Why Does Rho_b Blow Up (Griffiths Pg. 176)

[ehrenfest]

[SOLVED] Griffiths page 176

Homework Statement

Below example 4.4, Griffiths says "We cannot apply Gauss's law precisely at the surface of a dielectric, for rho_b blows up..."

Why does rho_b blow up?

Homework Equations

The Attempt at a Solution

 

[ehrenfest]

anyone?

 

[Meir Achuz]

That sentence is a bit convoluted, but G is probably referring to bound surface charge density, which corresponds to infinite volume charge density.
If a discussion in GIII confuses you, go back to GI, which has less of that confusing help.

 

[neelakash]

Hello ehrenfest,it is an interesting question and I wonder if I myself understood this point while I read this section a year or two ago.

The problem is with the definition of \rho_b=\ - \ div \ P and \sigma_b=\ P.n.UNDERSTAND that P is determined by \sigma_b and \rho_b

Note that when we say there is a quantity \rho_b and another quantity \sigma_b,we generally do not care if there is any sharp boundary between the charge distributions.I mean we do not say upto 0.005 cm below the surface we would call \sigma_b and below that we would call as \rho_b

As Griffiths gives hint the charge density is expected to be gradually fade out toward the boundary.This makes \rho_b a continuous function and hence P is also continuous.POLARIZATION GRADUALLY TAPERS OFF TO ZERO.This exculdes the existence of \sigma_b

But if there were a sharp boundary like I mentioned, polarization P would be discontinuos across that boundary near (on) the surface and by taking the divergence you will get \rho_b=\infty.

I hope the problem is now clear to you.

 

[ehrenfest]

This exculdes the existence of \sigma_b

I see. So you are saying that a uniformly polarized object of finite size is unphysical.

 

[Ben Niehoff]

Not quite. What he's referring to is that any surface charge distribution can be written as a volume charge distribution by means of delta functions. For example, the surface charge density of a uniformly charged sphere can be written

\rho(r, \theta, \phi) = \frac{Q}{4\pi a^2}\delta(r - a)

Naturally, this function blows up on the surface itself, just as the charge density (and divergence of E) blows up at a point charge.

 

[neelakash]

Not quite. What he's referring to is that any surface charge distribution can be written as a volume charge distribution by means of delta functions.

I meant not exactly this,but like this.Refer to Griffiths problem 1.45 (b).The derivative of a step function is a delta function.Likewise If there is a distinguishable sharp boundary like between core and cover (jacket) of a body,rho_b must blow up.

In reality you might have a uniform polarized object.But do not be too obsessed with the distinction of rho_b and sigma_b---they are the different faces of the SAME distribution and is not different like core and cover (jacket).