Electrodynamics PHYS7300 Spring 2004 Homework Assignment #1 Due: Jan 23rd, 2004 by 5:00 PM 1) Dipole Field. A charge of +q is placed on the z-axis at z = a. A charge of –q is placed on the zaxis at z = -a. Find an expression for the electric field using: a) Cartesian coordinates. b) Spherical coordinates. c) What happens in the limit that z is much greater than the separation of the charges, a? z
Solution: Refer to Fig. 1 for coordinate system layout and definitions.
G r+
+q @ z=+a
y
G x
θ
G
r− In Cartesian coordinates, the field point a zˆ is G x x = x xˆ + y yˆ + z zˆ. In Spherical coordinates, it is G x = r rˆ. I have also defined the vectors from each -q @ z=-a of the point charges as G G r+ = x − a zˆ, G G r− = x + a zˆ. Figure 1. Dipole arrangement for Problem #2. The magnitudes of these vectors are 1/ 2 G r+ = r+ = x 2 + y 2 + ( z − a ) 2 (Cartesian) 1/ 2 G r− = r− = x 2 + y 2 + ( z + a ) 2
G r+ = r+ G r− = r−
( ( = (r = (r
) )
) + 2ar cos θ )
2
+ a 2 − 2ar cos θ
1/ 2
2
+a
1/ 2
2
(Spherical, from the Law of Cosines).
For part (a), we write the electric field as
Homework #1, Solutions, pg. 1
G G G q ( x − azˆ ) ( x + azˆ ) − E= 4πε 0 xG − azˆ 3 xG + azˆ 3 q xxˆ + yyˆ + ( z − a) zˆ xxˆ + yyˆ + ( z + a) zˆ = − 4πε 0 r+ 3 r− 3 1 z−a z+a 1 1 q 1 x 3 − 3 xˆ + y 3 − 3 yˆ + 3 − 3 zˆ . r− 4πε 0 r+ r− r+ r− r+ 3/2 If z>>a, then r+ ≈ r− ≈ r , O((z ± a) ), so the x and y terms go away. We do not discard the O(z ± a) terms because that would remove all the physics from the problem, giving us the trivial result of zero. Instead, we get G 2aq E=− zˆ, 4πε 0 r 3 which is the normal dipole field equation. =
For part (b), it’s a bit more complicated because we have a charge configuration that is defined in Cartesian coordinates. There are two ways to tackle the problem: First Method: write the z-unit vector in terms of the spherical coordinate unit vectors. From Arfken, we find the spherical unit vectors in terms of the Cartesian rˆ = sin θ cos φ xˆ + sin θ sin φ yˆ + cosθ zˆ θˆ = cosθ cos φ xˆ + cosθ sin φ yˆ − sin θ zˆ
φˆ = − sin φ xˆ + cos φ yˆ. These can be inverted to give: xˆ = sin θ cos φ rˆ + cos θ cos φ θˆ − sin φ φˆ yˆ = sin θ sin φ rˆ + cosθ sin φ θˆ + cos φ φˆ zˆ = cos θ rˆ − sin θ θˆ. Then, we can write the vectors from each point charge as G r+ = rrˆ − azˆ = (r − a cosθ )rˆ + a sin θ θˆ G r = rrˆ + azˆ = (r + a cos θ )rˆ − a sin θ θˆ. −
Now, the electric field is simple to write: G G G q r+ r− E= − 4πε 0 r+ 3 r− 3 =
q r − a cos θ r + a cos θ − 4πε 0 r+ 3 r− 3
a sin θ a sin θ + rˆ + 3 r− 3 r+
θˆ .
Second Method: Find the potential field and then take the gradient (in spherical coordinates).
Homework #1, Solutions, pg. 2
q 1 1 − 4πε 0 r+ r− G 1 ∂Φ ˆ 1 ∂Φ ˆ ∂Φ E = −∇Φ = − rˆ − θ− φ. r ∂θ r sin θ ∂φ ∂r The last term is zero since neither r+ or r- depend on φ. The other differentiations are −1/ 2 ( 2r − 2a cos θ ) −1/ 2 ( 2r + 2a cos θ ) ∂Φ = − 3/ 2 3/ 2 ∂r r 2 + a 2 − 2ar cos θ r 2 + a 2 + 2ar cosθ Φ=
(
)
(
)
r − a cosθ r + a cosθ = − − . 3 3 r r + − −1/ 2 ( −2ar sin θ ) 1 ∂Φ 1 −1/ 2 ( 2ar sin θ ) = − 3/ 2 r ∂θ r r 2 + a 2 − 2ar cos θ 3/ 2 r 2 + a 2 + 2ar cosθ
(
)
(
)
a sin θ a sin θ = − + . 3 r− 3 r+ Clearly, this ends up giving the same answer as the first method did. When z>>a, again r+ and r- become approximately equal (and equal to r), and we get G q 2a cos θ 2a sin θ ˆ ˆ − + θ . E= r 3 4πε 0 r r3 Checking this answer: from the definition of the z unit vector we can rewrite this (in Cartesian coordinates) as G 2aq E=− zˆ, 4πε 0 r 3 which is the same as derived above.
Homework #1, Solutions, pg. 3
2) Jackson 1.3. Find the appropriate three-dimensional charge density for the following charge distributions: Solution: All of these problems involve taking a lower dimensional charge density and tacking on appropriate delta functions (and perhaps other terms) in order to convert to a threedimensional charge density. You will have to start by making an educated guess. Then, you check your guess (and perhaps modify it) by integrating the charge density over all space and seeing if you get Q as an answer. (a) A charge Q uniformly distributed over a spherical shell of radius R. We have a surface charge density in this problem, with units of C/m2. Given Q and R, we know σ = Q / 4π R 2 . This charge density is located only at one distance from the orgin—at r=R. At every other distance from the origin, there is no charge, so we know we are going to have a delta function in r of δ (r − R) . Since delta functions have units of one over the units of their argument, if we multiply the surface charge density by the delta function, we’ll have the correct units of C/m3. So, we guess that our answer is ρ (r ,θ , ϕ ) = σδ (r − R) We check this by integration: 3 2 2 ∫ ρ d x = ∫ σδ (r − R)r dr sin θ dθ dϕ = σ 4π ∫ δ (r −R)r dr
= 4πσ R 2 = Q. So, this solution checks out. (b) A charge per unit length λ uniformly distributed on a cylindrical surface with radius b. Now we have a charge distribution of λ = Q / L where L is the length of the cylinder. We also know that there is only a non-zero charge density when r = b (I’m using r as the cylindrical x-y distance coordinate instead of ρ since we use ρ to denote the charge density). This suggests that we need a delta function of δ (r − b) . The total charge density is going to be related to ρ (r , ϕ , z ) = λδ (r − b) , but a quick look at the units indicates this won’t be quite right (units are C/m2 not C/m3). In any case, we don’t have any more restrictions on the charge density, so we cannot just multiply by another delta function. Instead, let’s throw in a constant term that we’ll find by integrating ρ (r , ϕ , z ) = C λδ (r − b)
∫ ρd
3
x = C ∫ λδ (r − b)rdrdϕ dz = C λ 2π ∫ dz ∫ δ (r − b)rdr = 2πλ Cb ∫ dz = 2π bλCL = 2π bQC.
We need this to be equal to Q, so we know that C has to be 1/ 2π b , giving us our final answer of
Homework #1, Solutions, pg. 4
ρ (r , ϕ , z ) =
λ δ ( r − b) , 2π b
which has the correct units. We could have guessed something like this off the bat, because we required a term involving length and b is the only length scale in the problem. (c) In cylindrical coordinates, a charge Q spread uniformly over a flat circular disc of negligible thickness and radius R. We again have a surface charge density given by σ = Q / π R 2 . This charge is located completely on the z=0 plane, so we need a term δ ( z ) . There’s also the issue that the charge stops for radii greater than R. We can take care of that by dividing space into two regions: one with r ≤ R and one with r > R , with the later forced to have zero charge density. It seems then, that we have a good solution if we posit σδ ( z ) r ≤ R ρ (r , ϕ , z ) = r>R 0 Integrating to check gives R
3 ∫ ρ (r ,ϕ , z )d x = ∫ σδ ( z )rdrdϕ dz = 2πσ ∫ rdr ∫ δ ( z )dz 0
= 2πσ So, our guess is correct.
2
R = Q. 2
(d) The same thing, but using spherical coordinates. We still have the same surface charge density and we still have to divide space into the same two regions. The only real difference is going to be how we pin the charge down to be only located at z=0. In spherical coordinates, this can be done by noting that the z=0 plane is located at θ = π / 2 radians. So, we need a delta function of δ (θ − π / 2) , or, equivalently δ (cos θ ) , which is only true when cos θ = 0 , i.e., at θ = π / 2 . This delta function has no units (or at best inverse radians), so we’re going to be off by a factor of 1/length. Again, it’ll be related to R, since that’s the only lengthscale in the problem. To find it rigorously, we’ll tack on a constant factor for our charge density and solve for it by integration: Cσδ (cos θ ) r ≤ R ρ (r , θ , ϕ ) = 0 r>R
∫ ρ (r ,θ , ϕ )d
π
R
3
x = 2π Cσ ∫ r dr ∫ δ (θ − π / 2)dθ = 2π Cσ R 3 / 3 2
0
0
3
Q R 2 = RCQ 2 πR 3 3 3 ⇒C = 2R This gives us the corrected answer of = 2π C
Homework #1, Solutions, pg. 5
3σ δ (cos θ ) r ≤ R ρ (r , θ , ϕ ) = 2 R
which has the correct dimensions.
0
r>R
3) Jackson 1.4 Solution: In all these cases, we have spherical symmetry, which means that the electric field must point radially outward. This in turn dictates that we use spherical Gaussian G G ˆ = EA , where A is the surfaces to evaluate Gauss’ Law, so that E ⋅ nˆ = E and v∫ E ⋅ nda
area of the Gaussian sphere (for a sphere of radius r, A= 4π r 2 ). Furthermore, in all these cases, outside the sphere we get the same electric field, because the total charge, Q, is within the Gaussian surface, giving us EA = Q / ε 0 or 1 Q E= outside the sphere (r > a) . 4πε 0 r Inside the spheres we have different cases. First, we look at the uniform charge density case. Uniform charge density Within the sphere, the charge density must be ρ = Q / ( 4π a 3 / 3) . So, a Gaussian surface within the sphere will only enclose r
Q′ = 4π ∫ ρ r ′2 dr ′ = 0
4πρ 3 r3 r =Q 3 3 a
amount of charge. Gauss’ Law then gives G ˆ = E 4π r 2 = Q′ / ε 0 E v∫ ⋅ nda Q r3 Q r = 2 3 4π r ε 0 a 4πε 0 a 3 The E-field grows linearly with the radial distance. ⇒E=
Conducting sphere Within any conductor (in equilibrium), there is no charge (it all resides on the surface), and therefore E=0. Radially varying density The problem states that the charge density varies as r n . This is not quite enough information for the problem. We need to complete the problem by finding out what the charge density is. We use the same technique as in the previous problem: integrate the density and make sure it is equal to Q:
Homework #1, Solutions, pg. 6
ρ (r ,θ , ϕ ) = Cr n
∫ ρ (r ,θ , ϕ )d
a
3
x = Q = C ∫ r r dr sin θ dθ dϕ = 4π C ∫ r n + 2 dr = n 2
0
4π C n +3 a n+3
Q ( n + 3) 4π a n +3 Q ( n + 3) n ⇒ρ= r . 4π a n +3 Now, when we integrate this density within a Gaussian sphere of radius r, we get 4π Q ( n + 3) r n 2 Q ( n + 3) r n +3 Qr n +3 3 ′ ′ ′ d x r r dr ρ = = = ∫ a n +3 (n + 3) a n +3 4π a n +3 ∫0 for the amount of charge enclosed. It follows immediately, then, that the electric field is 1 1 Qr n +3 Q E= r n +1 = 2 n +3 n +3 4π r ε 0 a 4πε 0 a Note that this reduces to the Uniform case (n=0), as it should. ⇒C =
These E-fields are shown in the following Graph 4
Normalized E
n=-2
2
2
E=q/4πr Uniform n=2 0 0
1
2
Conducting(E=0)
r/a
3
4
Homework #1, Solutions, pg. 7
4) Jackson 1.5 (Hint: treat the potential as the product of two functions of r and use the equivalent of the chain rule for solving Poisson’s Equation.) Solution: We are given a potential of
1 α e −α r + , 4πε 0 r 2 and we are asked to find the charge distribution that gives rise to this potential. We will rewrite this potential as the product of two other potentials q Φ= φψ 4πε 0 q
Φ=
1 α φ = e −α r ; ψ = + . r 2 Note that Poisson’s equation gives the relationship between the potential and the charge density distribution: ∇ 2Φ = − ρ / ε 0 or ρ = −ε 0∇ 2 Φ. The Laplacian operator operating on a product of two functions is given by G G G ∇ 2 (φψ ) = ∇ ⋅∇(φψ ) = ∇ ⋅ φ∇ψ + ψ∇φ G G G G = φ∇ 2ψ + ∇φ ⋅∇ψ + ψ∇ 2φ + ∇ψ ⋅∇φ G G = φ∇ 2ψ + 2∇φ ⋅∇ψ + ψ∇ 2φ . Therefore, G G q G ρ (r ) = − φ∇ 2ψ + 2∇φ ⋅∇ψ + ψ∇ 2φ . 4π Recall that G1 rˆ ∇ = − 2 , and r r 1 G ∇ 2 = −4πδ ( r ) . r The Del operator operating on the constant part of ψ (α/2) is zero. Del operating on φ is G ∂ ∇e −α r = rˆ e −α r = −α e−α r rˆ. ∂r 1 ∂ ∂ 1 ∂ ∇ 2 e −α r = 2 r 2 e −α r = 2 −α r 2 e−α r r ∂r ∂r r ∂r
(
)
(
)
(
=−
α 2
(e
r Putting these all together we get
−α r
)
2α −α r 2r − α r 2 e −α r = α 2 − e . r
)
Homework #1, Solutions, pg. 8
G 4π q
ρ (r ) −
1 G 1 α 2 2α −α r −α r −α r = −4π e δ (r ) + + α − e + 2(−α e rˆ) ⋅ − 2 rˆ r r r 2 G α 2 α 3 2α α 2 2α −α r = −4πδ ( r ) + + − − e + 2 r2 r r 2 r
α 3 −α r G e . = −4πδ ( r ) e −α r + 2 We rewrite this as α3 G G ρ (r ) = qe−α rδ ( r ) − q e−α r . 8π This can be interpreted as a positive point charge (a proton) at the origin (the nucleus) and an azimuthally symmetric smeared out negative charge (an electron) surrounding it.
Homework #1, Solutions, pg. 9
