Confirming the dimension of induced charge density of a dielectric

patric44
Messages
308
Reaction score
40
Homework Statement
confirm the dimension of induced charge density of a dielectric ρ and σ
Relevant Equations
ρ = -1/4πk E.grad(k)
hi guys
our professor asked us to confirm the units of volume charge density ρ and also the surface charge density σ of a dielectric material given by
$$
\rho = \frac{-1}{4\pi k} \vec{E}\cdot\;grad(k)
$$
$$
\sigma= \frac{-(k-1)}{4\pi} \vec{E_{1}}\cdot\;\vec{n}
$$
I am somehow confused about the units, shouldn't the gradiant of k (the dielectric constant ) be dimensionless.
but that will leave ρ as the same units of E, which is not true as ρ =C/m^3.
can someone clarify
 
Physics news on Phys.org
##k## is dimensionless, but grad(##k##) is not.

Is your professor using Gaussian units?
 
thanks, by careful looking at the gradient operator i can see that it has a unit of 1/length
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...

Similar threads

Back
Top