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Introduction to Electrostatics
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:
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(using chain rule)
Can think of ∇ as a differential operator:
Now consider a vector field
The divergence is the flux ofemerging from an infinitesimalvolume, per unit volume:
dxdy
dz
Flux out of x faces:
Similarly for y, z => flux out of cube
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Dividing a finite region into infinitesimal blocks and noting pairwisecancellation in interior
=> Divergence Theorem:
The curl tells us about the circulation of the vector field:
(x,y) (x+dx, y)
(x+dx, y+dy)(x, y+dy)
z
Line integral:
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is the lineintegral of the vector field around an infinitesimal closed loop, ⊥ to x,y,z, per unit area
The x,y,zcomponentof
is the zcomponent of
Similarly for x, ycomponents
Dividing a finite surface into infinitesimal squares and noting pairwisecancellation 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 =
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( f, g, h are called scale factors)
For example, spherical coords
Again, for a scalar field
Also,
So, for spherical coords (f = 1, g = r, h = r sin ):
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SP 1.1
Divergence: Flux out of u faces: u u+du
Similarly for v, w =>
So, for spherical coords (f = 1, g = r, h = r sin ):
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Curl
(u,v)
(u, v+dv) (u+du, v+dv)
(u+du, v)
w
=> 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, andderivatives in curvilinear coords.
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Problem 1: Prove the following variant of Stokes's Theorem:
Stokes's Thm:
0
Must be true for arbitrary R => the integrals are equal.
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= unit normal vector
Now, we focus on electrostatics:
Coulomb's Law: (force on 1 by 2)
We adopt SI units; see Jackson appendix for conversion between SI andGaussian.
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, x
2)
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 D
2(x), are equal if:
for all wellbehaved functions f (x).
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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 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:
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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 integralform 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
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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 ofpoint 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 restof 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 unlessa negative infinite mass contribution arises from nonelectromagneticsource (“renormalization”).
What happens to at a charged surface?
Consider a small rectangular surface ⊥ to the charged surface:
l
hE
∥, 1
E∥, 2
as h 0 => tangential component ofis 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);
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Suppose we have a conductor held at fixed potential 0. We would like to
find everywhere outside the conductor. If we knew how charge distributeditself 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.
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Treatment of the boundary value problem is aided by use of Green'sIdentities.
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 Poissoneqn.
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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Green's 1st 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 conductorsplus 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 constanton 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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Consider a set of conductors. Conductor 1 has Q = 1 C and the rest have Q = 0.
1 2
3Q = 1 CQ = 0
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.
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 infinityis also acceptable.)
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Since the Poisson eqn is linear, a new potentialis 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 Q
j :
Cii are called capacitances. C
ij, i ≠ j, are called coeffs of induction.
The capacitance C of 2 conductors with Q2 = Q
1 and potential difference
V is defined by
The potential energy for the system of conductors is
As a special case, for a capacitor
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SP 1.3—1.9