# Homework Help: Laplacian for hyperbolic plates

1. Dec 23, 2017

### Physgeek64

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

2. Dec 23, 2017

### 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.

3. Dec 23, 2017

### Physgeek64

Thank you for all your help!!

4. Dec 23, 2017

### Physgeek64

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

5. Dec 23, 2017

### 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.

6. Dec 23, 2017

### Physgeek64

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

7. Dec 23, 2017

### TSny

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.

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 $\rho_b$?