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Jackson Classical Electrodynamics Solution Chapter1 Electrodynamics Universidad Nacional 44 pag.

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Chapter 1

Problem 1.1 Use Gauss’ theorem [and (1.21) if necessary] to prove the following: (a) Any excess charge placed on a conductor must lie entirely on its surface. (A conductor by definition contains charges capable of moving freely under the action of applied electric fields.) (b) A closed, hollow conductor shields its interior from fields due to charges outside, but does not shield its exterior from the fields due to charges placed inside it. (c) The electric field at the surface of a conductor is normal to the surface and has a magnitude σ/ǫ0 , where σ is the charge density per unit area on the surface.

Part a Inside the conductor, we must have E = 0, otherwise the charges inside would move. Application of Gauss’ law Z I 1 ρ(x) d 3 x E · n da = ǫ0 V S with a surface S just inside the surface of V yields ρ(x)=0 so excess charge must lie entirely on the surface of a conductor.

Part b With a charge outside the conductor, one can construct a surface S completely inside the conducting region and thus containing the hollow part. There is not electric flux through S and thus the inner surface always has zero surface charge density when no charge is contained within the hollow part. The effect of the external charge is to induce a non-zero surface charge density on the outer conductor surface in order to maintain E=0 inside the conductor. Since no net charge resides on the inner surface, E is still zero in the hollow part. Thus the conductor shields its entire interior from the fields of the external charge. If on the other hand a charge is placed within the hollow part, a corresponding charge of opposite sign is induced on the inner surface to keep E=0 in the 1

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CHAPTER 1.

Figure 1.1: Integration path used in 1.1c

conducting region. The charge induced on the inner surface obviously has to come from the conductor itself, so a total charge equal to the charge in the hollow part is induced on the outer surface. The fields from charges inside the conductor are thus still present outside the conductor.

Part c We prove the first part by the use of (1.21) I E · dl = 0. We use the integration path shown in Fig. 1.1. Inside the conductor E=0 so Z I Z 0 = E · dl = −∆l · 0 + +E · ∆l + ↑

↓

where the two integrals tend to zero as their path lengths are diminished. Any field E at the surface can be divided into components parallel and perpendicular to the surface: E = E⊥ + Ek and since ∆l is parallel to the surface, E⊥ · ∆l=0 and we are left with Ek =0 and we have shown that the electric field is normal to the surface. For the second part, introduce a Gauss box with on end inside the conductor and the other in vacuum. Again E=0 inside the conductor and we just showed that the electric field at the surface is normal to the surface. If we let the thickness across the surface tend to zero, there is no contribution to the electric flux through the sides of the box and the only contribution is from the box surface just outside the conductor surface. Thus from Gauss’ law E⊥ da =

σ da ǫ0

and E = σ/ǫ0 n.

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Problem 1.2 The Dirac delta function in three dimensions can be taken as the improper limit as α → 0 of the Gaussian function   1 D(α; x, y, z) = (2π)−3/2 α−3 exp − 2 (x2 + y 2 + z 2 ) 2α Consider a general orthogonal coordinate system specified by the surfaces u= constant, v= constant, w= constant, with length elements du/U, dv/V , dw/W in the three perpendicular directions. Show that δ (x − x′ ) = δ (u − u′ ) δ (v − v ′ ) δ (w − w′ ) · UV W by considering the limit of the Gaussian above. Note that as α → 0 only the infinitesimal length element need be used for the distance between the points in the exponent.

Solution I’m sorry to say that this problem is the only problem in Chapter 1 I have not been able to solve. If any of you mathematician out there can present me a simple solution, I would be most grateful.

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CHAPTER 1.

Problem 1.3 Using Dirac delta function in the appropriate coordinates, express the following charge distributions as three-dimensional charge densities ρ(x) (a) In spherical coordinates, a charge Q uniformly distributed over a spherical shell of radius R. (b) In cylindrical coordinates, a charge λ per unit length uniformly distributed over a cylindrical surface of radius b. (c) In cylindrical coordinates, a charge Q spread uniformly over a flat circular disc of negligible thickness and radius R. (d) The same as part (c), but using spherical coordinates.

Part a We work in spherical coordinates (r, θ, φ) and volume element d 3 x = r 2 d(cos θ) dφ dr . We can write charge density as ρ(x) = f (x)δ(r − R) = f (r)δ(r − R) = f (R)δ(r − R) where the factor f is still to be determined. An integration over all space should give the total charge Q: Z

ρ(x) d 3 x

= f (R)

f (R)

=

ρ(x)

=

Z

r 2 dr dΩ δ(r − R) = 4πf (R)R2 = Q ⇒

Q ⇒ 4πR2 Q δ(r − R) 4πR2

Part b Cylindrical coordinates (r, φ, z) and volume element d 3 x = ρ dρ dφ dz, see Fig. 1.2. We consider an arbitrary length L along the z-axis. ρ(x) Z

ρ(x) d 3 x

= f (x)δ(ρ − b) = f (b)δ(ρ − b) Z = f (b) ρ dρ dφ dz δ(ρ − b) = 2πbLf (b) = λL ⇒

f (b)

=

ρ(x)

=

λ ⇒ 2πb λ δ(ρ − b) 2πb

Part c Cylindrical coordinates (r, φ, z) and volume element d 3 x = ρ dρ dφ dz .

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Figure 1.2: Definition of variables

ρ(x) Z

= f (R)δ(z) Θ(R − ρ) Z Z Z = f (R) ρ dρ dφ dz δ(z) Θ(R − ρ) = 2πf (R) ρ dρ Θ(R − ρ) dzδ(z )

3

ρ(x) d x

= 2πf (R) f (R)

=

ρ(x)

=



ρ2 2

R

= 2πf (R)

0

R2 =Q ⇒ 2

Q ⇒ πR2 Q δ(z) Θ(R − ρ) πR2

Part d (r, θ, ρ), d 3 x = r 2 dr d(cos θ) dφ. This is the only example where the factor f will depend on one of the independent variables, r . ρ(x)

Z

= f (x)δ(cos θ) Θ(R − r) = f (r)δ(cos θ)Θ(R − r) Z ρ(x) d x = f (r) δ(cos θ) d(cos θ )Θ(R − r )r 2 dr dφ 3

The δ-function kills the d(cos θ) and the step function defines the limits on the r-integration to 0 ≤ r ≤ R. Z

ρ(x) d 3 x

Z

ρ(x) d 3 x

=

Z

R

f (r)r 2 dr dφ =

Z

R

[f (r )r] r dr dφ

0

0

Now, r dr dφ is an area element. Since the surface charge density is constant, the factor multiplying the area element must be a constant. Thus f (r )r can be moved outside the integration.

= 2πf (r)r

f (r)

=

ρ(x)

=

Z

R 0

r dr = πR2 f (r)r = Q ⇒

Q 1 ⇒ πR2 r Q 1 δ(cos θ)Θ(R − r ) πR2 r

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CHAPTER 1.

Problem 1.4 Each of three charged spheres of radius a, one conducting, one having a uniform charge density within its volume, and one having a spherically symmetric charge density that varies radially as r n (n > −3), has a total charge Q. Use Gauss’s theorem to obtain the electric fields both inside and outside each sphere. Sketch the behavior of the fields as a function of radius for the first two spheres, and for the third with n = −2, +2.

Solution In all cases, the electric field is radially symmetric so E = E(r )er . Furthermore, since all charge is found at r ≤ a, the electric field behavior for r > a is the same for all cases, namely the field from a point charge at the origin. (i) The sphere is conductive so Q resides on the surface only, see Sec. 1a, and σ = Q/4πa2 . According to Gauss’s theorem, E makes a sudden jump at r = a and drops off as if Q were centered at the origin. Thus Q Θ(r − a) 4πǫ0 r 2

E(r) =

(ii) Here Q is distributed evenly in the sphere so the constant space charge density is ρ = Q/ 43 πa3 . Application of Gauss’s theorem for r ≤ a yields E(r) · 4πr 2

=

E(r)

=

Q  r 3 4 3 Q πr = · ⇒ 4 ǫ0 a πa3 ǫ0 3 3 Q r 4πǫ0 a3

The behavior for r > a has already been mentioned above. (iii) Application of Gauss’s theorem for r ≤ a yields E(r) · 4πr 2 E(r)

Z

r

=

4π ǫ0

=

K r n+1 ǫ0 (n + 3)

0

r ′2 · Kr n dr ′ =

4πK r n+3 ⇒ ǫ0 n + 3

The constant K can be determined easily since the volume integral over all space of the charge distribution is the total charge Q. Plugging in the expression for K gives E(r)

=

Q  r n+3 4πǫ0 r 2 a

Fig. 1.3 shows sketches of the electric fields.

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Figure 1.3: Sketch of the electric field behavior of the charge distribution in Q Problem 1.4. The constant C is defined as C = 4πǫ 2 0a

Problem 1.5 The time-averaged potential of a neutral hydrogen atom is given by Φ=

αr q e−αr  1+ 2 4πǫ0 r

where q is the magnitude of the electronic charge, and α −1 = a0 /2, a0 being the Bohr radius. Find the distribution of charge (both continuous and discrete) that will give this potential and interpret your result physically.

Solution Obviously we have to employ Poisson’s equation (1.28) here. Φ only depends on r so we need the first term in the Laplacian in spherical coordinates (look inside the back cover). Note that the radial Laplacian can be written in two ways:   1 ∂2 1 ∂ 2 ∂ψ 2 r = (rψ) ∇ ψ= 2 r ∂r r ∂r2 ∂r For some reason, only the first version automatically gives you the discrete charge (the nucleus). It comes about when you manage to make one term look like ∇2 (1/r), which equals −4πδ (x). The second version is much easier, but you have to add a δ-function after doing the derivative. Here goes:      e−αr q 1 ∂ α2 −αr 1 −αr ∂ 2 + e −α r ∇2 Φ = − e r 4πǫ0 r 2 ∂r 2 ∂r r     ∂ 1 q 1 2 −αr −αr 2 ∂ α re − αe + r e−αr = 4πǫ0 r 2 ∂r ∂r r    1 α2 α3 2 −αr r e − αe−αr r 2 − 2 − 2re −αr + 2 2 r

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CHAPTER 1.

The second term in the square bracket is the −4πδ(x) and since it only contributes for r = 0, the factor multiplying it is just unity.  3  q α −αr ∇2 Φ = e − 4πδ(x) 4πǫ0 2   1 qα3 −αr = − qδ(x) − ⇒ e ǫ0 8π qα3 −αr e ρ(x) = qδ(x) − 8π If you try to write up the wave function for a hydrogen s-state and form the squared norm, you will find the second term is indeed the corresponding charge density.

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Problem 1.6 A simple capacitor is a device formed by two insulated conductors adjacent to each other. If equal and opposite charges are placed on the conductors, there will be a certain difference of potential between them. The ratio of the magnitude of the charge on one conductor to the magnitude of the potential difference is called the capacitance (in SI units it is measured in farads). Using Gauss’s law, calculate the capacitance of (a) two large, flat, conducting sheets of area A, separated by a small distance d; (b) two concentric conducting spheres with radii a, b (b > a); (c) two concentric conducting cylinders of length L, large compared to their radii a, b (b > a). (d) What is the inner diameter of the outer conductor in an air-filled coaxial cable whose center conductor is a cylindrical wire of diameter 1 mm and whose capacitance is 3 × 10−11 F/m? 3 × 10−12 F/m?

Solution The technique here is to first find the behavior of E as a function of Q, the magnitude of the charge on either conductor, and the geometric quantities describing the system from Gauss’s law (1.11): Z I 1 ρ(x) d 3 x E · n da = ǫ0 V S From this, one can find the potential difference V between the two conductors (1.20): Z B E · dl = −(ΦB − ΦA ) A

Part a Use a Gauss box with one end inside the conductor where E = 0 (this is the same as Problem 1.1c) and we get E = Q/Aǫ 0 . Then the potential difference is simply V = Ed = Qd/Aǫ 0 and capacitance C=

Q ǫ0 A = d V

Part b Here the electric field is only non-zero between the two shells E(r) V C

Q 1 4πǫ0 r 2 Z a   1 1 Q = − − E dr = 4πǫ0 a b b −1  Q 1 1 = = 4πǫ0 − b V a

=

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CHAPTER 1.

Part c

E(r) V C

Q 1 2πǫ0 L r Z a E dr = = −

=

=

Q b ln a 2πǫ L 0 b  −1 b Q = 2πǫ0 L ln V a

Part d With 2a = 1 mm and specified values for C/L, we can find 2b from   2πǫ0 2b = 2a exp C/L C/L = 3 × 10−11 F/m: 2b = 6.4 mm C/L = 3 × 10−12 F/m: 2b = 113 km! This final example is probably included to demonstrate the difficulties in changing the capacitance by geometric means only.

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a2

a1 Q

-Q

d-x 0

x

d

Figure 1.4: The situation in Problem 1.7. Notice that d ≫ a 1 , a2 .

Problem 1.7 Two long, cylindrical conductors of radii a 1 and a2 are parallel and separated by a distance d, which is large compared with either radius. Show that the capacitance per unit length is given approximately by −1  d C ⋍ πǫ0 ln a where a is the geometrical mean of the two radii. Approximately what gauge wire (state diameter in millimeters) would be necessary to make a two-wire transmission line with a capacitance of 1.2 × 10−11 F/m if the separation of the wires was 0.5 cm? 1.5 cm? 5.0 cm?

Solution The plan of attack is similar to that of the preceding solution. First, the geo√ metrical mean G is defined as G = n a1 a2 . . . an so here we have a = (a1 a2 )1/2 . The configuration of conductors i seen in Fig. 1.4 The electric field in all space in easily found from Gauss’s law:   Q −Q 1 − E(x) = 2πǫ0 L x d−x We take E(x) to be positive in the direction of the axis. Next we find the potential V :   Z a1 Q 1 Q 1 d − a1 d − a2 dx = + ln V = − a1 a2 x d−x 2πǫ0 L d−a2 2πǫ0 L Q d d ≫a1 ,a2 = ln πǫ0 L a We then find the capacitance per unit length:   C d −1 Q = πǫ0 ln = VL L a

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CHAPTER 1.

To determine the gauge wire diameter 2a corresponding to a given C/L=1.2 × 10−11 F/m and separation d, isolate a:   πǫ0 2a = 2d exp − C/L d = 5 mm: 2a= 1 mm d = 15 mm: 2a= 3 mm d = 50 mm: 2a= 10 mm What to say about these number? Well, 2a scales linearly in d but exponentially in C/L.

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Problem 1.8 (a) For the three capacitor geometries in Problem 1.6 calculate the total electrostatic energy and express it alternatively in terms of the equal and opposite charges Q and −Q placed on the conductors and the potential difference between them. (b) Sketch the energy density of the electrostatic field in each case as a function of the appropriate linear coordinate.

Solution At this point, the progression in the problems deviates from that of the text sections in that we bypass the sections on Green functions and move on to Section 1.11 on electrostatic energy. Three expressions for the electrostatic energy W are given; one emphasizing the charge distribution (1.53), one emphasizing the electric field (1.54), and one involving the capacitance Cij and the potentials of a system of conductors (1.62): 1 W = 2

Z

ρ(x)Φ(x) d 3 x =

ǫ0 2

Z

n

|E|2 d3 x =

n

1 XX Cij ViVj 2 i=1 j=1

We already have expressions for E involving Q and equations connecting Q and V from Problem 1.6 so the second form (1.54) is easily exploited. Even easier is the application of the third form, which reduces to W = 12 CV 2 when only two conductors are present.

Part a For the parallel-plate capacitor, C = ǫ0 A/d and V = Qd/Aǫ0 so W =

1 ǫ0 A 2 1 d Q2 V = 2 ǫ0 A 2 d

For the spherical capacitor,   1 1 Q − V = 4πǫ0 a b 4πǫ0 C = 1/a − 1/b 2πǫ0 1/a − 1/b 2 1 4πǫ0 V2 = V2 = Q W = 8πǫ0 2 1/a − 1/b 1/a − 1/b For the cylindrical capacitor, V

=

C

=

W

=

b Q ln 2πǫ0 L a 2πǫ0 L ln b/a πǫ0 L 2 ln b/a 2 Q V = 4πǫ0 L ln b/a

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CHAPTER 1.

Parallelplate Spherical Cylindrical

a

b,d

Figure 1.5: Sketches of the energy densities w in the three capacitor geometries.

Part b The energy density w is defined in (1.55) as w=

ǫ0 |E|2 2

We already found E in Problem 1.6. The three expressions are easily found: w

=

w

=

w

=

σ2 Q2 = , 0...