Laplacian for hyperbolic plates

[Physgeek64]

Homework Statement

Find a complete set of conditions on the constants a, b, c, n such that, for Cartesian coordinates (x, y, z), V = axn + byn + czn is a solution of Laplace’s equation $∇^2V = 0$. A mass filter for charged particles consists of 4 electrodes extended along the z direction, whose surfaces describe hyperbolae in the xy plane: $x^2 − y^2 = d^2$ for one pair, and $x^2 − y^2 = −d^2$ for the other pair, where d is the distance from the axis to the nearest point on an electrode surface. [Each hyperbola has two branches and thus describes two (diagonally opposite) electrodes.] A positive voltage $V_0$ is applied to the pair at $x^2 − y^2 = d^2$, and $−V_0$ is applied to the other pair. Taking the electrode length along the z axis to be effectively infinite, find the electric potential in the region between the electrodes.

Homework Equations

The Attempt at a Solution

So the boundary conditions i get are
$V(x,\sqrt{x^2-d^2},z)=V_0$
$V(x,\sqrt{x^2+d^2},z)=-V_0$
$\nabla^2V=0$

for solutions of the form $V=\summation a_nx^n+b_ny^n+c_nz^n$ $c_n=0$

When I use all of these boundary contains i find that we must only have odd powers of x and that

$\summation a_n x^n +b_n(x^2-d^2)^{\frac{n}{2}}=V_0$
$\summation a_n x^n +b_n(x^2+d^2)^{\frac{n}{2}}=-V_0$

but I can't see how to formulate a solution given these as my coefficients will depend on x .

Many thanks

 

[TSny]

Begin with the first task:

Find a complete set of conditions on the constants a, b, c, n such that, for Cartesian coordinates (x, y, z), V = axⁿ + byⁿ + czⁿ is a solution of Laplace’s equation $∇^2V = 0$.

 

[Physgeek64]

Begin with the first task:

$ax^{n-2}+by^{n-2} +cz^{n-2}=0$ ?
But this would imply
$ax^{n-2}=k_x$
$by^{n-2}=k_y$
$cz^{n-2}=k_z$
?

 

[TSny]

$ax^{n-2}+by^{n-2} +cz^{n-2}=0$ ?

OK

But this would imply
$ax^{n-2}=k_x$
$by^{n-2}=k_y$
$cz^{n-2}=k_z$
?

Is $k_x$ some constant? If so, how can you make $ax^{n-2}$ a constant?

 

[Physgeek64]

OKIs $k_x$ some constant? If so, how can you make $ax^{n-2}$ a constant?

Only if we have n=0?

 

[TSny]

Only if we have n=0?

If n = 0 then $ax^{n-2} = ax^{-2} = a/x^2$.

 

[Physgeek64]

If n = 0 then $ax^{n-2} = ax^{-2} = a/x^2$.

sorry n=2?

 

[TSny]

n=2?

Right.

There is no need to introduce the notation $k_x$, $k_y$, $k_z$ .

You found that you need the relation $ax^{n-2}+by^{n-2} +cz^{n-2}=0$ to be satisfied. And this must be satisfied for all points (x, y, z) in the region where Laplace's equation holds. Thus, the left hand side of this relation must be independent of the choice of (x, y, z). This can only happen for one value of n.

 

[Physgeek64]

Right.

There is no need to introduce the notation $k_x$, $k_y$, $k_z$ .

You found that you need the relation $ax^{n-2}+by^{n-2} +cz^{n-2}=0$ to be satisfied. And this must be satisfied for all points (x, y, z) in the region where Laplace's equation holds.Thus, the left hand side of this relation must be independent of the choice of (x, y, z). This can only happen for one value of n.

Right Then for the next part i have $V_0=ax^2+b(x^2-d^2)$ and $-V_0=ax^2+b(x^2+d^2)$

$2V_0=-2bd^2$
$b=-\frac{V_0}{d^2}$

and considering the point (d,0)
$a=\frac{V_0}{d^2}$ ?

 

[TSny]

There's more required for the first part. You've found that n must equal 2. Are there any restrictions on a, b, and c?

 

[Physgeek64]

There's more required for the first part. You've found that n must equal 2. Are there any restrictions on a, b, and c?

a+b+c=0 ?

 

[TSny]

a+b+c=0 ?

Yes. So, at this point, how would you write the function V(x, y, z) in the region where Laplace's equation holds?

 

[Physgeek64]

Yes. So, at this point, how would you write the function V(x, y, z) in the region where Laplace's equation holds?

V=ax^2+by^2-(a+b)z^2 ?

 

[TSny]

V=ax^2+by^2-(a+b)z^2 ?

OK.

Now, moving on to the particular problem with the hyperbolic electrodes, can you say anything about how V(x, y, z) should depend on z?

 

[Physgeek64]

OK.

Now, moving on to the particular problem with the hyperbolic electrodes, can you say anything about how V(x, y, z) should depend on z?

No dependence on z

 

[TSny]

No dependence on z

Good.

 

[Physgeek64]

Good.

$V=ax^2-ay^2$

$2V_0=-2bd^2$
$b=-\frac{V_0}{d^2}$

and considering the point (d,0)
$a=\frac{V_0}{d^2}$ ?

 

[TSny]

$V=ax^2-ay^2$

$2V_0=-2bd^2$
$b=-\frac{V_0}{d^2}$

and considering the point (d,0)
$a=\frac{V_0}{d^2}$ ?

That's a bit of a jump for me. Does the independence of V on z imply any restriction on the constants $a$ and $b$ ?

At this point, what's the simplest way to write V(x, y, z) in the region where Laplace's equation holds. By simplest, I mean write V with the least number of constants.

 

[Physgeek64]

That's a bit of a jump for me. Does the independence of V on z imply any restriction on the constants $a$ and $b$ ?

At this point, what's the simplest way to write V(x, y, z) in the region where Laplace's equation holds. By simplest, I mean write V with the least number of constants.

a=-b
$V=a(x^2-y^2)$

 

[TSny]

a=-b
$V=a(x^2-y^2)$

Yes. I now see that you had written $V=ax^2-ay^2$ in post #18, which I overlooked. Good. And you also correctly found the value of $a$.

So, what's your final answer for V(x, y, z)?

 

[Physgeek64]

Yes. I now see that you had written $V=ax^2-ay^2$ in post #18, which I overlooked. Good. And you also correctly found the value of $a$.

So, what's your final answer for the V(x, y, z)?

$V=\frac{V_0}{d^2}(x^2-y^2)$

 

[TSny]

Very good. Make sure that this does in fact satisfy the boundary conditions at all four electrodes and, of course, that it also satisfies Laplace's equation in the region between the electrodes.

 

[Physgeek64]

Very good. Make sure that this does in fact satisfy the boundary conditions at all four electrodes and, of course, that it also satisfies Laplace's equation in the region between the electrodes.

Thank you for all your help!

 

[Physgeek64]

Very good. Make sure that this does in fact satisfy the boundary conditions at all four electrodes and, of course, that it also satisfies Laplace's equation in the region between the electrodes.

Would you mind also helping with the next bit of the question

A solid, non-magnetic dielectric cylinder of radius R ≪ d and relative permittivity ǫr is placed between the electrodes, centred on the z axis. Sketch the distribution of surface polarization charge induced on the cylinder, and hence show that the cylinder does not acquire a dipole moment. By introducing a term of the form r−2 cos 2φ where appropriate (x = r cos φ, y = r sin φ), propose an approximate trial solution in cylin- drical coordinates for the potential inside and outside the cylinder, valid in the limit R/d → 0. Hence solve Laplace’s equation throughout the region between the electrodes in this limit.

For the first part I have
$\sigma=P \dot \hat{n}$
For a linear material
$P=\epsilon_0\chi_e E$
$E=-\nabla V = -\frac{2V_0}{d^2}x \hat{x} +\frac{2V_0}{d^2}y\hat{y}$
$P=\frac{2\epsilon_0\chi_eV_0}{d^2}(x\hat{x}+y\hat{y})$
$\sigma = P \dot \hat{n} = \frac{2\epsilon_0\chi_eV_0}{d^2}(x\hat{x}+y\hat{y}) \dot (cos\theta \hat{x} + sin\theta \hat{y})$
using $x=Rcos\theta$ and $y=Rsin\theta$
$\sigma = \frac{2\epsilon_0\chi_eV_0R}{d^2}$

Which is uniform all the way around so no dipole moment

 

[TSny]

The first part of the problem asks for a (rough?) sketch of the surface charge distribution. Then, use this sketch to deduce zero net dipole moment of this charge distribution. (That's my interpretation of what they want, anyway.)

Note that a uniform, nonzero $\sigma$ would imply a nonzero net charge on the cylinder.

 

[Physgeek64]

The first part of the problem asks for a (rough?) sketch of the surface charge distribution. Then, use this sketch to deduce zero net dipole moment of this charge distribution. (That's my interpretation of what they want, anyway.)

Note that a uniform, nonzero $\sigma$ would imply a nonzero net charge on the cylinder.

I don't see how you could sketch it without working out $\sigma$

Could we not have a non zero volume charge though? To satisfy the conservation of charge

 

[TSny]

I don't see how you could sketch it without working out $\sigma$

Sketch E-field lines in the region of the origin. Look for symmetry. This should help you see any symmetry of the surface charge density.

Could we not have a non zero volume charge though? To satisfy the conservation of charge

Usually, it is assumed that the susceptibility $\chi_e$ is constant inside the material (i.e., does not depend on position). From $\bf D = \epsilon_0 \left( 1 + \chi_e \right) \bf E$, deduce that $\nabla \cdot \bf D$ $= \epsilon_0 \left( 1 + \chi_e \right)\nabla \cdot \bf E$. But we know $\nabla \cdot \bf D$ $= 0$ (assuming no free volume charge density $\rho_f$). Hence, $\nabla\cdot \bf E$ $= 0$. And this implies zero total volume charge density $\rho_f + \rho_b$. Hence, what can you conclude about any bound charge density $\rho_b$?