.pdf)

my equation form is: $u' +\frac {2}{r} u =- \rho _ 0 \frac {r}{R \varepsilon _ 0 }$
$u' +p(r) u = g(r)$

so I've reached that the integrating factor is $\mu (r) = e^{\int p(r)dt} = r^2$
and the answer for $u$ contains an integral:

$$u(r)= \frac {\int \mu(r) \cdot g(r) dr}{\mu(r)}$$
and another integral to calculate the potential:
$$ \psi (r) = \int u(r) dr $$
what are my integration bounderies? are they in both of the cases 0 to R?
or just the second one? and why?

thank you.

 

[Kosta1234]

.

 

[PeroK]

Hi.

I've the following charge density: $\rho = \rho_0 \frac {r}{R}$
I'm getting a trouble to calculate the potential inside a sphere of radius R located in the center of axis with the given charge density (using poisson equation):

the Laplacian in spherical coordinates is: $\frac {1}{r^2} \frac {\partial}{\partial r} (r^2 \cdot \partial \frac {\psi}{\partial r}) = - \frac {\rho _ 0 r}{R \varepsilon _0 }$
(the components theta and phi are canceled because of the symmetry of the problem.

so outside of the sphere it's pretty easy:
$$ \frac {1}{r^2} \frac {\partial}{\partial r} (r^2 \cdot \partial \frac {\psi}{\partial r}) = $$
is a solution of form $$ \psi = \frac {a}{r} + b $$
and $b = 0$ because the potential is infinity is zero. I can find out the $a$ using $a = kq = k \int dq dV$
so that $\psi = \frac {\pi k \rho _ R^3}{r}$ where $r>R$ .

inside
So I'm getting little bit confused to know the solution form of the equation:
$$ \frac {1}{r^2} \frac {\partial}{\partial r} (r^2 \cdot \partial \frac {\psi}{\partial r}) = - \frac {\rho _ 0 r}{R \varepsilon _0 } $$

Why not integrate that equation, rather than differentiating the LHS?

 

[Kosta1234]

Actually I tried that, but I didn't knew what I'm doing:

$$ \frac {\partial}{\partial r } (r^2 \cdot \frac {\partial \psi}{\partial r})=- \frac {\rho_0 r^3}{R \varepsilon _0} $$
$$ \int \partial \psi = \int_0^R ( \frac {1}{r^2} \int_0^R - \frac {\rho_0 r^3}{R \varepsilon _0} dr) dr $$
I'm getting confused because of the two integrals and their bounderies.

 

[PeroK]

Actually I tried that, but I didn't knew what I'm doing:

$$ \frac {\partial}{\partial r } (r^2 \cdot \frac {\partial \psi}{\partial r})=- \frac {\rho_0 r^3}{R \varepsilon _0} $$
$$ \int \partial \psi = \int_0^R ( \frac {1}{r^2} \int_0^R - \frac {\rho_0 r^3}{R \varepsilon _0} dr) dr $$
I'm getting confused because of the two integrals and their bounderies.

You're overthinking this. Note that since you have already by symmetry shown that the potential is a function of $r$ only, you could replace the partial derivatives with ordinary derivatives.

In any case, the integral of $r^3$ is $\frac{r^4}{4} + C$ the last time I tried it!

And the integral of $\frac{d}{dr}f(r)$ is $f(r) + C$.

 

[Kosta1234]

Well I ment to write:
$$ \int d \psi = \int_0^R ( \frac {1}{r^2} \int_0^R - \frac {\rho_0 r^3}{R \varepsilon _0} dr) dr $$ If that what you ment.


also, here is what I'm getting:

$$ \psi = - \int \frac {1}{r^2} ( \cdot \frac {\rho_0 r^4}{4R \varepsilon _0} + C ) dr $$

** Do I've to work with bounderies to r ($r \in [0.R]$ ) or to add constant. in the next integral I will get a second constant

$$ \psi = - \int \frac {1}{r^2} ( \cdot \frac {\rho_0 r^4}{4R \varepsilon _0} + C ) dr $$
$$ \psi = - \int \frac {1}{r^2} (\frac {\rho_0 r^4}{4R \varepsilon _0} + \frac {C}{r^2} dr )$$
$$ \psi = - \int ( \frac {\rho_0 r^2}{4R \varepsilon _0} + \frac {C}{r^2} dr )$$
$$ \psi = - ( \frac {\rho_0 r^3}{12R \varepsilon _0} - \frac {C}{r} ) + D$$
$$ \psi = \frac {C}{r} - \frac {\rho_0 r^3}{12R \varepsilon _0} +D $$

so I now have two constants..

checking what happening in $r = R$ will not solve this to me.. so where I'm wrong?

 

[PeroK]

Well I ment to write:
$$ \int d \psi = \int_0^R ( \frac {1}{r^2} \int_0^R - \frac {\rho_0 r^3}{R \varepsilon _0} dr) dr $$ If that what you ment.


also, here is what I'm getting ...

You are really making this difficult. For example:

Actually I tried that, but I didn't knew what I'm doing:

$$ \frac {\partial}{\partial r } (r^2 \cdot \frac {\partial \psi}{\partial r})=- \frac {\rho_0 r^3}{R \varepsilon _0} $$

From this you should be able simply to write down that:

$$r^2 \cdot \frac {\partial \psi}{\partial r} = - \frac {\rho_0 r^4}{4R \varepsilon _0} + C$$

And, then integrate again. You can use physical constraints and boundary conditions to determine constants like $C$.

With integrals of the form $r^n$, you can just write down the answers.

Note also that inside the sphere, the solution must be valid for $r \rightarrow 0$, hence in this case you must have $C = 0$.

 

[Kosta1234]

thank you.