Gauss law, having problem with spherical load distribution, me

In summary: V\,where 'int' stands for 'integrate'. So in this case, you would use the following equation:Q=\int_{\text{all...}}\rho\,dV\,where 'int' stands for 'integrate'. So in this case, you would use the following equation:Q=\int_{\text{all...}}\rho\,dV=3*10^-17\,C\,The answer to your question is that a fraction of the total load is contained within the region r<R/2.
  • #1
kliker
104
0
A region in space containing cargo which is distributed spherically so that the stocking load ρ is given by the equotations:

ρ = a for r<=R/2
ρ = 2a(1-r/R) for R/2<=r<=R
ρ=0 for r>=R

The total charge Q is 3*10^-17 C, the radius R of the overall distribution is 2*10^-14 m and a is a constant which has dimensions of C / m^3. Determine the constant a as
function of Q and R and the numerical value. b) Using the
Gauss's law find an expression for the measure of the electric field as a function of distance r
from the center of distribution. Do this separately for each of the three areas.
Make sure that your results agree with the borders of the areas. What fraction of
total load is contained within the region r<R/2?i spent like 3 hours to solve this, but no success
 
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  • #2
kliker said:
A region in space containing cargo which is distributed spherically so that the stocking load ρ is given by the equotations:

ρ = a for r<=R/2
ρ = 2a(1-r/R) for R/2<=r<=R
ρ=0 for r>=R

The total charge Q is 3*10^-17 C, the radius R of the overall distribution is 2*10^-14 m and a is a constant which has dimensions of C / m^3. Determine the constant a as
function of Q and R and the numerical value. b) Using the
Gauss's law find an expression for the measure of the electric field as a function of distance r
from the center of distribution. Do this separately for each of the three areas.
Make sure that your results agree with the borders of the areas. What fraction of
total load is contained within the region r<R/2?


i spent like 3 hours to solve this, but no success

How is charge density, [tex]\rho[/tex], related to total charge [tex]Q[/tex] for a spherical charge? If you can find that relation, the first part is solved by inserting your values of [tex]\rho[/tex] in that relation.

What is Gauss's Law? More accurately, what is the equation for Gauss's Law? If you know the equation, you should be able to insert the charge density into this and find the answer.
 
  • #3
jdwood983 said:
How is charge density, [tex]\rho[/tex], related to total charge [tex]Q[/tex] for a spherical charge? If you can find that relation, the first part is solved by inserting your values of [tex]\rho[/tex] in that relation.

What is Gauss's Law? More accurately, what is the equation for Gauss's Law? If you know the equation, you should be able to insert the charge density into this and find the answer.

thanks for the answer

i know that ρ = Q/(4/3pir^3) right?

if i take r as R/2 and find ρ, how can i find a? i mean for R/2 we have two different equotations, then

if i take a gausian surface the Qencl would be Qencl = ρ * V where V = 4/3pir^3 then E would be E = Qencl/eo*Qencl

i really think it's something like this but I am not 100% sure

ps: this is from the book "university physics by d young"
 
  • #4
kliker said:
thanks for the answer

i know that ρ = Q/(4/3pir^3) right?

if i take r as R/2 and find ρ, how can i find a? i mean for R/2 we have two different equotations, then

if i take a gausian surface the Qencl would be Qencl = ρ * V where V = 4/3pir^3 then E would be E = Qencl/eo*Qencl

i really think it's something like this but I am not 100% sure

ps: this is from the book "university physics by d young"

You have three areas where the charge density exists:

[tex]
\rho=\left[\begin{array}{ll}a &\,\,\,\,\,r<\frac{R}{2} \\ a\left(1-\frac{r}{R}\right)&\,\,\,\,\,\frac{R}{2}\leq r\leq R \\ 0&\,\,\,\,\,r>R\end{array}\right.
[/tex]

I would solve the middle one for [tex]a[/tex] in terms of [tex]\rho[/tex]:

[tex]
\rho=a\left(1-\frac{r}{R}\right)\rightarrow a=\frac{\rho}{1-\frac{r}{R}}
[/tex]

Then replace [tex]\rho[/tex] with your total charge that you have above. The electric field equation you have is wrong, you should have:

[tex]
E_r=\frac{\rho}{\varepsilon_0}=\frac{Q_{enc}}{V\varepsilon_0}
[/tex]
 
  • #5
thank you again, this is actually what i did before, the answer in the book for a question is a = 8Q/5piR^3

but i keep finding a = 6Q/piR^3
 
Last edited:
  • #6
jdwood983 said:
I would solve the middle one for [tex]a[/tex] in terms of [tex]\rho[/tex]:

[tex]
\rho=a\left(1-\frac{r}{R}\right)\rightarrow a=\frac{\rho}{1-\frac{r}{R}}
[/tex]

Then replace [tex]\rho[/tex] with your total charge that you have above. The electric field equation you have is wrong, you should have:

Why would you do that? :confused:
 
  • #7
gabbagabbahey said:
Why would you do that? :confused:

hello, gabbagabbahey, what would you do in this case? I'm really stuck, this is like the last exercise i need to solve in order to finish my assignment

thanks in advance
 
  • #8
kliker said:
i know that ρ = Q/(4/3pir^3) right?

No, that is only when the charge is uniformly distributed over a sphere of radius 'r'...More generally, the volume charge density is defined according to the equation [itex]dq=\rho dV[/itex], where [itex]dq[/itex] is the amount of charge located in the infinitesimal volume element [itex]dV[/itex]...to get the total charge of any distribution, you need to integrate over all space (or at least a volume that encloses the entire charge distribution):

[tex]Q=\int_{\text{all space}}\rho(\textbf{r})dV[/tex]
 
  • #9
gabbagabbahey said:
No, that is only when the charge is uniformly distributed over a sphere of radius 'r'...More generally, the volume charge density is defined according to the equation [itex]dq=\rho dV[/itex], where [itex]dq[/itex] is the amount of charge located in the infinitesimal volume element [itex]dV[/itex]...to get the total charge of any distribution, you need to integrate over all space (or at least a volume that encloses the entire charge distribution):

[tex]Q=\int_{\text{all space}}\rho(\textbf{r})dV[/tex]

can you explain me what ρ(r) and "all space" are?

from the integral i found that Qw = Q*ln(V) where V = 4/3pir^3
 
  • #10
Haven't you encountered volume integrals before? Have you taken 2nd year calculus?
 
  • #11
gabbagabbahey said:
Haven't you encountered volume integrals before? Have you taken 2nd year calculus?

no i don't think so, I am studying computer science 1 year, sorry for my bad english I'm from russia
 
  • #12
You really need to have taken a multivariable calculus course to understand electrodynamics...what course is this problem for, and what textbook are you studying from? Are you familiar with spherical coordinates?
 
  • #13
gabbagabbahey said:
You really need to have taken a multivariable calculus course to understand electrodynamics...what course is this problem for, and what textbook are you studying from? Are you familiar with spherical coordinates?

it's for physics, this exercise (23.27) is from the book university physics by Hugh D. Young

where can i find these courses? i really need to understand and solve this by tomorrow

Thanks
 
  • #14
kliker said:
it's for physics, this exercise (23.27) is from the book university physics by Hugh D. Young

That text has a few examples of volume integrals in spherical coordinates, I suggest you look at them.
 
  • #15
gabbagabbahey said:
That text has a few examples of volume integrals in spherical coordinates, I suggest you look at them.

thank you for the answers, i ve studied everything in there

after some time i found this AWESOME guy http://www.youtube.com/watch?v=nvq_4OFp_IE&NR=1

explaining different examples, i watched all his videos, and i think i understand what i have to do now

if i find any problems i ll update this thread

thanks everyone
 
  • #16
gabbagabbahey said:
Why would you do that? :confused:

I would've done that because the answer requested the electric field in the three different areas. I was using that as just a suggestion to solve the problem, but maybe I didn't say it was the complete answer but it is for sure a starting point.

Also, Young's book is an algebra based text (at least the copy I have is), so I'm assuming that he doesn't know anything about integrals, hence not suggestion using one.
 
  • #17
after some search

because the distribution is not done uniformly

dq = ρ * dv

Qencl = integral(ρ*dv)

Qencl = integral(ρ*4*pi*r^2)dr

what would ρ be here though? in the above video the man says it's k*r but shouldn't it be Q/(4/3*pi*r^3)?

im stuck again...

life can be sucky sometimes :(
 
  • #18
You are given [itex]\rho[/itex] in your problem statement...
 
  • #19
gabbagabbahey said:
You are given [itex]\rho[/itex] in your problem statement...

yes i used it, but again i can't find the correct answer which is

a = 8Q/(5*pi*R^3)
 
  • #20
How about showing me your steps then...
 
  • #21
gabbagabbahey said:
How about showing me your steps then...

integral(0,R/2)ρ*4πr^2dr where ρ i put α from the first equotation, then it's like α*4π*integral(0,R/2)r^2

so Qenclosed will be equals to α*4π*(R^3/8/2)

Qencl = α*4π*(R^3/4) (1)

then i find how much is Q and compare it with Qencl

for Q i just take the integral from 0 to R

and it is Q = α * 4π*(R^3/3)

so Qencl = 3/4*Q

from (1) we have α = (3/4)*Q/4π*(R^3/4)

α = 3*Q/4*π*R^3

sorry for not using latex, i really can't understand how to use it

Also i really appreciate the help you guys are giving me here, thanks
 
  • #22
Why are you only integrating from zero to R/2?

[tex]Q=\int_0^\infty \rho(r)4\pi r^2dr=\int_0^{R/2} (\alpha)4\pi r^2dr+\int_{R/2}^R \left(2\alpha\left(1-\frac{r}{R}\right)\right)4\pi r^2dr+\int_R^\infty (0)4\pi r^2dr[/tex]
 
  • #23
WOW

THANKS MY FRIEND

WHAT A STUPID MISTAKE I MADE

to calculate Q it took me like 10 minutes, cause it was TOO long, but i FINALLY found the desired result

how can i ++++rep you?

THANKS AGAIN
 
  • #24
You're welcome!:smile:

kliker said:
how can i ++++rep you?

I suppose if you felt so inclined, you could vote for me as homework helper of the year in the annual poll that is posted in 'general discussion', come December:wink: (Shamelessly stumps for votes:biggrin:)
 

1. What is Gauss law?

Gauss law is a fundamental law in electromagnetism that describes the relationship between electric charges and electric fields. It states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space.

2. How does Gauss law apply to spherical load distribution?

Gauss law can be used to calculate the electric field created by a spherical charge distribution. By using a spherical Gaussian surface, the flux through the surface can be related to the enclosed charge and the resulting electric field can be determined.

3. What is the equation for Gauss law?

The mathematical representation of Gauss law is ∮SE⃗ · dA⃗ = Qenc0, where ∮SE⃗ · dA⃗ is the electric flux through a closed surface S, Qenc is the total charge enclosed by the surface, and ε0 is the permittivity of free space.

4. How is Gauss law related to Coulomb's law?

Gauss law is a generalization of Coulomb's law, which only applies to point charges. Gauss law applies to any type of charge distribution, including continuous distributions such as spherical load distributions. Coulomb's law can be derived from Gauss law by considering a point charge as a sphere with a very small radius.

5. What is the significance of Gauss law in electromagnetism?

Gauss law is an important law in electromagnetism as it allows us to relate the electric field to the distribution of electric charges. It also helps us to solve complex problems involving electric fields by simplifying the calculations through the use of Gaussian surfaces. Additionally, Gauss law is a fundamental law that forms the basis for other important laws and principles in electromagnetism.

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