- #1

Kosta1234

- 46

- 1

- 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

$$ \frac {1}{r^2} \frac {\partial}{\partial r} (r^2 \cdot \partial \frac {\psi}{\partial r}) = $$

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 ## .

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.

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.

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