Calculate potential form poisson equation

In summary,**The potential inside a sphere of charge density ##\rho## is given by the Laplacian in spherical coordinates:$$ \frac {1}{r^2} \frac {\partial}{\partial r} (r^2 \cdot \partial \frac {\psi}{\partial r}) = - \frac {\rho _ 0 r}{R \varepsilon _0 }.Inside the sphere, the equation becomes trivial:$$ u' +\frac {2}{r} u =- \rho _ 0 \frac {r}{R \varepsilon _0}.
  • #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 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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  • #2
.
 
  • #3
Kosta1234 said:
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?
 
  • #4
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.
 
  • #5
Kosta1234 said:
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##.
 
  • #6
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?
 
  • #7
Kosta1234 said:
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:

Kosta1234 said:
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##.
 
  • #8
thank you.
 

Related to Calculate potential form poisson equation

1. What is the Poisson equation?

The Poisson equation is a mathematical equation used to describe the behavior of electric fields in a given region. It relates the electric potential to the charge distribution in that region.

2. How is the potential calculated from the Poisson equation?

The potential can be calculated by solving the Poisson equation using mathematical methods such as the finite difference method or the finite element method. These methods involve discretizing the region into smaller elements and solving the equation for each element.

3. What are the units of the potential in the Poisson equation?

The potential in the Poisson equation has units of volts (V) or joules per coulomb (J/C). This represents the amount of energy required to move a unit of charge from one point to another in the electric field.

4. Can the Poisson equation be used for any charge distribution?

Yes, the Poisson equation can be used for any charge distribution, as long as the charge distribution is known. It is a fundamental equation in electrostatics and is used to model a variety of physical systems, such as capacitors, conductors, and semiconductor devices.

5. What are some real-world applications of the Poisson equation?

The Poisson equation has many practical applications, including in the design of electronic devices, such as transistors and integrated circuits. It is also used in the field of fluid dynamics to model the flow of fluids in a given region. Additionally, it is used in physics research to study the behavior of electric fields in different systems.

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