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Phys. 505 Electricity and Magnetism Fall 2003 Prof. G. Raithel Problem Set 1
Problem 1.3 5 Points
a): By symmetry, the solution must be of the form ρ(x) = ρ(r) = Qδ(r − R)f , with a constant f to be specified by the condition
Q =
ρ(r)4πr^2 dr =
Qδ(r − R)f 4 πr^2 dr = Qf 4 πR^2. (1)
Thus, ρ(r) = Qδ 4 (πRr− 2 R ).
b): By symmetry, the solution must be of the form ρ(x) = ρ(r) = λδ(r − b)f , with a constant f to be specified by the condition
λ =
ρ(r)2πrdr =
λδ(r − b)f 2 πrdr = 2λf πb. (2)
Thus, ρ(r) = λδ( 2 rπb− b).
c): By symmetry, the solution must be of the form ρ(x) = ρ(r, z) = δ(z)Θ(R − r)f (r). There, Θ is the step function, and f (r) a function specified by a normalization condition that describes how much charge is supposed to be on a ring with radius r and radial thickness dr (for r < R):
Q (^2) πRπrdr 2 =
z
ρ(r, z)2πrdzdr =
z
δ(z)f (r)2πrdzdr = f (r)2πrdr. (3)
Thus, ρ(r, z) = Qδ(z)Θ( πR 2 R −r).
d): By symmetry, the solution must be of the form ρ(x) = ρ(r, θ) = δ(cos θ)Θ(R − r)f (r), where f (r) is a function specified by a normalization condition that describes how much charge is supposed to be on a shell with radius r and radial thickness dr (for r < R):
Q (^2) πRπrdr 2 =
cos θ,φ
ρ(r, θ)r^2 dφd cos θ dr = 2π
cos θ
δ(cos θ)f (r)r^2 d cos θ dr = 2πf (r)r^2 dr. (4)
Thus, ρ(r, cos θ) = Qδ(cos πR^ θ)Θ( (^2) r R−r).
Problem 1.5 5 Points
Method 1: Consider first the case r > 0. Then, the expression
ρ−(r) = −≤ 0 ∆Φ(r) = − 4 qπ ∆ exp(− r αr)(1 + αr 2 ) (5)
is well defined, and yields
ρ−(r) = − 4 qπ^1 r ∂ r^2 exp(−αr)(1 + αr 2 ) = − qα
3 8 π exp(−αr)^.^ (6)
For r → 0, it is Φ(r) → (^4) π≤q 0 r and ρ+(r) = − 4 qπ ∆ (^1) r = qδ^3 (r) = q 4 δπr(r) 2.
Obviously, ρ−(r) accounts for the electron and ρ+(r) for the proton, and the total charge density is the sum
ρ = q(δ^3 (r) − α 8 π^3 exp(−αr)) = 8 qπ ( 2 δ r( 2 r )− α^3 exp(−αr)).
Method 2: Write exp(− r αr)(1 + αr 2 ) as a product of functions f (r) = (^1) r and g(r) = exp(−αr)(1 + αr 2 ), and use the product rule ∆(f g) = f ∆g + g∆f + 2 ∇f · ∇g. All derivatives are found to be well behaved at r = 0, except ∆f , which equals − 4 πδ^3 (r). Considering that, the calculation confirms the result of method 1.
You may verify that the total charge
ρ(r)d^3 r = 0, which must be the case.
Problem 1.11 5 Points
R
x
R
y
dx
dy’
dy
dz
x
y
z
E-Field
A
C
B
A’ B’
C’
dx’
Problem 1.11, Method 1 (enlightning but a little cumbersome):
The point of interest is located at the origin of a cartesian coordinate system the xy-plane of which defines the tangent plane to the curved surface (at the point of interest). Further, the directions of the x- and y-axes are such that locally the curved surface can be described as the set of points with coordinates S(x, y) = xˆx + yyˆ − ( (^2) R^1 x x^2 + (^2) R^1 y y^2 )ˆz. It has been discussed in class that it is always possible to find orthogonal coordinate axes on the tangent plane that allow such a parametrization of the curved surface. For the situation depicted in the figure, it would be Rx > 0 and Ry > 0. Due to the orthogonality of ˆx and yˆ, a surface element with side lengths dx and dy on the curved surface has the area da = dx dy.
At a location S(x, y) on the curved surface, a normal vector to the plane is ∂ ∂xS × ∂ ∂yS = (^) Rxx xˆ + (^) Ryy yˆ + ˆz. Using this fact, it can be seen that an area element da′^ = dx′^ dy′^ located at a height dz above da, constructed as shown in the figure, will have the property that the “sidewalls” of a pillbox with da′^ and da as upper and lower surfaces are orthogonal to the curved surface. Since the electric field, depicted in red in the figure, also is orthogonal to the surface, the electric flux through the sidewalls is zero. Due to Gauss’s law, the fluxes through da and da′^ must be equal and opposite. Considering that the field is (anti)parallel to both vectors da and da′, it follows that E′da′^ = Eda, with E and E′^ being the field magnitudes on da and da′, respectively.
Simple geometry apparent in the figure yields dx′^ = dx(1+ (^) Rdzx ) and dy′^ = dy(1+ (^) Rdzy ). Thus, for infinitesimally small dx, dy and dz it is da′^ = (1 + (^) Rdzx + (^) Rdzy )da, and
E′^ = E dada′ = E(1 + (^) Rdz x
)−^1 = E(1 − (^) Rdz x
− (^) Rdz y
Thus,
E
E′^ − E
dz =
E
∂E
∂n =^ −
Rx^ +^
Ry
q.e.d. (10)
Note: for the case depicted in the figure, both principal radii are > 0, and consequently ∂E∂n < 0, as expected.
Problem 1.11, Method 3: (fast, less enlightning):
Using the coordinate system in the figure, at location S(x, y) on the curved surface the normal vector to the plane is ∂ ∂xS × ∂ ∂yS = (^) Rxx ˆx + (^) Ryy ˆy + ˆz. Since we are only interested in the local behavior in the vicinity of the origin, x ø Rx and y ø Ry , and the length of the normal vector is 1, with corrections of order (x/Rx)^2 and (y/Ry )^2. Thus, the electric field on the surface has, locally, a form E(x, y) = E
( (^) x Rx ˆx^ +^
y Ry ˆy^ + ˆz
with locally constant magnitude E. At the origin, we therefore have
∂Ex ∂x =^
E
Rx^ and^
∂Ey ∂y =^
E
Ry^.^ (13)
Noting that the field is normal to the curved surface, it also is
∂Ez ∂z =^
∂E
∂z =^
∂E
∂n (14)
From ∇ · E = 0 it then follows
∂Ez ∂z =^
∂E
∂n =^ −
( ∂E
x ∂x +^
∂Ey ∂y
= −E
Rx^ +^
Ry
and
E
∂E
∂n =^ −
Rx^ +^
Ry
q.e.d. (16)
Problem 1.12 5 Points
We consider a volume V limited by a conducting surface ∂V. For volume and surface charge densities ρ(x) and σ(x) the potential is Φ(x), while for ρ′(x) and σ′(x) it is Φ′(x).
It is ρ(x) = −≤ 0 ∆Φ(x), and ∂ ∂nΦ = (^) ≤σ 0 (with ˆn pointing from the volume into the conductor). Corresponding equations for the primed quantities apply. The reciprocation theorem follows from these facts and Green’s 2nd identity:
V
ρΦ′d^3 x +
∂V
σΦ′da =
−≤ 0
V
(∆Φ)Φ′d^3 x + ≤ 0
∂V
∂n Φ
′da =
(by Green′s 2nd identity) −≤ 0
V
Φ(∆Φ′)d^3 x + ≤ 0
∂V
′
∫^ ∂n da^ =
V
ρ′Φd^3 x +
∂V
σ′Φda q.e.d.
