# Calculate potential form poisson equation

• Kosta1234

#### 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.

Last edited:

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?

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.

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##.

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?

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##.

thank you.