How Do Charge Distributions Affect a Spherical Dielectric Shell?

fkf
Messages
9
Reaction score
0

Homework Statement


Consider a spherical dielectric shell so that ε = ε_0ε_r for a < r < b and ε = ε_0 for 0 < r < a. If a charge Q is placed at the center of the shell, find

a) P for a < r < b
b) ρ_pv for a < r b
c) ρ_ps at r = a and r = b

Homework Equations


ρ_pv = -div(P)
ρ_ps = P \cdota_n


The Attempt at a Solution


I've managed to solve a) with the answer
Q/(4*pi*r^2) * (ε_r-1)/ε_r​
which is correct (checked the answers).

I've also solved b) which is ρ_pv = 0 since ρ_pv = -div(P).

I have trouble to find ρ_ps at r = a and r = b. The answer states that
-Q/(4*pi*a^2) * (ε_r-1)/ε_r and -Q/(4*pi*b^2) * (ε_r-1)/ε_r​
respectively. Where is that negative sign coming from?
 
Physics news on Phys.org
On the r = a I would guess that we get a negative sign since we're having a unit vector a_r which goes outward from the origin. And since the normal to the surface when r = a is the -a_r I would guess that the negative sign when r = a is correct. But when r = b, the unit vector from P and a_r are booth positive, so why negative there?
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top