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Potential inside a grounded spherical conducting shell due to point dipole

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JayKo
#1
Feb10-09, 10:32 AM
P: 128
1. The problem statement, all variables and given/known data
Suppose a point dipole is located at the centre of a grounded spherical conducting shell. Find the potential and electric field at points inside the shell. (Hint: Use zonal harmonics that are regular at the origin to satisfy the boundary conditions on the shell.)


2. Relevant equations

since the metal is grounded then only negative charges are left on the surface.

V(r)=[tex]\frac{1}{4\pi\epsilon}[/tex][tex]\sum\frac{q}{r}[/tex]

3. The attempt at a solution


i'm not sure the effect of spherical shell on the potential at any points inside the points, anyone can offer some insight thanks?
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gabbagabbahey
#2
Feb10-09, 11:06 AM
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You seem to be attempting to use Coulomb's law; but that is a bad idea. The dipole will induce some unknown charge density onto the shell...correct? How can you possibly use Coulomb's law when you don't know what that charge density is?

Instead, use the hint provided. Where does the potential obey Laplace's equation? What is the general solution to Laplace's equation using spherical harmonics? Think of appropriate boundary conditions and apply them to that solution.
JayKo
#3
Feb11-09, 01:54 AM
P: 128
well, i just informed by professor the point dipole at the origin will have the potential of [tex]\frac{1}{4\pi\epsilon}\frac{p*cos\theta}{r^{2}}[/tex] inside the sphere (p=dipole moment).

the boundary condition at the shell is V=0. and i need 1 more boundary condition to solve the laplace equation, which in this case is the solution is of the form, ([tex]A*_{l}r^{l}+\frac{B_{l}}{r^{l+1}}[/tex]) [tex]P_{l}*cos\theta[/tex]. (the latex has some problem, but the way the theta term in denominator is not meant to be there ad should be shifted outside to be the nominator) just not sure what is the next boundary condition as i cant choose r=0 as the potential fx will blow up. any ideas how to kick start the next step, thanks

jambaugh
#4
Feb11-09, 02:36 AM
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Potential inside a grounded spherical conducting shell due to point dipole

I don't know if this helps but consider that since the shell is conducting and grounded the field outside should be zero as should be the potential. Thus the superposition of the fields due to the charge distribution on the sphere and the dipole inside should cancel outside the sphere.

You don't need to solve for the charge distribution on the sphere explicitly you need only determine the component of the potential it must be producing to cancel the potential of the dipole. You can then use the fact that the potentials of both sphere and internal dipole must tend to zero at r->infinity. [Edit: Correction! Not necessarily the potentials but the fields i.e. gradient of the potentials must approach 0 at r->infinity.]
gabbagabbahey
#5
Feb11-09, 07:55 AM
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Quote Quote by JayKo View Post
well, i just informed by professor the point dipole at the origin will have the potential of [tex]\frac{1}{4\pi\epsilon}\frac{p*cos\theta}{r^{2}}[/tex] inside the sphere (p=dipole moment).

the boundary condition at the shell is V=0. and i need 1 more boundary condition to solve the laplace equation, which in this case is the solution is of the form, ([tex]A*_{l}r^{l}+\frac{B_{l}}{r^{l+1}}[/tex]) [tex]P_{l}*cos\theta[/tex]. (the latex has some problem, but the way the theta term in denominator is not meant to be there ad should be shifted outside to be the nominator) just not sure what is the next boundary condition as i cant choose r=0 as the potential fx will blow up. any ideas how to kick start the next step, thanks
Would the potential blow up at the origin if there was no dipole there? If not, then the blow up at the origin is due entirely to the dipole potential and so you can say that the potential due to just the shell must be of the form:

[tex]V(r,\theta)=\sum_{l=0}^{\infty}A_l r^l P_l (\cos\theta)[/tex]

And so the potential due to the shell and the dipole is of the form:

[tex]V(r,\theta)=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}+\sum_{l= 0}^{\infty}A_l r^l P_l (\cos\theta)[/tex]

JayKo
#6
Feb11-09, 08:33 AM
P: 128
Quote Quote by gabbagabbahey View Post
Would the potential blow up at the origin if there was no dipole there? If not, then the blow up at the origin is due entirely to the dipole potential and so you can say that the potential due to just the shell must be of the form:

[tex]V(r,\theta)=\sum_{l=0}^{\infty}A_l r^l P_l (\cos\theta)[/tex]

And so the potential due to the shell and the dipole is of the form:

[tex]V(r,\theta)=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}+\sum_{l= 0}^{\infty}A_l r^l P_l (\cos\theta)[/tex]

thanks for the help.
potential at the origin is zero if no dipole, i supposed.
good reasons, but here comes the questions.
since the potential is of the function (radius,theta) we need 4 equations (boundary conditions), and what would that be? the only condition i know off is r=R (at the shell) V=0.
gabbagabbahey
#7
Feb11-09, 08:43 AM
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Quote Quote by JayKo View Post
thanks for the help.
potential at the origin is zero if no dipole, i supposed.
good reasons, but here comes the questions.
since the potential is of the function (radius,theta) we need 4 equations (boundary conditions)
Really? Why would you need 4 BCs?

The general solution only has 2 unknown constants [itex]A_l[/itex] and [itex]B_l[/itex] (Which is the way it should be since it is the solution to a 2nd order differential equation!)

The fact that the potential due to the shell is bounded at r=0 allowed you to determine the [itex]B_l[/itex] values.

Now just apply the other boundary condition [tex]V(r=R,\theta)=0[/itex] to find the [itex]A_l[/itex] values.
JayKo
#8
Feb11-09, 08:46 AM
P: 128
Quote Quote by jambaugh View Post
I don't know if this helps but consider that since the shell is conducting and grounded the field outside should be zero as should be the potential. Thus the superposition of the fields due to the charge distribution on the sphere and the dipole inside should cancel outside the sphere.

You don't need to solve for the charge distribution on the sphere explicitly you need only determine the component of the potential it must be producing to cancel the potential of the dipole. You can then use the fact that the potentials of both sphere and internal dipole must tend to zero at r->infinity. [Edit: Correction! Not necessarily the potentials but the fields i.e. gradient of the potentials must approach 0 at r->infinity.]
i see, well, is it possible to assume r->infinity, V=0. for the dipole? as i need to establish the boundary condition to solve for the coefficient of A.thanks

[you need only determine the component of the potential it must be producing to cancel the potential of the dipole] how do i determine the potential of the shell itself? assuming i get your meaning.
JayKo
#9
Feb11-09, 08:52 AM
P: 128
Quote Quote by gabbagabbahey View Post
Really? Why would you need 4 BCs?

The general solution only has 2 unknown constants [itex]A_l[/itex] and [itex]B_l[/itex] (Which is the way it should be since it is the solution to a 2nd order differential equation!)

The fact that the potential due to the shell is bounded at r=0 allowed you to determine the [itex]B_l[/itex] values.
wait, i don't get it. [tex]A*_{l}r^{l}+\frac{B_{l}}{r^{l+1}} [/tex] you see, when r=0, the terms blow up. anything to the power of zero is still zero.how to determine [tex]B_l[/tex] ?
gabbagabbahey
#10
Feb11-09, 08:53 AM
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Quote Quote by JayKo View Post
i see, well, is it possible to assume r->infinity, V=0. for the dipole? as i need to establish the boundary condition to solve for the coefficient of A.thanks

The solution [tex]V(r,\theta)=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}+\sum_{l= 0}^{\infty}A_l r^l P_l (\cos\theta)[/tex] is only valid inside the sphere; the boundaries are therefor at r=0 and r=R not r=infinity.

In order to find the [itex]A_l[/itex]'s apply the condition at the shell [tex]V(r=R,\theta)=0[/tex]. You still haven't actually used that condition!
JayKo
#11
Feb11-09, 09:03 AM
P: 128
Quote Quote by gabbagabbahey View Post
The solution [tex]V(r,\theta)=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}+\sum_{l= 0}^{\infty}A_l r^l P_l (\cos\theta)[/tex] is only valid inside the sphere; the boundaries are therefor at r=0 and r=R not r=infinity.

In order to find the [itex]A_l[/itex]'s apply the condition at the shell [tex]V(r=R,\theta)=0[/tex]. You still haven't actually used that condition!
oh i see, i m left with determining the coefficient, [itex]A_l[/itex]. i suppose the method is called fourier trick by david griffith? right?

which means

[tex]\sum_{l=0}^{\infty}A_l r^l P_l (\cos\theta)=-\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}[/tex]

and then multiply both sides by sin(theta)? and integrate it
gabbagabbahey
#12
Feb11-09, 09:06 AM
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Quote Quote by JayKo View Post
wait, i don't get it. [tex]A*_{l}r^{l}+\frac{B_{l}}{r^{l+1}} [/tex] you see, when r=0, the terms blow up. anything to the power of zero is still zero.how to determine [tex]B_l[/tex] ?
All the [itex]B_l[/itex]s must be zero except for [itex]B_1[/itex]---which corresponds to the potential of the dipole which is the only contribution which should be allowed to "blow up" at the origin. If any of the other [itex]B_l[/itex]s were non-zero, you would have other terms where you end up dividing by zero at the origin.

The reason why the [itex]B_1[/itex] term corresponds to the dipole, is because of the form of the potential of the dipole

[tex]V_{dip}=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}=\frac{1}{4\p i\epsilon_0}\frac{p*P_1(\cos\theta)}{r^{1+1}}\implies B_1=\frac{p}{4\pi\epsilon_0}[/tex]
gabbagabbahey
#13
Feb11-09, 09:10 AM
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Quote Quote by JayKo View Post
oh i see, i m left with determining the coefficient, [itex]A_l[/itex]. i suppose the method is called fourier trick by david griffith? right?
Sort of, one method is to use a "Legendre trick" and multiply each side of the equation by [tex]P_m(\cos\theta)\sin\theta d\theta[/tex] and integrate from 0 to pi.

But a much easier method is to notice that [tex]\cos\theta=P_1(\cos\theta)[/tex] and simply compare coeffecients.
gabbagabbahey
#14
Feb11-09, 09:12 AM
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Quote Quote by JayKo View Post
which means

[tex]\sum_{l=0}^{\infty}A_l r^l P_l (\cos\theta)=-\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}[/tex]
You mean:

[tex]\sum_{l=0}^{\infty}A_l R^l P_l (\cos\theta)=-\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{R^{2}}[/tex]

where capital R is the radius of the shell, right?
JayKo
#15
Feb11-09, 09:19 AM
P: 128
Quote Quote by gabbagabbahey View Post
All the [itex]B_l[/itex]s must be zero except for [itex]B_1[/itex]---which corresponds to the potential of the dipole which is the only contribution which should be allowed to "blow up" at the origin. If any of the other [itex]B_l[/itex]s were non-zero, you would have other terms where you end up dividing by zero at the origin.

The reason why the [itex]B_1[/itex] term corresponds to the dipole, is because of the form of the potential of the dipole

[tex]V_{dip}=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}=\frac{1}{4\p i\epsilon_0}\frac{p*P_1(\cos\theta)}{r^{1+1}}\implies B_1=\frac{p}{4\pi\epsilon_0}[/tex]
how you come about this equation?[tex]V_{dip}=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}=\frac{1}{4\p i\epsilon_0}\frac{p*P_1(\cos\theta)}{r^{1+1}}[/tex]
this is from the [itex]B_l[/itex] term, i got it now. thanks
it should be this right?

[tex]V_{dip}=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}=\frac{B_L*p* P_1(\cos\theta)}{r^{1+1}}[/tex]

hence comparing it and we get [itex]=B_l=\frac{p}{P_l*4*pi*epsilon}[/itex]
gabbagabbahey
#16
Feb11-09, 09:26 AM
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Quote Quote by JayKo View Post
how you come about this equation?[tex]V_{dip}=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}=\frac{1}{4\p i\epsilon_0}\frac{p*P_1(\cos\theta)}{r^{1+1}}[/tex]
i dont get it, as it is not the form of zonal harmonic. why?
It is exactly in the form of a zonal harmonic....

What is [tex]\frac{B_1}{r^{1+1}}P_1(\cos\theta)[/tex] if [tex]B_1=\frac{p}{4\pi\epsilon_0}[/tex]?

What is [tex]\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}P_l(\cos\theta)[/tex] if [tex]B_l=\left\{ \begin{array}{lr} 0, & l\neq1\\ \frac{p}{4\pi\epsilon_0}, & l=1\end{array}[/tex]?

How is that not in the form of a zonal harmonic?
JayKo
#17
Feb11-09, 09:31 AM
P: 128
Quote Quote by gabbagabbahey View Post
Sort of, one method is to use a "Legendre trick" and multiply each side of the equation by [tex]P_m(\cos\theta)\sin\theta d\theta[/tex] and integrate from 0 to pi.

But a much easier method is to notice that [tex]\cos\theta=P_1(\cos\theta)[/tex] and simply compare coeffecients.
here why [tex]P_1=1[/itex]? isn't it = r?
gabbagabbahey
#18
Feb11-09, 09:35 AM
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Quote Quote by JayKo View Post
here why [tex]P_1=1[/itex]? isn't it = r?
From page 138, Table 3.1 in Griffiths (3rd edition), [tex]P_1(x)=x[/tex]....so [tex]P_1(\cos\theta)=\cos\theta[/tex]

When someone writes [tex]P_l(\cos\theta)[/tex], they mean [itex]P_l[/itex] as a function of [itex]\cos\theta[/itex] not [itex]P_l[/itex] times [itex]\cos\theta[/itex]......perhaps that is the source of your confusion?


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