# Potential inside a grounded spherical conducting shell due to point dipole

• JayKo
In summary: You need four boundary conditions to solve for the potential at the origin. You can assume that the potential is zero at the origin, and solve for the potential due to the shell and the dipole.
JayKo

## Homework Statement

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.)

## Homework Equations

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

V(r)=$$\frac{1}{4\pi\epsilon}$$$$\sum\frac{q}{r}$$

## The Attempt at a Solution

http://www.sci.sdsu.edu/classes/physics/phys196/ferguson/media/Image24.gif

i'm not sure the effect of spherical shell on the potential at any points inside the points, anyone can offer some insight thanks?

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.

well, i just informed by professor the point dipole at the origin will have the potential of $$\frac{1}{4\pi\epsilon}\frac{p*cos\theta}{r^{2}}$$ 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, ($$A*_{l}r^{l}+\frac{B_{l}}{r^{l+1}}$$) $$P_{l}*cos\theta$$. (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 can't choose r=0 as the potential fx will blow up. any ideas how to kick start the next step, thanks

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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.]

JayKo said:
well, i just informed by professor the point dipole at the origin will have the potential of $$\frac{1}{4\pi\epsilon}\frac{p*cos\theta}{r^{2}}$$ 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, ($$A*_{l}r^{l}+\frac{B_{l}}{r^{l+1}}$$) $$P_{l}*cos\theta$$. (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 can't 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:

$$V(r,\theta)=\sum_{l=0}^{\infty}A_l r^l P_l (\cos\theta)$$

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

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

gabbagabbahey said:
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:

$$V(r,\theta)=\sum_{l=0}^{\infty}A_l r^l P_l (\cos\theta)$$

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

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

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.

JayKo said:
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 $A_l$ and $B_l$ (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 $B_l$ values.

Now just apply the other boundary condition $$V(r=R,\theta)=0[/itex] to find the $A_l$ values. jambaugh said: I don't know if this helps but consider that since the shell is conducting and grounded the field outside should be zero as should be the potential. Thus the superposition of the fields due to the charge distribution on the sphere and the dipole inside should cancel outside the sphere. You don't need to solve for the charge distribution on the sphere explicitly you need only determine the component of the potential it must be producing to cancel the potential of the dipole. You can then use the fact that the potentials of both sphere and internal dipole must tend to zero at r->infinity. [Edit: Correction! Not necessarily the potentials but the fields i.e. gradient of the potentials must approach 0 at r->infinity.] i see, well, is it possible to assume r->infinity, V=0. for the dipole? as i need to establish the boundary condition to solve for the coefficient of A.thanks [you need only determine the component of the potential it must be producing to cancel the potential of the dipole] how do i determine the potential of the shell itself? assuming i get your meaning. gabbagabbahey said: Really? Why would you need 4 BCs? The general solution only has 2 unknown constants $A_l$ and $B_l$ (Which is the way it should be since it is the solution to a 2nd order differential equation!) The fact that the potential due to the shell is bounded at r=0 allowed you to determine the $B_l$ values. wait, i don't get it. [tex]A*_{l}r^{l}+\frac{B_{l}}{r^{l+1}}$$ you see, when r=0, the terms blow up. anything to the power of zero is still zero.how to determine $$B_l$$ ?

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

In order to find the $A_l$'s apply the condition at the shell $$V(r=R,\theta)=0$$. You still haven't actually used that condition!

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

In order to find the $A_l$'s apply the condition at the shell $$V(r=R,\theta)=0$$. You still haven't actually used that condition!

oh i see, i m left with determining the coefficient, $A_l$. i suppose the method is called Fourier trick by david griffith? right?

which means

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

and then multiply both sides by sin(theta)? and integrate it

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JayKo said:
wait, i don't get it. $$A*_{l}r^{l}+\frac{B_{l}}{r^{l+1}}$$ you see, when r=0, the terms blow up. anything to the power of zero is still zero.how to determine $$B_l$$ ?

All the $B_l$s must be zero except for $B_1$---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 $B_l$s were non-zero, you would have other terms where you end up dividing by zero at the origin.

The reason why the $B_1$ term corresponds to the dipole, is because of the form of the potential of the dipole

$$V_{dip}=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}=\frac{1}{4\pi\epsilon_0}\frac{p*P_1(\cos\theta)}{r^{1+1}}\implies B_1=\frac{p}{4\pi\epsilon_0}$$

JayKo said:
oh i see, i m left with determining the coefficient, $A_l$. 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 $$P_m(\cos\theta)\sin\theta d\theta$$ and integrate from 0 to pi.

But a much easier method is to notice that $$\cos\theta=P_1(\cos\theta)$$ and simply compare coeffecients.

JayKo said:
which means

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

You mean:

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

where capital R is the radius of the shell, right?

gabbagabbahey said:
All the $B_l$s must be zero except for $B_1$---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 $B_l$s were non-zero, you would have other terms where you end up dividing by zero at the origin.

The reason why the $B_1$ term corresponds to the dipole, is because of the form of the potential of the dipole

$$V_{dip}=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}=\frac{1}{4\pi\epsilon_0}\frac{p*P_1(\cos\theta)}{r^{1+1}}\implies B_1=\frac{p}{4\pi\epsilon_0}$$

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

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

hence comparing it and we get $=B_l=\frac{p}{P_l*4*pi*epsilon}$

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JayKo said:
how you come about this equation?$$V_{dip}=\frac{1}{4\pi\epsilon_0}\frac{p*\cos\theta}{r^{2}}=\frac{1}{4\pi\epsilon_0}\frac{p*P_1(\cos\theta)}{r^{1+1}}$$
i don't get it, as it is not the form of zonal harmonic. why?

It is exactly in the form of a zonal harmonic...

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

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

How is that not in the form of a zonal harmonic?

gabbagabbahey said:
Sort of, one method is to use a "Legendre trick" and multiply each side of the equation by $$P_m(\cos\theta)\sin\theta d\theta$$ and integrate from 0 to pi.

But a much easier method is to notice that $$\cos\theta=P_1(\cos\theta)$$ and simply compare coeffecients.

here why $$P_1=1[/itex]? isn't it = r? JayKo said: here why [tex]P_1=1[/itex]? isn't it = r? From page 138, Table 3.1 in Griffiths (3rd edition), [tex]P_1(x)=x$$...so $$P_1(\cos\theta)=\cos\theta$$

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

gabbagabbahey said:
It is exactly in the form of a zonal harmonic...

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

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

How is that not in the form of a zonal harmonic?
ok,i got it now. $$B_l=\frac{p}{4\pi\epsilon_0}$$ and that l=1 by comparing the power r on both sides. but legendre polynomial $$P_1(x)=x$$ here how come B_l =1?

gabbagabbahey said:
From page 138, Table 3.1 in Griffiths (3rd edition), $$P_1(x)=x$$...so $$P_1(\cos\theta)=\cos\theta$$
i m using the same book and referring to it now.thanks. ok i got it. cause i misunderstand the bracket (cos theta) as times instead of function of cosine theta.thanks a lot.

btw, a personal question if you don't mind? what you working as now?

gabbagabbahey said:
You mean:

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

where capital R is the radius of the shell, right?

yupe,big R.
from the above equation, i come to this:

$$\sum_{l=0}^{1}A_1 =-\frac{1}{4\pi\epsilon_0}\frac{p}{R^{3}}$$
right?

JayKo said:
yupe,big R.
from the above equation, i come to this:

$$\sum_{l=0}^{1}A_l =-\frac{1}{4\pi\epsilon_0}\frac{p}{R^{3}}$$
right?

That doesn't quite follow...don't you mean

$$A_l=\left\{ \begin{array}{lr} -\frac{1}{4\pi\epsilon_0}\frac{p}{R^{3}}, & l=1 \\0, & l\neq 1\end{array}$$

?

gabbagabbahey said:
That doesn't quite follow...don't you mean

$$A_l=\left\{ \begin{array}{lr} -\frac{1}{4\pi\epsilon_0}\frac{p}{R^{3}}, & l=1 \\0, & l\neq 1\end{array}$$

?
oh i see, cause i not sure when it is not 1, the A_l is actually 0.now i got it thanks. so basically that is the end of this question. thanks again. gute nacht

## 1. What is a grounded spherical conducting shell?

A grounded spherical conducting shell is a hollow spherical object made of a material that allows electric charge to flow freely. It is connected to the ground, which means that any excess charge on the shell will be immediately neutralized by the ground.

## 2. What is a point dipole?

A point dipole is a hypothetical arrangement of two equal and opposite electrical charges, with negligible distance between them. It is used to model the behavior of electric fields in situations where the distance between charges is very small compared to the distance from the observer.

## 3. What is the potential inside a grounded spherical conducting shell due to a point dipole?

The potential inside a grounded spherical conducting shell due to a point dipole is zero. This is because the grounded shell is connected to the ground, which acts as a charge sink, neutralizing any excess charge. In other words, the electric potential inside the shell is constant and equal to the potential of the ground, which is zero.

## 4. How does the distance from the point dipole affect the potential inside the grounded spherical conducting shell?

The distance from the point dipole does not affect the potential inside the grounded spherical conducting shell. This is because the charge on the dipole is considered to be concentrated at a single point, and the potential inside a conducting shell is constant regardless of the distance from the charge.

## 5. Can a grounded spherical conducting shell shield external electric fields?

Yes, a grounded spherical conducting shell can shield external electric fields. This is because the excess charges on the shell will rearrange themselves in such a way that the electric field inside the shell is zero. This phenomenon is known as electrostatic shielding and is commonly used in electronic devices to protect sensitive components from external electric fields.

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