Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Electrostatics: Bound Charges in Dielectrics, Study notes of Physics

An explanation of bound charges in dielectric materials, integrating the effect of all the dipoles using polarization p. Gauss's law is used to recognize free and bound charges, and the relationship between the electric field, polarization, and dielectric constant is discussed.

Typology: Study notes

Pre 2010

Uploaded on 08/18/2009

koofers-user-jun
koofers-user-jun 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
PH 316 Bound Charge in Dielectrics 10/6/05 MJM (worksheet on 2nd page)
This is intended as a brief summary of some of what we have been up to.
Vdipole = kp cos /r2, or V dipole = k (pr)/r3 .
In a dielectric material, we integrate to get the effect of all the dipoles. We use the polarization P (the
dipole moment per unit volume, so that a small batch of dipoles is represented by P .
The integral starts out as
V(r) = k d P(r-r')/|r-r'|r-r'|r-r'|3 .
We showed that ' 1/|r-r'|r-r'|r-r'| = (r-r')/|r-r'|r-r'|r-r'|3 . ( ' is the gradient with respect to r'.).
This gave
V(r) = k d P[ ' 1/|r-r'|r-r'|r-r'|] => => k da bound /|r-r'|r-r'|r-r'| + k d bound/|r-r'|r-r'|r-r'|.
where b = Pn^, and b = - 'P . (some vector identities were used)
From gauss's law, we had o E = , and now we recognize both free charge and bound charge. Free
charge could exist on conductors, or it might exist as net charge within a dielectric, where teeny charges
got put into molten dielctric, then hardened. Bound charge is the ends of dipoles, either sticking out the
surface of a dielectric, or as a net charge (a batch of dipole ends where P is spreading out (diverging). So
o E = f + b = f + ( -P) .
Then (oE + P) = f and then D = f, where D = oE + P .
In a linear dielectric P = oE, and then
D = o (1+)E = E = r o E ,
where is the dielectric constant, is the electric susceptibility (I'm omitting the subscript for now)
and r is the 'relative' dielectric constant, the ration of to o .
Applying the divergence theorem to a tiny pillbox, we showed that D is continuous across a boundary
between dielectrics (there must be no free charge right at the boundary), and from curl E = 0, and Stokes
theorem, that E|r-r'||r-r'| is continuous. (It turns out that the potential V is also continuous across a boundary.
Why isn't E continuous? From o E = f + b and a tiny pillbox, we find that E would jump at the
boundary if there was bound charge present at the boundary: E 2 - E 1 = b/o, even if no free charge
was present.
(Next page for a short worksheet).
Name___________________________________ Box __________
1
pf2

Partial preview of the text

Download Electrostatics: Bound Charges in Dielectrics and more Study notes Physics in PDF only on Docsity!

PH 316 Bound Charge in Dielectrics 10/6/05 MJM ( worksheet on 2nd page ) This is intended as a brief summary of some of what we have been up to. Vdipole = kp cos /r^2 , or V (^) dipole = k ( pr )/r^3. In a dielectric material, we integrate to get the effect of all the dipoles. We use the polarization P (the dipole moment per unit volume, so that a small batch of dipoles is represented by P . The integral starts out as V(r) = k  d P( r-r' )/|r-r'|r-r'|r-r'|^3. We showed that  ' 1/|r-r'|r-r'|r-r'| = ( r-r' )/|r-r'| r-r' |r-r'|^3. (  ' is the gradient with respect to r'.). This gave V(r) = k  d P[  ' 1/|r-r'|r-r'|r-r'|] => => k  da bound /|r-r'| r-r' |r-r'| + k  d bound/|r-r'| r-r' |r-r'|. where b = Pn^ , and b = -  'P. (some vector identities were used) From gauss's law, we had o  E = , and now we recognize both free charge and bound charge. Free charge could exist on conductors, or it might exist as net charge within a dielectric, where teeny charges got put into molten dielctric, then hardened. Bound charge is the ends of dipoles, either sticking out the surface of a dielectric, or as a net charge (a batch of dipole ends where P is spreading out (diverging). So o  E = f + b = f + ( -  P ). Then (o E + P ) = f and then  D = f, where D = o E + P. In a linear dielectric P = oE, and then D = o (1+) E =  E = r o E , where  is the dielectric constant,  is the electric susceptibility (I'm omitting the subscript for now) and r is the 'relative' dielectric constant, the ration of  to o. Applying the divergence theorem to a tiny pillbox, we showed that D is continuous across a boundary between dielectrics (there must be no free charge right at the boundary), and from curl E = 0, and Stokes theorem, that E|r-r'||r-r'| is continuous. (It turns out that the potential V is also continuous across a boundary. Why isn't E continuous? From o  E = f + b and a tiny pillbox, we find that E would jump at the boundary if there was bound charge present at the boundary: E 2 - E 1 = b/o, even if no free charge was present. (Next page for a short worksheet). Name___________________________________ Box __________ 1

A sphere of radius R carries a polarization. P = C r , where C is a constant. [ This is basically Prob. 4.10, p. 169 ] a) Calculate the bound charges b and b. b) Find the electric field inside the sphere (think Gauss's law) c) Find the electric field outside the sphere 2