Download Introduction to Electrostatics

Introduction to Electrostatics
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Reading: Jackson 1.1 through 1.9, 1.11
Griffiths Ch 1 and Appendix A
Brief review of vector calculus
Consider a scalar field
The gradient
tells us how F varies on small displacements
=> ∇F points in dir. of max increase of F; |∇F| = rate of increase of F along this dir.
In Cartesian coords:
(using chain rule)
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Can think of ∇ as a differential operator:
Now consider a vector field
The divergence
is the flux of
emerging from an infinitesimal
volume, per unit volume:
dz
dx
Flux out of x faces:
Similarly for y, z => flux out of cube
dy
Dividing a finite region into infinitesimal blocks and noting pair­wise
cancellation in interior 3
=> Divergence Theorem:
The curl
tells us about the circulation of the vector field:
(x, y+dy)
(x+dx, y+dy)
z
(x,y)
Line integral:
(x+dx, y)
The x,y,z­component
is the line
of
integral of the vector field around an infinitesimal closed loop, ⊥ to x,y,z, per unit area
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is the z­component of
Similarly for x, y­components
Dividing a finite surface into infinitesimal squares and noting pair­wise
cancellation in interior => Stokes's Theorem:
Now consider curvilinear coord systems (still in 3D Euclidean space,
orthonormal at each point).
Line element
(For those who took relativity: f, g, h =
( f, g, h are called scale factors) 5
For example, spherical coords
Again, for a scalar field
Also, So, for spherical coords (f = 1, g = r, h = r sin ):
SP 1.1
Divergence: Flux out of u faces:
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u
u+du
Similarly for v, w =>
So, for spherical coords (f = 1, g = r, h = r sin ):
Curl
(u, v+dv)
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(u+du, v+dv)
w
(u,v)
(u+du, v)
=> w­component of
Similarly for u, v­components =>
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So, for spherical coords (f = 1, g = r, h = r sin ):
See front and back covers of Jackson for vector identities, theorems, and
derivatives in curvilinear coords.
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Problem 1: Prove the following variant of Stokes's Theorem:
= unit normal vector
Stokes's Thm:
0
Must be true for arbitrary R => the integrals are equal.
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Now, we focus on electrostatics:
Coulomb's Law:
(force on 1 by 2)
We adopt SI units; see Jackson appendix for conversion between SI and
Gaussian.
Electric field:
(field due to a point charge q1 at
)
(field due to a continuous charge distribution)
Point charges can be treated as a distribution using the Dirac delta function.
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In 1D:
if the region of integration (x1, x2)
includes x = a; zero otherwise
Specifically:
if region includes x = a
Clearly (x­a) is undefined at x=a, so it really is not a function.
We can try to express it as a limit, e.g.:
But
does not exist.
This limit does exist!
Thus, (x) makes sense when it appears as part of an integrand—this is the only context in which it should be used.
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With
for a collection of point charges,
Two expressions involving delta functions, D1(x) and D2(x), are equal if:
for all well­behaved functions f (x).
Problem 2: Show that
when k is any non­zero constant.
For arbitrary f (x) :
Dirac delta function in curvilinear coords:
assuming
is not a degenerate point, i.e., it is not characterized by more than one set of coord values.
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Examples of degenerate points: origin in plane polar coords (multiple )
z­axis in spherical and cylindrical coords
(again, multiple )
At a point with multiple values of coord w :
Examples
Cylindrical coords: point
Not on z­axis:
On z­axis:
Check:
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Spherical coords: point
Not on z­axis:
using cos  rather than  :
On positive z­axis:
using cos  rather than  :
The origin (degenerate in both  and ):
Note: dimension of delta function = inverse of dimension of argument.
Dimension of
is length­3
(r) is length­1
() is rad­1 (dimensionless)
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Examples involving delta function in curvilinear coords
Find the charge density
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for the following situations:
1) Spherical coords, charge Q uniformly distributed over a spherical shell with radius R:
2) Cylindrical coords, charge per unit length  uniformly distributed over a cylindrical surface of radius b:
3) Cylindrical coords, charge Q spread uniformly over a flat circular disk of negligible thickness and radius R:
4) Same as (3), but in spherical coords:
Check:
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Coulomb's Law plus linear superposition of electric fields yields integral
form of Gauss's Law:
Divergence Thm yields the differential form:
Coulomb's Law also yields
=> existence of scalar potential:
Work done on moving a charge against the field:
Potential energy of a charge q = q 
SP 1.2
Potential energy of a collection of point charges
Start with one point charge q1, located at
; its potential
Bring next charge, q2, in from infinity to point
Work done against field of 1 is
Bring in a 3rd charge, q3, from infinity to
Work done = etc.
=> potential energy = total work done
For a continuous charge dist :
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Vector identity:
Taking the bounding surface at infinity, the first term vanishes
(E ∝ r­2 ,  ∝ r­1 ).
=> energy density in the field
Red flag: Field energy density is non­negative, but energy of a set of
point charges can be negative (e.g., 2 charges of opposite sign).
What's going on?
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In the integral form of W,
is the total potential at , due to the rest
of the charge dist plus the charge at . For a truly continuous charge dist,
this is the same as the  due to the rest of the dist, since the amount of charge at a point vanishes. For a point charge,
at the location of the point charge => an infinite "self­energy" that was not included in the energy of the point charge collection (i.e., the charges were taken as already assembled).
Note: infinite self­energy => infinite mass of a point particle unless
a negative infinite mass contribution arises from non­electromagnetic
source (“renormalization”).
What happens to
l
at a charged surface?
Consider a small rectangular surface ⊥ to the charged surface:
as h  0
h
E∥, 2
E∥, 1
=> tangential component of
is continuous across the surface

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Gaussian pillbox:

In a conductor, free charges move in response to an applied
Charge flows until
inside conductor (
of induced charges
cancels applied field).
=> any net and induced charge resides on the surface
 = const in a conductor, since
External
is perpendicular to conductor's surface (since tangential component of
is continuous across surface and
= 0 inside);
Suppose we have a conductor held at fixed potential 0. We would like to find  everywhere outside the conductor. If we knew how charge distributed
itself on the surface of the conductor, we could use
But, can we find
without knowing
Yes! We'll find a differential eqn for
and apply boundary conditions.
For regions where  = 0,
Let's verify that
satisfies the Poisson eqn.
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= 0 everywhere except r = 0, where r­2 is undefined.
Treatment of the boundary value problem is aided by use of Green's
Identities.
Start with the Divergence Thm:
= normal derivative on S, directed
outward from within V)
(Green's First Identity)
(interchanging  and )
Subtracting eqns =>
(Green's Theorem)
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;
(1.36)
Note:
1. For surface at , this reduces to our original result for
(assuming
falls off faster than R­1).
2. If  = 0 in V,  anywhere in V depends only on  and
on S.
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Green's 1st Identity leads to powerful conclusions re. the uniqueness of solns to electrostatic boundary value problems. Specifically, problems with Dirichlet or Neumann boundary conditions have unique solns to the Poisson
eqn.
Dirichlet:  is specified everywhere on the bounding surface.
Neumann:
is specified everywhere on bounding surface.
Proof: Suppose there are 2 different solns, 1 and 2
In V:
On S: Dirichlet:
Neumann:
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st
Green's 1 Identity with  =  = U:
0
product = 0 for both Dirichlet and Neumann
=>  is unique to within an additive constant.
Suppose the boundary of V consists of the surface of a set of conductors
plus a sphere at ∞.
Dirichlet:  is specified on each conductor (e.g., by connecting a
battery btwn the conductor and ground)
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Neumann:
is specified on each conductor.
We don't know !
Suppose we know total charge on each conductor. Is soln unique? In this case, for each conductor surface,
The potentials can be brought out of the integrals since they are constant
on each conductor surface, and the resulting integral = Q by Gauss's Law.
So, again, U = const =>  is unique to within an additive constant.
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Thus, the soln to Poisson's eqn is unique for a region bounded by conductors with either specified potentials or specified charges. (A boundary at infinity
is also acceptable.)
Consider a set of conductors. Conductor 1 has Q = 1 C and the rest have Q = 0.
2
Q = 0
1
Q = 1 C
3
Q = 0
This has a unique soln for the potential,
The surface charge dist,
is also uniquely determined. Now alter the charge on conductor 1 from 1 C to  C.
Do the charge dists change?
Suppose the dists are all multiplied by :
Q remains 0 on 2 and 3.
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Since the Poisson eqn is linear, a new potential
is a solution.
Since the charge on each conductor is specified, the soln is unique.
So: NO—the charge dists don't change.
If all conductors but conductor 1 have Q = 0, then  ∝ Q1.
Since the Poisson eqn is linear, the sum of 2 solns is a soln.
In particular, we can add the
obtained for the case where
(1) all but conductor 1 have Q = 0 and (2) all but conductor 2 have Q = 0.
Again, the soln is unique for these boundary conditions.
Thus, the potential at the surface of the ith conductor, Vi , is given by
The pij are called coefficients of capacity and depend only on the geometry of the conductors.
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Inverting the eqns for Vi in terms of Qj :
Cii are called capacitances. Cij, i ≠ j, are called coeffs of induction.
The capacitance C of 2 conductors with Q2 = ­Q1 and potential difference
V is defined by
The potential energy for the system of conductors is
As a special case, for a capacitor
SP 1.3—1.9