Uncharged conductor inside an insulating shell

[guyvsdcsniper]

Homework Statement

An uncharged conducting sphere of radius a is coated with a thick
insulating shell (dielectric constant r ) out to radius b. This object is now placed in
an otherwise uniform electric field Eo. Find the dipole moment of the conductor.

Relevant Equations

p = ∫rσ(r)dA.

I worked this problem out in griffiths and my work checks out for for the potentials, b.c. and the coefficients. I will post the solutions just because my work is a little harder to read.

What I am having trouble finding is the dipole moment of the conductor.

I know the formula for dipole moment can be written as p = ∫rσ(r)dA.

Im stuck on how to find σ or if this formula is the right approach.

 

[kuruman]

If you know the potentials, you know the fields. The free surface charge density is equal to the discontinuity of the normal component of the electric displacement vector D at the conductor-dielectric boundary.

 

[guyvsdcsniper]

If you know the potentials, you know the fields. The free surface charge density is equal to the discontinuity of the normal component of the electric displacement vector D at the conductor-dielectric boundary.

So is it pretty much this? The electric field that i got for the dielectric is confirmed in griffiths.
I know that E inside the conductor is 0.

So the dipole moment would be equal to:
p = ∫rεE(r)*rdrdθ ?

 

[kuruman]

You had it right the first time. The dipole moment is $$\mathbf{p}=\int \sigma(\theta)\mathbf{r}'dA'.$$It's a surface integral - there is no $dr.$

 

[guyvsdcsniper]

You had it right the first time. The dipole moment is $$\mathbf{p}=\int \sigma(\theta)\mathbf{r}'dA'.$$It's a surface integral - there is no $dr.$

ahh. I wrote one thing but was thinking something different.

so the dA should be,

$$\mathbf{p}=\int \sigma(\theta)\mathbf{r}'r^2sin(\theta)d(\theta)d(\phi).$$
, right?

And with the electric field point in the z direction, the equatorial plane would have 0 free charges, so
$$\mathbf{r}'$$ would equal $$z\vec z$$ which I could then say equals $$rcosθ\vec z$$? and then I can just integrate?

 

[kuruman]

Not quite. $\mathbf{r'}$ is the position vector of area element $dA'$ on the surface of the conductor. How would you write $\mathbf{r'}$ in Cartesian unit vector notation? Please use LaTeX to do that.

 

[guyvsdcsniper]

Not quite. $\mathbf{r'}$ is the position vector of area element $dA'$ on the surface of the conductor. How would you write $\mathbf{r'}$ in Cartesian unit vector notation? Please use LaTeX to do that.

It should look like this?

$\vec r$ = $x\vec x + y\vec y +z\vec z$

 

[kuruman]

It should look like this?

$\vec r$ = $x\vec x + y\vec y +z\vec z$

What makes it a vector on the surface of a sphere of radius $a$ which is the conductor?

 

[guyvsdcsniper]

What makes it a vector on the surface of a sphere of radius $a$ which is the conductor?

Sorry, mistake.
$\vec r$ = $x\hat x + y\hat y +z\hat z$

This is what you mean correct?

 

[kuruman]

That's not it. A vector on the surface of the conductor has magnitude equal to the radius $a$ of the conductor at any value of its Cartesian components $x$, $y$ and $z$. The magnitude of the vector that you have written is not $a$, it is $r=\sqrt{x^2+y^2+z^2}$.

Hint: Look up the radial unit vector $\hat r$ here. Then consider that any radial vector can be written as $\mathbf{r}=r~\hat r$.

 

[Delta2]

Dipole moment can also be a volume or a line integral depending on the kind of charge density.

 

[kuruman]

Dipole moment can also be a volume or a line integral depending on the kind of charge density.

Sure, but here we have a spherical conductor.

 

[Delta2]

Sure, but here we have a spherical conductor.

Yes right, for conductors in static equilibrium and at least theoretically, all the charge can be only on their (inner or outer) surface so we have to deal with a surface charge density.

 

[kuruman]

Yes right, for conductors in static equilibrium and at least theoretically, all the charge can be only on their (inner or outer) surface so we have to deal with a surface charge density.

Formally, one can also use the volume integral $\mathbf {p}=\int \mathbf {r'}\rho(\mathbf {r'})dV'$ to find the dipole moment by writing the volume charge density in terms of a Dirac delta function $\rho(\mathbf r')=\sigma(\theta',\phi')\delta(r'-a)$. The result is the same.

 

[guyvsdcsniper]

That's not it. A vector on the surface of the conductor has magnitude equal to the radius $a$ of the conductor at any value of its Cartesian components $x$, $y$ and $z$. The magnitude of the vector that you have written is not $a$, it is $r=\sqrt{x^2+y^2+z^2}$.

Hint: Look up the radial unit vector $\hat r$ here. Then consider that any radial vector can be written as $\mathbf{r}=r~\hat r$.

I did a similar problem, where it was two insulators with bound charges stuck to each surfaces. When finding the dipole moment for one of the surfaces, I followed these general steps

And what happens with these problems is the equatorial plane has no charge leaving just the z component.

I don't see why the problem of this post is any different and i can't use
$\vec r$ = $x\hat x + y\hat y +z\hat z$, considering they problems are nearly identical

 

[kuruman]

As far as I can tell, you have not derived the surface charge density from the potentials because the derivation is not posted here. You just assumed that it is $\sigma(\theta) =\sigma_0 \cos\!\theta$ because that was the case in another problem. The fact that it is zero in the equatorial plane is an insufficient reason to write it in this form. The form $\sigma(\theta) =\sigma_0(1- \sin\!\theta$) is also zero in the equatorial plane. What makes this expression incorrect and your expression correct?

Furthermore, I am not convinced that you understand how to find the three components of the dipole moment given an arbitrary surface charge density $\sigma(\theta,\phi)$ and that is what I am trying to help you to do. The choice is yours: (a) do it by analogy to another problem without really understanding why and when it works or (b) learn how to do it right and gain the understanding and confidence that you can perform these tasks without guessing.

 

[guyvsdcsniper]

As far as I can tell, you have not derived the surface charge density from the potentials because the derivation is not posted here. You just assumed that it is $\sigma(\theta) =\sigma_0 \cos\!\theta$ because that was the case in another problem. The fact that it is zero in the equatorial plane is an insufficient reason to write it in this form. The form $\sigma(\theta) =\sigma_0(1- \sin\!\theta$) is also zero in the equatorial plane. What makes this expression incorrect and your expression correct?

Furthermore, I am not convinced that you understand how to find the three components of the dipole moment given an arbitrary surface charge density $\sigma(\theta,\phi)$ and that is what I am trying to help you to do. The choice is yours: (a) do it by analogy to another problem without really understanding why and when it works or (b) learn how to do it right and gain the understanding and confidence that you can perform these tasks without guessing.

It may have come off that i was trying to use the fact that $\sigma(\theta) =\sigma_0 \cos\!\theta$ in another problem but that wasn't my intention. I was using it as a reference to show how my integral should look.

Why can't I make the argument that this sphere is surrounded by a uniform field. If the positive and negative charges contribute equally at the midpoint, they would cancel each other so that the electric field is zero on the midpoint. And if I am in an x,y,z space, with z pointing in the direction of the field, then that means the x-y plane would have an electric field and σ of zero, so $\vec r$ = $x\hat x + y\hat y +z\hat z$ would not depend on the x and y components

 

[kuruman]

It may have come off that i was trying to use the fact that $\sigma(\theta) =\sigma_0 \cos\!\theta$ in another problem but that wasn't my intention. I was using it as a reference to show how my integral should look.

Fair enough. You still have to show that in this problem, the surface charge density is indeed given by $\sigma(\theta) =\sigma_0 \cos\!\theta$ and find an expression for $\sigma_0$ in terms of the given quantities, namely $E_0$, $a$ and $\varepsilon_r$.

And if I am in an x,y,z space, with z pointing in the direction of the field, then that means the x-y plane would have an electric field and σ of zero, so $\vec r$ = $x\hat x + y\hat y +z\hat z$ would not depend on the x and y components

I don't understand what you are trying to say here. I agree with the first part of the statement. You can certainly make the azimuthal symmetry argument which says that the z-direction is special and therefore the electric potential and field vectors are independent of the spherical angle $\phi$, i.e. they are functions of the colatitude angle $\theta$ only.

What confuses me is the second part about $\vec r$. It is a position vector with Cartesian components $x$, $y$ and $z$ and can be anywhere in space. Thus, it depends on what values you choose for these components. Furthermore, the electric field $\mathbf{E}(\mathbf{r})$ depends on these components. Specifically, its $x$ and $y$ components are zero in the $xy$-plane only and non-zero at points off that plane. Because of the symmetry, in a plane at constant $z$, $E_x(x,y,z)=E_y(x,y,z)$ and that's all that symmetry buys you.

 

[guyvsdcsniper]

Fair enough. You still have to show that in this problem, the surface charge density is indeed given by $\sigma(\theta) =\sigma_0 \cos\!\theta$ and find an expression for $\sigma_0$ in terms of the given quantities, namely $E_0$, $a$ and $\varepsilon_r$.I don't understand what you are trying to say here. I agree with the first part of the statement. You can certainly make the azimuthal symmetry argument which says that the z-direction is special and therefore the electric potential and field vectors are independent of the spherical angle $\phi$, i.e. they are functions of the colatitude angle $\theta$ only.

What confuses me is the second part about $\vec r$. It is a position vector with Cartesian components $x$, $y$ and $z$ and can be anywhere in space. Thus, it depends on what values you choose for these components. Furthermore, the electric field $\mathbf{E}(\mathbf{r})$ depends on these components. Specifically, its $x$ and $y$ components are zero in the $xy$-plane only and non-zero at points off that plane. Because of the symmetry, in a plane at constant $z$, $E_x(x,y,z)=E_y(x,y,z)$ and that's all that symmetry buys you.

Ok I can accept that. So from post #10, are you saying I should use:

as $\vec r$.?

 

[kuruman]

Ok I can accept that. So from post #10, are you saying I should use:
View attachment 301032 as $\vec r$.?

That's exactly what I am saying. Now you can write the position vector to a point on the conducting sphere as $\mathbf{r}=asin\theta\cos\!\varphi~\hat x+asin\theta\sin\!\varphi~\hat y+a\cos\theta~\hat z$, and do three separate integrals to find $p_x$, $p_y$ and $p_z$. You should expect the first two to be zero but you will also see how this method works in case you have a situation where they are not zero.

 

[guyvsdcsniper]

That's exactly what I am saying. Now you can write the position vector to a point on the conducting sphere as $\mathbf{r}=asin\theta\cos\!\varphi~\hat x+asin\theta\sin\!\varphi~\hat y+a\cos\theta~\hat z$, and do three separate integrals to find $p_x$, $p_y$ and $p_z$. You should expect the first two to be zero but you will also see how this method works in case you have a situation where they are not zero.

That makes a lot of sense. I see what you were saying now, I was being really stubborn about going about it the other way.

And so dA should be,
$dA = R^2sin\theta d\theta d\phi$ since its a surface integral?

 

[kuruman]

That makes a lot of sense. I see what you were saying now, I was being really stubborn about going about it the other way.

And so dA should be,
$dA = R^2sin\theta d\theta d\phi$ since its a surface integral?

Yes, except that the given radius of the sphere is $a$ not $R$.