| Topic: |
Science > Physics |
| User: |
"Gaganpreet Singh Arora" |
| Date: |
06 Jun 2005 12:51:20 PM |
| Object: |
Stuck on a problem for two days |
I don't know where I am going wrong in this electrostatics problem so
could you please help me and point out my mistake.
A spherically symmetrical but nonuniform volume distribution of charge
produces an electric field E = K*(r^4), directed radially outward from
the centre of the sphere. r is the radial distance from the centre, K
is a constant. What is the volume charge density of the sphere?
I imagined a spherical (shell) Gaussian surface of radius 'x' and
thickness dx in the shell. Then for that (I hope the assumption is
true) I assumed that the charge is concentrated at the centre (as is
the case in spherical distribution)
dq = (a) 4 (pi) (x^2) dx
a = density , pi = 3.14...
now Electric field at a distance r
dE = dq/[4 (pi) e (r^2)] e = permittivity (epsilon not)
dE = a (x^2) dx/[e (r^2)] (put value of dq)
dE (r^2) e = a (x^2) dx
now i have a and x but i have to integrate with respect to x. so i am
stuck here
Another thing I tried to do was since i know the formula of electric
field for sphere (its for a symmetrical charge distribution, i know it
won't work here.)
formula = ar/3e.
ar/3e = K r^4
a = 3 K e r^4 (which is exactly half of that the answer that
i have to get)
I tried two ways more also but i end up with the same stuff as in
first.
So what do i do now?
.
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Stuck on a problem for two days |
06 Jun 2005 01:22:50 PM |
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Gaganpreet Singh Arora wrote:
I don't know where I am going wrong in this electrostatics problem so
could you please help me and point out my mistake.
A spherically symmetrical but nonuniform volume distribution of charge
produces an electric field E = K*(r^4), directed radially outward from
the centre of the sphere. r is the radial distance from the centre, K
is a constant. What is the volume charge density of the sphere?
The easiest way to solve that is by using div E = 4 pi rho (Gauss' law
in differential form!), where one should express the divergence
operator in spherical coordinates.
If you don't know that, you have to use the integral form.
Essentially, it says for a spherical charge distribution that the
electric field at r is given by:
E(r) = 4 pi/r^2 int_0^r rho(r') r'^2 dr'
You know what E(r) is, so you can calculate backwards (essentially
getting rid of the integral by taking a suitable derivative) in order
to get rho.
I imagined a spherical (shell) Gaussian surface of radius 'x' and
thickness dx in the shell. Then for that (I hope the assumption is
true) I assumed that the charge is concentrated at the centre (as is
the case in spherical distribution)
dq = (a) 4 (pi) (x^2) dx
a = density , pi = 3.14...
What density do you mean here? The density in the shell, i.e. a=a(x)?
Is dq supposed to mean the charge inside the shell?
now Electric field at a distance r
dE = dq/[4 (pi) e (r^2)] e = permittivity (epsilon not)
dE = a (x^2) dx/[e (r^2)] (put value of dq)
dE (r^2) e = a (x^2) dx
now i have a and x but i have to integrate with respect to x. so i am
stuck here
Especially since you don't know a(x), since that's what was asked for...
Another thing I tried to do was since i know the formula of electric
field for sphere (its for a symmetrical charge distribution, i know it
won't work here.)
formula = ar/3e.
ar/3e = K r^4
a = 3 K e r^4 (which is exactly half of that the answer that
i have to get)
No, that doesn't make sense at all.
I tried two ways more also but i end up with the same stuff as in
first.
So what do i do now?
See above.
Bye,
Bjoern
.
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| User: "Edward Green" |
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| Title: Re: Stuck on a problem for two days |
06 Jun 2005 04:40:11 PM |
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Bjoern Feuerbacher wrote:
Gaganpreet Singh Arora wrote:
A spherically symmetrical but nonuniform volume distribution of charge
produces an electric field E = K*(r^4), directed radially outward from
the centre of the sphere. r is the radial distance from the centre, K
is a constant. What is the volume charge density of the sphere?
The easiest way to solve that is by using div E = 4 pi rho (Gauss' law
in differential form!), where one should express the divergence
operator in spherical coordinates.
Surely not.
If you don't know that, you have to use the integral form.
Yes. We know by symmetry that the field is radial, and we know its
value as a function of r. Therefore, taking appropriate units, we know
the amount of charge enclosed within a sphere of radius r, and
differentiating, the amount enclosed inside a shell between r and dr.
Finally we divide by 4pi to get the volume charge density.
Essentially, it says for a spherical charge distribution that the
electric field at r is given by:
E(r) = 4 pi/r^2 int_0^r rho(r') r'^2 dr'
You know what E(r) is, so you can calculate backwards (essentially
getting rid of the integral by taking a suitable derivative) in order
to get rho.
I guess that's what you meant, but that's so simple, I wouldn't dream
of trying to express the divergence in spherical coordinates!
.
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Stuck on a problem for two days |
07 Jun 2005 05:49:02 AM |
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Edward Green wrote:
Bjoern Feuerbacher wrote:
Gaganpreet Singh Arora wrote:
A spherically symmetrical but nonuniform volume distribution of charge
produces an electric field E = K*(r^4), directed radially outward from
the centre of the sphere. r is the radial distance from the centre, K
is a constant. What is the volume charge density of the sphere?
The easiest way to solve that is by using div E = 4 pi rho (Gauss' law
in differential form!), where one should express the divergence
operator in spherical coordinates.
Surely not.
If you don't know that, you have to use the integral form.
Yes. We know by symmetry that the field is radial, and we know its
value as a function of r. Therefore, taking appropriate units, we know
the amount of charge enclosed within a sphere of radius r, and
differentiating, the amount enclosed inside a shell between r and dr.
Finally we divide by 4pi to get the volume charge density.
Essentially, it says for a spherical charge distribution that the
electric field at r is given by:
E(r) = 4 pi/r^2 int_0^r rho(r') r'^2 dr'
You know what E(r) is, so you can calculate backwards (essentially
getting rid of the integral by taking a suitable derivative) in order
to get rho.
I guess that's what you meant, but that's so simple, I wouldn't dream
of trying to express the divergence in spherical coordinates!
If you have that already available (and you can find this easily
in many math books for physicists), it's the fastest way.
Bye,
Bjoern
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