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a real man's physics question...

Question: find energy of uniform charge filled sphere containing total charge Ze.


radius of sphere = a

charge ze contained

2. Relevant equations

by gauss: E= (ze)*r/4pi(epsilon)*a^3 at any point within the sphere.

and

dv=r^2*sin(theta)*dTheta*dphi*dr

Energy density= 0.5*epsilon*E^2

3. The attempt at a solution

letting energy= integral(energy density)dv

with limits: r= 0-a, theta= 0-pi, phi=0- 2pi

i get wrong answer:

(ze)^2/40epsilon*pi*a ration between top and bottom, should be 3(ze)^2/20epsilon*pi*a any ideas? I suspect this initial approach wrong?

many thanks!

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niceperson
Question: find energy of uniform charge filled sphere containing total charge Ze.


radius of sphere = a

charge ze contained

2. Relevant equations

by gauss: E= (ze)*r/4pi(epsilon)*a^3 at any point within the sphere.

and

dv=r^2*sin(theta)*dTheta*dphi*dr

Energy density= 0.5*epsilon*E^2

3. The attempt at a solution

letting energy= integral(energy density)dv

with limits: r= 0-a, theta= 0-pi, phi=0- 2pi

i get wrong answer:

(ze)^2/40epsilon*pi*a ration between top and bottom, should be 3(ze)^2/20epsilon*pi*a any ideas? I suspect this initial approach wrong?

many thanks!

An energy is taken from infinity, you need the sum of 2 integrals, 1 of them the energy outside of the sphere from infinity to a with a normal coulomb field and 1 inside the sphere which is the one you have.
Reply 2
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niceperson
OP
Is the second integral for energy outside: ze*integrate(a-infinity)ze/4pi(epsilon)r^2

bucause this is not giving corect answer still ie: ze^2/4pi(epsilon)*a

when added to (ze)^2/40epsilon*pi*a this is still 11/40...not 3/20
niceperson
Is the second integral for energy outside: ze*integrate(a-infinity)ze/4pi(epsilon)r^2

bucause this is not giving corect answer still ie: ze^2/4pi(epsilon)*a

when added to (ze)^2/40epsilon*pi*a this is still 11/40...not 3/20

did you remember the factor of 0.5?
Reply 4
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niceperson
OP
For which integral, the energy density one?
niceperson
For which integral, the energy density one?

they're both energy densities (or should be). I was particularly thinking the one I told you to add.
Reply 6
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niceperson
OP
I see so the second integral should be:

: 0.5*epsilon* integrate(a-infinity) (ze/4pi(epsilon)*r^2)^2 dr
niceperson
I see so the second integral should be:

: 0.5*epsilon* integrate(a-infinity) (ze/4pi(epsilon)*r^2)^2 dr

I think so yes. I don't know what that gives you.
what level of physics is this?
^^ A Level/1st Year University as its fairly basic integration once you have the fields since the fields are spherically symmetric, i.e. they have no angular dependance.



I was just wondering what motivated the choice of the thread title?
Reply 10
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niceperson
OP
sorry... correction it should be dv not dr. will try this
^^ A Level/1st Year University.


thats reassuring:biggrin:
F1/someone

attempt



can you point out my mistake?


edit: And this is not A-level physics, at least not with the triple integral:p: (although a very basic one)
nota bene
F1/someone

attempt



can you point out my mistake?


edit: And this is not A-level physics, at least not with the triple integral:p: (although a very basic one)

if you're trying to do this question then the mistake is your limits are wrong. its 0 to a inside the sphere, when you have r on the top and a coulomb field of 1/r^2 from a to infinity. That's why it is the sum of 2 integrals.
nota bene
F1/someone

attempt



can you point out my mistake?


edit: And this is not A-level physics, at least not with the triple integral:p: (although a very basic one)



Exactly the point I made, it would be a waste of ink to write down a triple integral for this problem seeing as the field is spherically symmetric. You can immediately write the 4π 4 \pi out the front and only consider the radial equation. I did it at A Level, as well as the analogous gravitational examples.

As F1F pointed out, there is a discontinuity at the boundary on the sphere due to the field changing from a radially linear field to a radial inverse square law, so you need to sum over the inside and outside contributions.
Reply 15
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niceperson
OP
It worked!!!! In general then to find energy/potential of charge systems we integrate the energy density to infinity in sections for the different levels of E.

question:

In the textbook potential outside the sphere is: q/4pi*epsilon*r

and inside q(3a^2/2-r^2/2)/4pi*epsilon*a^3

why can we not just integrate these expressions inside and to infinity?
F1 fanatic
if you're trying to do this question then the mistake is your limits are wrong. its 0 to a inside the sphere, when you have r on the top and a coulomb field of 1/r^2 from a to infinity. That's why it is the sum of 2 integrals.

Ah, yes. I've got the limits right, but the r's wrong because I haven't started revising electricity yet:redface: (already did the first integral, which was simple) Thanks!
niceperson
It worked!!!! In general then to find energy/potential of charge systems we integrate the energy density to infinity in sections for the different levels of E.

question:

In the textbook potential outside the sphere is: q/4pi*epsilon*a

and inside q(3a^2/2-r^2/2)/4pi*epsilon*a^3

why can we not just integrate these expressions inside and to infinity?

simply put, provided you take the limits into account then you can do it with potentials. You've made life harder for yourself in a way by considering energy densities but as you found out it gives the same answer, so you didn't break physics which is always good news.
Reply 18
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niceperson
OP
but integrating electric fields from 0 to infinity

V=integration(0-a)ze*r/4*pi*e*a^3 dr + integration(a-infinity)ze/4*pi*e*r^2 dr

= ze/8*pi*e*a + ze/4*pi*e*a which is NOT correct

when substitued into in E= 1/2 VQ

with this approach what am i missing?
Inside the sphere the integrand should be proportional to r4 r^4 since the E field is linear inside so when squared it becomes r2 r^2 and there is a r2 r^2 from the spherical polar coordinate volume element. Outside the sphere the integrand should be proportional to 1r2 \frac{1}{r^2} since there is a 1r4 \frac{1}{r^4} contribution from E^2 and r2 r^2 from the coordinate choice:


Energy=ϵ02⋅Z2e216π2ϵ02a6⋅4π∫0ar4dr⏟=a55+ϵ02⋅Z2e216π2ϵ02⋅4π∫a∞drr2⏟=1a=3Z2e220πϵ0a \displaystyle \mathrm{Energy} = \frac{\epsilon_0}{2}\cdot\frac{Z^2 e^2}{16 \pi^2 \epsilon_0^2 a^6}\cdot 4\pi \underbrace{\int_{0}^{a} r^4 dr}_{= \frac{a^5}{5}} + \frac{\epsilon_0}{2}\cdot \frac{Z^2 e^2}{16 \pi^2 \epsilon_0^2}\cdot 4\pi \underbrace{\int_{a}^{\infty} \frac{dr}{r^2}}_{=\frac{1}{a}} = \frac{3 Z^2 e^2}{20 \pi \epsilon_0 a}

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