EMc2.pdf for electric and magnetic fields

electric field

1. Ch. 2 The electric potential
0
Path A = Path B an electrostatic field can be expressed by
The sum of electric fields induced by charges
Path independent
Path independent
16

2. 1 statvolt= 1 erg/esu= 300 volt
Example: Find the potential associated with the electric field
described in Fig. 2.3, the components of which are Ex = Ky,
Ey= Kx, Ez= 0, with K a constant.
0

3. Let , , be some continuous, differentiable function of the coordinates.
The direction of the vector ∇f at any point is the direction
in which one must move from that point to find the most
rapid increase in the function f .

5. 2.5 Potential of a charge distribution
Superposition must work for potentials as well as fields. If we have
several sources, the potential function is simply the sum of the potential
Functions. In the case of a discrete distribution of source charges,
where is the distance from the charge to the point (, , .
For a continuous distribution,
To see how this goes in
the
case of the infinitely long charged wire,
let us arbitrarily locate the reference
point 1 at a distance 1 from the wire.
Then to carry a charge from P1 to any
other point P2 at distance r2 requires
the work per unit charge,
20

6. The flat disk of radius in Fig. 2.9 carries a positive charge
spread over its surface with the constant density σ, in C/m2.
For we can approximate the potential as follows:
21

7. the energy of a system of discrete point charges
the work required to assemble a charge distribution , , and the
potential , , of that distribution:
22

8. Comparing this with the potential at the center of the disk, σa/2e0, we see
that, as we should expect, the potentialfalls off from the center to the
edge of the disk. The electric field, therefore, must have an outward
component in the plane of the disk. That is why we remarked earlier that
the charge, if free to move, would redistribute itself toward the rim.
23

9. 24
the potential at is
1, φ(r, θ) depends on q and l only through their product,
p ≡ ql.
2. φ(r, θ) is proportional to 1/r2, in contrast with the 1/r
dependence for a point-charge potential.
3. there is angular dependence in , , in contrast
with the point-charge potential.

10. A general charge distribution also has a
quadrupole term in the potential that goes
(where lis some length scale of the system), and
an octupole term that goes like .
the electric field,
E=−∇φ, associated
with the
dipole potential

11. the equation for the constant potential curves
is the radius associated with the angle θ =0.
The slope of a given curve at a given point,
relative to the local ˆr and ˆθ basis vectors at
that point, is dr/r dθ.

13. Now consider the E field. We will do things in reverse order here, first finding the slope
of the tangent, and then using that to find the equation of the field-line curves. The
slope of the tangent is immediately obtained from the Erand Eθcomponents

14. Consider a finite volume V of some shape, the surface of which we
shall denote by S. We are already familiar with the notion of the total
flux Femerging from S. It is the value of the surface integral of F
extended over the whole of S:
=
0
The meaning of div F can be expressed in this
way: div F is the flux out of , per unit of
volume, in the limit of infinitesimal .

15. For the electric field E, we have Gauss’s law,

16. Obviously, similar statements must apply to the other pairs of sides.
That is, the net flux out of the box is DxDzDy(∂Fy/∂y) through the
sides parallel to the x-z plane and DyDzDx(∂Fx/∂x) through the sides
parallel to the y-z plane. Thus the total flux out of the little box is

19. A stable position for a charged particle must be one where the potential φ is either
lower than that at all neighboring points (if the particle is positively charged) or higher
than that at all neighboring points (if the particle is negatively charged). Clearly neither
is possible for a function whose average value over a sphere is always equal to its value
at the center, according to Theorem 2.1.

20. Gauss’s law
divergence theorem
Integral vs. differential
(physics)
(mathematics)
MKS
cgs
How electric charge density, electric
potential, and electric field are related.

21. Let us consider the line integral of some vector fieldF(x, y, z), taken
around a closed path, some curve C that comes backto join itself.
circulation
The limitof the ratio of circulation to patch areais

22. =

23. The curlmeter: Exploring an electric field with this device, we
would find, wherever curlEis not zero, a tendency for the
wheel to turn around the shaft. With a spring to restrain
rotation, the amount of twist could be used to indicate the
torque, which would be proportional to the component
of the vector curlEin the direction of the shaft. If we can find
the direction of the shaft for which the torque is maximum
and clockwise, that is the direction of the vector curl E.
In an electrostatic field E, the curlmeterwill always read
zero! That follows from a fact we have already learned;
namely, in the electrostatic field the line integral of E around
any closed path is zero.
If E equals the negative gradient of a potential function φ
(which is the case for any electrostatic field E), then

26. 41

27. Triboelectric effect
Neural signal
transduction

28. The Hodgkin-Huxley Modeldt
tdV
CtItI
mionicm
)(
)()( 
The total membrane currentlea kKNaio n ic
IIII  KNa, ,
][
][
log
5.61
 i
S
S
z
E
in
ex
i Ionic reversal potential due to electric force])()[),((
iii
EtVttVGI 
Hodgkin and Huxley assume that])([
4
KKK
EtVnGI  ])([
3
NaNaNa EtVhmGI 
n describes the fraction of activated potassium channels
m describes the fraction of activated sodium channels
h describes the fraction of inactivated sodium channels

29. For a two-state modelnn
n
n
1

 n
nn
nn
nVnV
dt
dn





)()1)((
These rate constants are approximated by80/
10/)10(
125.0)(
]1[100
10
)(
V
n
Vn
eV
e
V
V







 )(/
0)()(
Vt
n
ennntn

  n

V (mV) V (mV)
activation/inactivation
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