Most of chapter 1 will be covered as needed, but not all at the beginning

Binomial series (needed on Friday or Monday)
dot product (needed for Thurs HW)
 

surface area element in each one

volume element in each one

general displacement dr in each one

Collect student class work for the above

• F12 = k q1 q2 / r^2, where k = 9 x 10^9 Nm^2/C^2 = 1/(4 pi epsilon_o)
• E(at 2 due to 1) =F12/q2 = k q1/r^2
• E due to a point charge q = kq/r^2
• E due to distinct charges - sum over terms, take account of x,y, z components
• collect in-class calculation

• dE due to an infinitesimal charge dq  = k dq/r^2
• dq could be lambda dx for a 1-D charge distribution
• dq could be sigma da for a charge distributed over an area
• dq could be rho dv for a volume distribution of charge
• E due to line of charge - set up an integral in one variable (students in class)
• E due to uniform surface layer of charge/area
• uniform charge density on a disc, find E on axis - set up in class
• discuss limit as z ->0 so point approaches center of the disc, and z-> infinity
• HW - E on axis of a spherical cap of radius R and angle theta0, at a distance h from the center of the cap on the axis
• E on axis of a cylinder of uniform charge/volume - set up an integral in two variables ( collected)
• Collect student class work for the integral for E on axis of a cylinder of charge


 

 

Later (see below) we will do the electric potential on the axis of a cylindrical shell of charge

Homework due today

• Prob 1.3 (dot product) p. 7
• Prob 1.4 (cross product) p. 7
• Prob 2.6 (electric field on axis of a uniformly charged disk, started in class) p. 64

In the text, Griffiths uses a script r [p. 60 and other places].
The sketch below indicates I will be using s instead of script r.

• gradient of a scalar function
• dr = r^ dr + theta^ r d theta +phi^ r sin theta d phi (spherical polar)
• in class: use df = grad f . dr to work out the gradient in xyz.
• start from df = grad f . dr and work out the gradient operator in cylindrical and spherical coordinates [collect in class 1/2]
• in class, work out d/dx (1/s) [see p. 60, where his script r is my s]. [collect in class 1/2]
• note that  grad (1/s) = -s/s^3

Formulate the integral for E in terms of the gradient of a scalar function, the electric potential V. One then calculates V and takes the negative gradient to obtain the electric field.

• constant number of lines of force leave each coulomb of charge. 4 pi k per coulomb
• E = lines per unit area passing through the surface.
• lines sliding along the surface don't count
• method is simpler than Gauss's law, but very same underlying basis where symmetry exists
• calculate the approximate electric field at 2.3 m from the axis of a uniformly-charged right circular cylinder (at equal distances from either end of the cylinder) of radius 1.0 m and length 380 m. The charge density in the cylinder is 4.7 x 10-9 C/m^3. [collect in class]

• prob 1.13, p. 15 gradient of r^n
• E on axis of a charged disc of radius R
• show all steps in integration, with a good skecth
• show what happens as z->0. Identify the answer with a familiar result.
• show what happens as z-> infinity. E->0 is unsatisfactory. Show how a familiar result is obtained
• E on axis of a cylinder of uniform charge density rho, length L, radius R.
• Let z->0 and show that a proper result is obtained
• Let z-> infinity and show that the proper result is obtained.
• The following two problems are postponed until a later assignment
• prob 1.20 p. 22 product rules for grad
• probs 1.26a and 1.27 p. 27

Friday September 8 and Monday September 11

Laplace's equation: laplacian of V = - rho/epsilon

Electric conductors

• net charge all at the surface
• E inside a static solid conductor is zero
• charges free to move in a conductor
• charges are not moving, and they would move if E<>0
• E must be perpendicular to surface
• charges are free to move along the surface
• charges are at rest
• charges would move along the surface if E(tangent to surface)<>0
• relation between sigma and E at the surface
• E inside =0, so E exists only outside the surface
• Children's Faraday view: E = lines/area = (4 pi k sigma da)/da = 4 pi sigma
• Adult view: use gaussian surface to show 4 pi sigma =Esurface
• E due to an infinite sheet of charge, chg/area = sigma
• there is electric flux above and below the sheet
• Faraday: E = lines/area = 1/2 (4 pi k sigma da)/da = 2 pi sigma
• Gauss: calculate flux out both top and bottom
• force on a charge at the surface depends on Esurface/2
• remove a tiny area A
• tiny area A acts like an infinite sheet when one is super-close to it
• with tiny area in place, E inside =0, just beneath center of tiny area
• E(tiny area) = 2 pi k sigma
• therefore E due to the rest of the conductor surface must also equal 2 pi k sigma
• and so, the force on the tiny area is its charge times E from the rest of the conductor

Homework Due (date not yet set)

• Calculate E on the axis of a cylinder of length L and radius R, uniform charge density rho, at a distance z>L/2 from the center of the cylinder. This must be set up as an integral in two variables and the result presented in terms of rho, k, L, R, and z.
• There is a spherical surface of radius R and polar angle alpha, having a uniform charge density sigma. Set up an integral in theta (the angle from the polar axis) for the electric field on the axis of this spherical 'cap', at a distance h>R from the center of the cap.

The first homework problem above has you calculate the electric potential on the axis of a cylindrical shell of charge. This is the basis a primitive model of a charged wire, based on 6 cylindrical elements, only 3 of which are independent. The six sections all have the same potential, so you can write down 3 equations in 3 unknowns to determine charge density on each element. Get Maple or something to solve this. You should definitely find that the charge density is higher at the ends than in the middle.

At some point in the quarter we will go into the lab and check on the charge distribution on a long charged conducting rod.

Laplace's equation

• derive it
• taylor series leads to iterative solution
• simple, simple spreadsheet of a thin sheet of conductor lets us look at charge near the edge and in the middle

The line of 1000 v values is supposed to represent a line (or more properly a sheet coming out of the page) of conductor. The iterative solution to laplace's equation is shown surrounding this sheet (in the middle of a grid which runs about 50 cells horizontally and about 100 cells vertically.

Estimate the ratio of the surface electric field perpendicular to the sheet at the end to the surface electric field perpendicular to the sheet in the middle. Take the length and width of each cell to be 0.05 m.

Estimate the ratio of the charge density in the middle of the sheet to the charge density at the end of the sheet.
 

Separating the variables in Laplace's equation. In 2-d, one gets periodic and exponential solutions. A series solution is often called for, with coefficients required.

Homework for week of September 25

• You were to work out with laptops in class example 3.3 starting on p. 127. This amounted to using the coeffficients, and obtaining a graph like Fig 3.19 and one like 3.18
• You were also to work out the coefficients for a problem where the potential V in the y-direction went from 0 to 1 and back down to zero while y was going from 0 to pi.
• These were due by the end of the week (or maybe the middle of the week)
• I'll still take these for a while longer.

Image charges

• Image of a charge above a grounded conducting plane is as far below the plane as the charge is above it, in a mirror image position. The image is of opposite sign to the charge above the plane.
• this satisfies the condition that E at the plane is perpendicular to the surface
• this satisfies the condition that V=0 everywhere on the surface of the plane
• the potential and field on the 'real' side of the plane can be calculated using both the real charge and the image
• the surface charge distribution on the plane may be calculated from E due to both charges
• Image of a charge near a grounded conducting sphere
• the image location is such that R^2 = ab, where a is the distance of the charge and b is the distance of the image from the center of the sphere. R is the sphere radius
• the image is of opposite sign and smaller in magnitude than the original charge, in the ratio of two distances
• for a charge near a conducting sphere, one additional image charge is needed, at the center of the sphere

Electric dipoles and the multipole expansion

• an electric dipole has +q and -q separated by a distance d. d points from - to + charge.
• electric dipole moment = p = qd.
• V at a displacement r from the dipole is k (p dot r)/r^3, as shown on p. 146 (and in class)
• This makes sense, since only two vectors are around, p and r, and V is a scalar.
• We worked out in class that E = -kp/r^3 -k(p dot r)r/r^5 [this is problem 3.33, p. 155]
• The multipole expansion is done in section 3.4 starting on p. 146.
• It could be written as V0+V1+V2, where
• V0 is the total charge,
• V1 is the potential due to the electric dipole moment of the charge distribution
• V2 is the potential due to the electric quadrupole moment of the charge distribution
• etc.
• The expansion rests on 1/sqrt(1+x^2-2x cos(theta)) = sum on nof x^n P_n(theta), where P_n is the legendre polynomial of order n.

Homework problems due Tuesday October 3, 2000

• (Very easy) use Maple to show that 1/sqrt(1+x^2-2x cos(theta)) = sum on nof x^n P_n(theta), by simply asking for an expansion in x. You must compare the coefficients with those in the book and demonstrate they are the same as the P_n.
• Problem 4.8 p. 165 (not very hard either)
• Quadrupole moment problem (looks messy but is not messy or horrible)
• a) Show that the quadrupole moment can be written as the integral of (2z^2 -x^2 -y^2)/2 over the charge distribution. [see P_2(theta) and note z=r cos theta]
• for an ellipsoidal distribution of charge, rho is constant over a bounding surface of (x/a)^2+(y/b)^2+(z/c)^2 = 1. Let a=b for the case we wish to consider
• b) show that the total charge is rho times the volume of the ellipsoid, which is 4/3 pi a^2b
• change variables to x' = x/a, y' = y/b, z'=z/c. This changes the bounding surface to x'^2+y'^2+z'^2 = 1, a sphere of unit radius. the volume element changes too, such that dx=a dx', etc. Then one only integrates over the volume of the unit sphere
• c) Show that the quadrupole moment is given by QM = rho 4 pi a^2 b (c^2-a^2)/15.
• again change variable as done above. You will be left with things like the integral of x^2 over the unit sphere. But for the unit sphere the integral of x^2 is 1/3 of the integral for r^2 because the integrals of x^2 and y^2 and z^2 over the unit sphere are all the same, and r^2 = x^2+y^2+z^2

Homework due on Monday October 9, 2000

• Problem 3.8 (to do with image charges and a sphere) p. 126.
• A 1.5 microcoulomb charge is 0.085 m from a grounded, conducting infinite plane. Directly below the charge is a hemispherical conducting bulge, part of the grounded conducting plane. Determine the force exerted on the 1.5 microcoulomb plane if the radius of the bulge is 0.050 m. [This requires placing the correct number of image charges, then figuring out the force.]
• Problem 4.7 in Griffiths, p. 165. [Try bringing the charges in from infinity one at a time. The work to bring in either with a uniform field present is infinite, but the net amount of work is finite.]

Homework due on Thursday October 19, 2000

• Problem 4.18 p. 184
• Problem 5.11 p. 220
• Problem 5.46 p. 249

Homework due on Monday October 23, 2000

• 5.25 p. 239
• 6.1 p. 259

Homework due on Tuesday November 7, 2000

• problem 9.11, p. 382
• maple solution to non-reflective coating

Homework due Friday November 10, 2000 Due by end of class Friday, no later.

• Plot like Fig 9.17, one of these for each polarization direction, (each counts as one problem)
• Plot for TE
• Plot for TM
• Work out the condition for odd TE modes in a slab waveguide of thickness 2a.
• Determine the lowest two TE modes for a slab waveguide 10 microns thick at a vacuum wavelength of 0.5 microns . The indices are 1.55 inside and 1.3 outside