Solution Booklet
Rwitaban Goswami
Contents
1 Vector Calculus 3
2 Curvilinear Coordinates 10
3 Helmholtz Theorem, Delta Functions 19 4 Electrostatic Fields, Potentials, Energy 29
5 Conductors & Image Charges 35
6 Solutions to Laplace Equation 43
7 Dipoles & Multipole expansion 54
8 Biot-Savart Law 61
9 Dielectrics 72
10 Magnetodynamics 79
1
Vector Calculus
1.1. Calculate the curl and divergence of the following vector functions. If the curl turns out to be zero, construct a scalar function φ of which the vector field is the gradient:
(c) Hx= x2− z2 ; Hy = 2 ; Hz = 2xz Solution: ~ H(x, y, z) = (x2− z2, 2, 2xz) (1.1.20) Curl: ∇ × ~H = ˆı ˆ kˆ ∂ ∂x ∂ ∂y ∂ ∂z x2− z2 2 2xz (1.1.21) = 0ˆı− (2z + 2z)ˆ+ 0ˆk = −4z ˆ (1.1.22) Div: ∇ · ~H = ∂ ∂x(x 2 − y2) + ∂ ∂y(2) + ∂ ∂z(2xz) (1.1.23) = 2x + 0 + 2x = 4x (1.1.24)
1.2. Calculate the Laplacian of the following functions: (a) Ta = x2+ 2xy + 3z + 4 Solution: ∇ · ∇Ta = ∂2T a ∂x2 + ∂2T a ∂y2 + ∂2T a ∂z2 (1.2.1) = 2 + 0 + 0 (1.2.2) = 2 (1.2.3)
(b) Tb = sin(x) sin(y) sin(z)
Solution: ∇ · ∇Tb = ∂2T b ∂x2 + ∂2T b ∂y2 + ∂2T b ∂z2 (1.2.4)
= −3 sin(x) sin(y) sin(z) (1.2.5) (1.2.6) (c) Tc= e−5xsin(4y) cos(3z) Solution: ∇ · ∇Tc= ∂2T c ∂x2 + ∂2T c ∂y2 + ∂2T c ∂z2 (1.2.7)
= 25e−5xsin(4y) cos(3x)− 16e−5xsin(4y) cos(3x) (1.2.8) − 9e−5xsin(4y) cos(3x) (1.2.9)
(d) ~v = x2ˆı + 3xz2ˆ− 2xz ˆk
Solution:
∇2~v =∇2v
xˆı +∇2vyˆ +∇2vzˆk (1.2.11)
= 2 ˆı + 6x ˆ (1.2.12)
1.3. Test the Stokes’ theorem for the vector field ~v = xyˆi + 2yzˆj + 3zˆk using a triangular area with vertices (0, 0, 0), (0, 2, 0) and (0, 0, 2).
Solution: x y z (0,0,0) (0,2,0) (0,0,2) Γ d~S
According to Stokes theorem: ˛ Γ ~v· d~Γ = ¨ S∇ × ~v · d~S (1.3.1) where
~v is a vector field (given in our case)
S is the surface (of the triangle) with a given orientation (defined by orientation of Γ) Γ is the boundary of the surface with a given orientation (orientation specified in figure) Let us work on the RHS first. For that, we need d~S. We can simply observe that d~S = dy dz ˆı. But there might be more complicated surfaces in the future (or in your exams), so we need a more rigorous way of finding d~S. We can use our MA knowledge (existant or not) to parametrize the triangle surface as
~r (y, z) = 0 ˆı + y ˆ + z ˆk; 0 ≤ y ≤ 2, 0 ≤ z ≤ 2 − y (1.3.2) This parametrization is very simple because our surface is really a part of the yz plane. Now d~S = ~ry× ~rz = (dy ˆ)× (dz ˆk) = dy dz ˆı. Why did we choose ~ry× ~rz and not ~rz× ~ry? Because
the former leads to the orientation we want. Now since d~S has only an ˆı component, we only need the ˆı component of ∇ × ~v, since it’s being dotted with d~S in the RHS.
(∇ × ~v)ˆı=
∂
∂y(3zk)− ∂
RHS = ˆ y=2 y=0 ˆ z=2−y z=0 −2y dz dy (1.3.4) =−2 ˆ y=2 y=0 2y− y2dy (1.3.5) =−2(4 − 8 3) =− 8 3 (1.3.6)
Now we need LHS. Γ can be parametrized as ~ r1(t) = 0 ˆı + t ˆ + (2− t) ˆk; 0 ≤ t ≤ 2 =⇒ d~r1(t) = dt ˆ− dt ˆk (1.3.7) ~ r2(t) = 0 ˆı + 0 ˆ + t ˆk; 0 ≤ t ≤ 2 =⇒ d~r2(t) = dt ˆk (1.3.8) ~ r3(t) = 0 ˆı + t ˆ + 0 ˆk; 0≤ t ≤ 2 =⇒ d~r3(t) = dt ˆ (1.3.9)
Each of ~r1, ~r2, ~r3 is a parametrization of one side of the triangle.
LHS = ˆ 0 2 ~v· (ˆ− ˆk) dt + ˆ 0 2 ~v· ˆk dt + ˆ 2 0 ~v· ˆdt (1.3.10) = ˆ 2 0 vz− vydt− ˆ 2 0 vzdt + ˆ 2 0 vydt (1.3.11) = ˆ 2 0 3(2− t)k − 2t(2 − t) dt (1.3.12) − ˆ 2 0 3tk dt (1.3.13) + ˆ 2 0 2t· 0 dt (1.3.14) =−2 ˆ 2 0 2t− t2dt (1.3.15) =−8 3 = RHS (1.3.16)
1.4. Compute the unit normal vector ˆn to the ellipsoidal surfaces defined by constant values of Φ(x, y, z) = V x 2 a2 + y2 b2 + z2 c2 . What is ˆn when a = b = c?
Solution: The unit normal vector to a surface specified by the equation Φ(x, y, z) = C is given by the gradient direction∇Φ
When a = b = c, ˆn = √x ˆı+y ˆ+z ˆk
x2+y2+z2 = ˆr, which is obvious because the ellipsoids become
spheres.
1.5. A force defined by ~F = A(y2ˆi + 2x2ˆj) is exerted on a particle which is initially at the origin
of the co-ordinate system. A is a positive constant. We transport the particle on a triangular path defined by the points (0,0,0), (1,0,0), (1,1,0) in the counterclockwise direction.
(a) How much work does the force do when the particle travels around the path? Is this a conservative force? Solution: x z y (0,0,0) (1,0,0) (1,1,0) Γ
We need the work done by the force = ¸ΓF~ · d~Γ. Recall MA105 in which a curve was parametrized. We need to parametrize Γ in order to obtain the integral. It might not be absolutely necessary in this example, but starting off this way would make other harder examples easier to solve.
Γ can be parametrized as ~ r1(t) = t ˆı + 0 ˆ + 0 ˆk; 0≤ t ≤ 1 =⇒ d~r1(t) = dt ˆı (1.5.1) ~ r2(t) = 1 ˆı + t ˆ + 0 ˆk; 0 ≤ t ≤ 1 =⇒ d~r2(t) = dt ˆ (1.5.2) ~ r3(t) = t ˆı + t ˆ + 0 ˆk; 0≤ t ≤ 1 =⇒ d~r3(t) = dt ˆı + dt ˆ (1.5.3)
Each of ~r1, ~r2, ~r3 is a parametrization of one side of the triangle.
W = ˆ 1 0 ~ F · ˆıdt + ˆ 1 0 ~ F · ˆdt + ˆ 0 1 ~ F · (ˆı+ ˆ) dt (1.5.4) = ˆ 1 0 Fxdt + ˆ 1 0 Fydt− ˆ 1 0 Fx+ Fydt (1.5.5) = A ˆ 1 0 02+ 2· 12− t2− 2t2dt (1.5.6) = A ˆ 1 0 2− 3t2dt (1.5.7) = A (1.5.8)
(b) The particle is placed at rest right at the origin. Is this a stable situation? Give any argument (mathematical, physical, intuitive) to justify the stability (or instability) of this situation.
Solution: For the particle to be in a stable equilibrium, the force it experiences upon slight displacement in any direction has to be opposite to the displacement. So it can be visualized that if we draw a very small sphere around the origin, the flux of ~F will be negative. So∇ · ~F should be negative at origin as well.
∇ · ~F = 0 + 0 = 0 (1.5.9) So the particle is not in stable equilibrium.
1.6. The area bounded by the curve r = 2R cos θ has a surface charge density σ(r, θ) = σ0Rr sin4θ.
What is the total amount of charge?
Solution: x y θ = 0 2R θ = π 4 θ = π2 θ = −π 4 θ =−π 2 r = 0 r = 2R cos θ
The curve is a circle, as we can see, with −π
2 ≤ θ < π 2 Charge = ¨ circle σ dA = σ0 ˆ θ=π 2 θ=−π 2 ˆ r=2R cos θ 0 r Rsin 4θ r dr dθ (1.6.1) = 8σ0R 2 3 ˆ θ=π 2 θ=−π 2 cos3θ sin4θ dθ (1.6.2) = 8σ0R 2 3 ˆ sin θ=1 sin θ=−1
sin4θ− sin6θ d sin θ (1.6.3)
= 8σ0R 2 3 ( 2 5− 2 7) (1.6.4) = 32σ0R 2 105 (1.6.5)
1.7. Suppose that the height of a certain hill (in feet) is given by
(a) Where is the top of Sameer Hills located, and how high is it?
Solution: The top of Sameer hills will correspond to a maxima of h(x, y). And we know that ∇h = ~0 at the maxima. So we need to find points satisfying ∇h = ~0.
∇h = ∂h ∂xˆı + ∂h ∂y ˆ (1.7.1) = 10(2y− 6x + 14)ˆı+ 10(2x − 8y + 10) ˆ (1.7.2) ∴∇h = 0 =⇒ 2y − 6x + 14 = 0; 2x − 8y + 10 = 0 (1.7.3) =⇒ (x, y) = (3, 2) (1.7.4) h(3, 2) = 710 (1.7.5)
So the top of Sameer hills is located 3 km east and 2 km north of the closest town, and it is at a height of 710 feet
(b) How steep is the slope (in feet per km) at a point 1 km north and 1 km east of Hostel 16? In what direction is the slope steepest, at that point?
Solution:
∇h(1, 1) = 100ˆı+ 40 ˆ (1.7.6) So the steepness of the slope at (1, 1) is |∇h(1, 1)| = 20√29 and the slope is steepest at |∇h(1,1)|∇h(1,1) = √1
29(5 ˆı + 2 ˆ)
1.8. The gradient operator ∇ behaves like a vector in “ some sense”. For example, divergence of a curl (∇.∇ × ~A = 0) for any ~A, may suggest that it is just like ~A. ~B× ~C being zero if any two vectors are equal. Prove that∇ × ∇ × ~F =∇(∇. ~F )− ∇2F . To what extent does this look like~
the well known expansion of ~A× ~B× ~C ?
Solution: We will be using Einstein summation convention. According to this convention: 1. The whole term is summed over any index which appears twice in that term
2. Any index appearing only once in the term in not summed over and must be present in both LHS and RHS
Don’t be confused by this. The convention only is to simply omit the P symbol. You can mentally insert the P symbol to make sense of what the summation will look like
Now, as may have been done in the lectures: ~
A· ~B = AiBi (1.8.1)
~
A× ~B = ijkAjBkeˆi (1.8.2)
where ˆei is the unit vector in i direction
When theP symbol is reinserted, this simply means P
Now we will represent each component of ∇, ∂ ∂i by ∂i, ∇ ×∇ × ~F= ijk∂j ∇ × ~F k eˆi = ijkklm∂j∂lFmeˆi = kijklm∂j∂lFmeˆi (1.8.3) = (δilδjm− δimδjl)∂j∂lFmeˆi (∵ ijkilm = δjlδkm− δjmδkl) (1.8.4) = (∂m∂iFm− ∂l∂lFi) ˆei (∵ δijAj = Ai) (1.8.5) = ∂i(∂mFm) ˆei− ∂l∂l(Fieˆi) (1.8.6) =∇(∇ · ~F )− ∇2F~ (1.8.7) We know that ~A× ~B × ~C = ~B( ~A· ~C)− ( ~A· ~B) ~C. If we write ~A = ∇, ~B = ∇, ~C = ~F , we get ∇ × ∇ × ~F = ∇(∇ · ~F )− (∇ · ∇) ~F =∇(∇ · ~F )− ∇2F . So this serves as a fake proof of~
the identity
2
Curvilinear Coordinates
2.1. Consider a planar surfaceS parallel to xy plane bounded by a closed curve C. Consider a vector field
~
F = 0.5(−y ˆı+ x ˆ)
Prove that the value of the integral of this vector field along the curve C is exactly equal to the area of the planar surface S. Using this find the area inside the curve x23 + y
2 3 = 1
Solution: We have to prove some line integral to be equal to some area integral. Stokes’ theorem immediately springs to mind (this kind of application of Stokes’ theorem on a plane is another theorem called Green’s theorem).
˛ C ~ F · d~l = ¨ S∇ × ~ F · dˆn (2.1.1) = ¨ S ˆ k· dx dy ˆk (2.1.2) = ¨ S dx dy (2.1.3) = Area(S) (2.1.4) (2.1.5) Thus we can use this to find the area enclosed inside any curve. Let C be the curve defined by x23 + y23, and S be the region enclosed by this curve. Let us parametrize C using the
paremeter t
C := { (cos3t, sin3t)| 0 ≤ t < 2π } (2.1.6)
Area(S) = ˛ C ~ F · d~l (2.1.9) = 1 2 ˆ 2π t=0
(− sin3t)(−3 cos2t sin t) + (cos3t)(3 sin2t cos t) dt (2.1.10)
= 3 2 ˆ 2π t=0 sin2t cos2t dt (2.1.11) = 3 8 ˆ 2π t=0 sin22t dt (2.1.12) = 3 16 ˆ 4π t=0 sin2t dt (2.1.13) = 3 4 ˆ π t=0 sin2t dt (2.1.14) = 3π 8 (2.1.15)
2.2. A ship sails from the southernmost point of India (6.75oN, 93.84oE) to the southernmost point
of Africa (34.5oS, 20.00oE) following the shortest possible path.
(a) Given that the radius of the earth is 6400 km, what is the distance it has covered? Solution: We cannot directly calculate the distance. We need the angle subtended by the path on Earth’s centre. For that we need to convert to Cartesian coordinates
~
r1 = R sin θ1cos φ1ˆı + R sin θ1sin φ1ˆ + R cos θ1ˆk = R(−0.067ˆı− 0.991 ˆ+ 0.118 ˆk)
(2.2.1) ~
r2 = R sin θ2cos φ2ˆı + R sin θ2sin φ2ˆ + R cos θ2ˆk = R(0.774 ˆı− 0.282 ˆ− 0.566 ˆk)
(2.2.2) Where, R = 6400 km θ1 = 90o− 6.75o = 83.25o θ2 = 90o+ 34.5o = 124.5o φ1 = −93.84o φ2 = −20o
Let θ0 be the angle between ~r
(b) If instead of sailing, one had travelled in an aeroplane - by what percentage would the shortest possible distance change?
Solution: The change in distance will only be due to the change in R, as R will be increased by the cruising altitude of an aeroplane ≈ 10 km. So percentage change =
10
6400 = 0.15%
2.3. Compute the divergence of the vector field given by:
~v = r cos θ ˆr + r sin θ ˆθ + r sin θ cos φ ˆφ
Check the divergence theorem for this using the volume of an inverted hemisphere of radius R, resting on the xy plane and centered at the origin.
Solution: Let us first calculate the divergence of ~v ∇ · ~v = h 1 rhθhφ ∂(vrhθhφ) ∂r + ∂(hrvθhφ) ∂θ + ∂(hrhθvφ) ∂φ (2.3.1) where hr = Scale factor of r = 1 hθ = Scale factor of θ = r
hφ = Scale factor of φ = r sin θ
=⇒ ∇ · ~v = 1 r2sin θ ∂(r3sin θ cos θ) ∂r + ∂(r2sin2θ) ∂θ + ∂(r2sin θ cos φ) ∂φ (2.3.2) = 1 r2sin θ 3r
2sin θ cos θ + 2r2sin θ cos θ
− r2sin θ sin φ
(2.3.3)
= 5 cos θ− sin φ (2.3.4)
Now we need to check the divergence theorem for the hemisphere
According to divergence theorem ˚ V ∇ · ~v dV = ‹ S ~v· d~S (2.3.5) Let us first calculate the LHS. For that we need dV . We know that for spherical polar coordinates, dV = hrhθhφdr dθ dφ = r2sin θ dr dθ dφ LHS = ˚ V ∇ · ~v dV = ˆ φ=2π φ=0 ˆ θ=π 2 φ=0 ˆ r=R r=0
(5 cos θ− sin φ) r2sin θ dr dθ dφ (2.3.6)
= R 3 3 ˆ φ=2π φ=0 ˆ θ=π 2 θ=0
(5 sin θ cos θ− sin θ sin φ) dθ dφ (2.3.7) = 5πR
3
3 (2.3.8)
For the RHS we need d~S. But the surface has 2 parts, the hemispherical cap S1 and
the base S2. Now for spherical coordinates you know that d ~S1 = d ~Sr = hθhφdθ dφ ˆr =
r2sin θ dθ dφ ˆr. You can also find this using parametrization. We can see (using visualization
or parametrization) that d ~S2 = r dr dφ ˆθ. ~v S1 = ~v r=R
= R cos θ ˆr + R sin θ ˆθ + R sin θ cos φ ˆφ (2.3.9) ~v S2 = ~v θ=π 2 = r ˆθ + r cos φ ˆφ (2.3.10) =⇒ RHS = ‹ S1 ~v· d ~S1+ ‹ S2 ~v· d ~S2 = ˆ φ=2π φ=0 ˆ θ=π 2 θ=0 (R cos θ) R2sin θ dθ dφ (2.3.11) + ˆ φ=2π φ=0 ˆ r=R r=0 (r) r dr dθ (2.3.12) = ˆ φ=2π φ=0 ˆ θ=π 2 θ=0 R3sin θ cos θ dθ dφ (2.3.13) + ˆ φ=2π φ=0 ˆ r=R r=0 r2dr dθ (2.3.14) = πR3+ 2πR 3 3 = 5πR3 3 = LHS (2.3.15)
compute ˛
C
~ F · ~dr
C is any positively oriented curve on the surface of the circular cylinder of radius 1 with it’s axis along the +ve y axis and with one face as the xy plane with O as centre.
Solution: x y z C1 C S
Let C1 be the unit circle on the z = 0 plane with the same orientation as C and S is the
surface on the cylinder enclosed between C and C1, with the normal direction such that it
matches the orientation ofC Now, by Stokes theorem
− ˛ C1 ~ F · d~l + ˛ C ~ F · d~l = ¨ S∇ × ~ F · dˆn (2.4.1) But, ∇ × ~F = 2 ˆk (2.4.2) =⇒ ¨ S∇ × ~ F · dˆn = 0 (2.4.3) Since dˆn· ˆk = 0 =⇒ ˛ C ~ F · d~l = ˛ C1 ~ F · d~l (2.4.4) Thus the line integral of any closed curve on the cylinder (which goes around the cylinder) is equal to the line integral of the unit circle at z = 0 plane centered at origin. Let us find this line integral
Paremetrizing C1 using the variable t
C1 := { (cos t, sin t, 0) | 0 ≤ t < 2π } (2.4.5) =⇒ d~l = dxˆı+ dy ˆ (2.4.6) =− sin tˆı+ cos t ˆ (2.4.7) =⇒ ˛ C1 ~ F · d~l = ˆ 2π t=0
(sin t)(− sin t) + (3 cos t)(cos t) dt (2.4.8)
2.5. If ~a and ~b are constant vectors, φ(~r) = (~a× ~r) · (~b × ~r) is the potential over all space, then find the electric field and charge density in spherical co-ordinates.
[Hint: ~E =−∇φ, ρ = 0∇ · ~E]
Solution: First let us simplify φ(~r)
φ(~r) = (~a× ~r) · (~b × ~r) (2.5.1) = ~a· (~r ×~b × ~r) ∵ ~a×~b · ~c = ~a ·~b × ~c (2.5.2) = ~a· ((~r · ~r)~b − (~b · ~r)~r) The bac-cab rule (2.5.3) = r2(~a·~b) − (~a · ~r)(~b · ~r) (2.5.4)
Now to find ~E ~
E =−∇φ (2.5.5)
=−(~a ·~b)∇r2+∇(~a · ~r)(~b · ~r) ∵ (~a·~b) is const, we pull it out (2.5.6) =−(~a ·~b) 1
hr
∂r2
∂r r +ˆ ∇(~a · ~r)(~b · ~r) Evaluate ∇ in spherical (2.5.7) =−2r ˆr(~a ·~b) + (~b · ~r)∇(~a · ~r) + (~a · ~r)∇(~b · ~r) ∵ scalar product rule of gradient
(2.5.8) = −2~r(~a ·~b) + (~b · ~r)~a + (~a · ~r)~b ∵ ~a, ~b const (2.5.9) Now to find ρ
ρ = 0∇ · ~E (2.5.10)
=−20(~a·~b)∇ · ~r + 0∇ · ((~b · ~r)~a + (~a · ~r)~b) (2.5.11)
=−20(~a·~b)
1 hrhθhφ
∂rhθhφ
∂r + 0∇ · ((~b · ~r)~a + (~a · ~r)~b) Evaluate ∇· in spherical
(2.5.12) =−60(~a·~b) + 0((~b· ~r)∇ · ~a + ~a · ∇(~b · ~r)) (2.5.13)
+ 0((~a· ~r)∇ ·~b +~b · ∇(~a · ~r)) ∵ product rule of div (2.5.14)
=−60(~a·~b) + 0(~a· ∇(~b · ~r) +~b · ∇(~a · ~r)) (2.5.15)
=−60(~a·~b) + 0(~a·~b +~b · ~a) ∵ scalar product rule of gradient
(2.5.16)
= −40(~a·~b) (2.5.17)
2.6. A vector field is given by
~v = ay ˆi + bx ˆj where a, b are constants.
i. Plane polar
Solution: Let us take the anti-clockwise line integral. We first need to write ~v in terms of polar coordinates.
ˆı ˆ =cos θ − sin θ sin θ cos θ ˆr ˆ θ (2.6.1)
~v = ay ˆı + bx ˆ = (ay cos θ + bx sin θ) ˆr + (−ay sin θ + bx cos θ) ˆθ (2.6.2) = r cos θ sin θ(a + b) ˆr + r(−a sin2θ + b cos2θ) ˆθ (2.6.3)
y x r θ For a circle, r = R =⇒ d~Γ = dr ˆr + r dθ ˆθ = r ˆθ + dr dθrˆ dθ = r dθ ˆθ (2.6.4) ˛ Γ ~v· d~Γ = ˆ θ=2π θ=0
R2(−a sin2θ + b cos2θ) dθ = (b− a)πR2 (2.6.5)
ii. Cartesian Solution: y x y =−√R2− x2 y =√R2− x2
Let us divide the circle into two parts y≥ 0 & y < 0. For a circle, y = ±√R2− x2
˛ Γ ~v· d~Γ = ˆ x=−R x=R (a√R2− x2ˆı + bx ˆ)· (ˆı− √ x R2− x2 ˆ) dx (2.6.7) + ˆ x=R x=−R (−a√R2− x2ˆı + bx ˆ)· (ˆı+ √ x R2− x2) dxˆ (2.6.8) = ˆ x=R x=−R−2a √ R2− x2+ 2bx 2 √ R2− x2 dx (2.6.9) = (b− a)πR2 (2.6.10)
(b) Imagine a right circular cylinder of length L with its axis parallel to the z axis standing on the circle. Use cylindrical co-ordinate system to show that the Stokes theorem is valid over its surface.
Solution:
x y z
L
According to Stokes theorem, ¨ S ∇ × ~v · d~S = ˛ Γ ~v· d~Γ (2.6.11) Let us find ∇ × ~v ∇ × ~v = ˆı ˆ ˆk ∂ ∂x ∂ ∂y ∂ ∂z ay bx 0 = (b− a) ˆk = (b − a) ˆz (2.6.12)
We need to find d ~S1 of the curved part of the cylinder. We can find it out using
parametrization (Ugh! Not MA 105 again!):
d ~S2 = ρ dρ dφ ˆz LHS = ¨ S∇ × ~v · d~S = ˆ φ=2π φ=0 ˆ z=L 2 z=L 2 :0 (b− a) ˆz · ˆρdφ dz (2.6.17) + ˆ φ=2π φ=0 ˆ r=R r=0 (b− a) ˆz · ˆzρ dρ dφ (2.6.18) = (b− a)πR2 = RHS (2.6.19)
2.7. Although the gradient, divergence and curl theorems are the fundamental integral theorems of vector calculus, it is possible to derive a number of corollaries from them. Show that:
(a) ´V(∇T ) dV = ¸ST d~S [Hint: Let ~v = ~c T where ~c is a constant, in the divergence theorem]
Solution: ˚ V ∇ · ~v dV = ‹ S ~v· d~S (2.7.1) =⇒ ˚ V ∇ · (~c T ) dV = ‹ S (~c T )· d~S (2.7.2) =⇒ :0 ˚ V T (∇ · ~c) dV + ~c · ˚ V (∇T ) dV = ~c · ‹ S T d~S (2.7.3) =⇒ ˚ V (∇T ) dV = ‹ S T d~S (2.7.4)
3
Helmholtz Theorem, Delta Functions
3.1. Consider a vector field ~F (~r) which dies faster than 1r as r → ∞, show the following results (a) Using Helmholtz theorem as discussed in Lecture 5, show that ~F (~r) may be written as
~ F (~r) =−∇ 1 4π ˚ V ∇0· ~F (~r0) |~r − ~r0| dτ0+∇ × 1 4π ˚ V ∇0× ~F (~r0) |~r − ~r0| dτ0
(b) Derive the same expression for ~F (~r) using ~ F (~r) = ˚ V dτ0F (~r~ 0)δ3(~r− ~r0)
boundary of the integral is to be understood at∞ Hint: Use the following
(i) −4πδ3(~r− ~r0) = ∇2 1 |~r−~r0| (ii) ∇ × ∇ × ~v = ∇∇ · ~f − ∇2f~ (iii) ∇|~r−~r1 0| =−∇0 1 |~r−~r0| (iv) ∇ ×|~r−~rF (~~ r00)| =− ~F (~r)× ∇ 1 |~r−~r0| (v) Q2.7b Solution: ~ F (~r) = ˚ V dτ0F (~r~ 0)δ3(~r− ~r0) (3.1.6) =− 1 4π ˚ V dτ0F (~r~ 0)∇2 1 |~r − ~r0| Using (i) (3.1.7) =− 1 4π∇ 2 ˚ V ~ F (~r0) |~r − ~r0|dτ 0 ∵ ~r0 is constant wrt ∇ (3.1.8) =− 1 4π∇ ∇ · ˚ V ~ F (~r0) |~r − ~r0|dτ 0 !! (3.1.9) + 1 4π∇ × ∇ × ˚ V ~ F (~r0) |~r − ~r0|dτ 0 !! Using (ii) (3.1.10) =− 1 4π∇ ˚ V ~ F (~r0)· ∇ 1 |~r − ~r0| dτ0 (3.1.11) + 1 4π∇ × ˚ V ~ F (~r0)× ∇ 1 |~r − ~r0| dτ0
= 1 4π∇ ˚ V∇ 0· F (~r~ 0) |~r − ~r0| ! dτ0 ! (3.1.15) − 1 4π∇ ˚ V ∇0· ~F (~r0) |~r − ~r0| dτ 0 ! (3.1.16) − 1 4π∇ × ˚ V∇ × ~ F (~r0) |~r − ~r0| ! dτ0 ! (3.1.17) + 1 4π∇ × ˚ V ∇ × ~F (~r0) |~r − ~r0| dτ 0 !
∵ product rules of div and curl (3.1.18) = 1 4π∇ :0 ‹ S ~ F (~r0) |~r − ~r0| · d ~S0 ! (3.1.19) − 1 4π∇ ˚ V ∇0· ~F (~r0) |~r − ~r0| dτ 0 ! (3.1.20) + 1 4π∇ × :0 ‹ S ~ F (~r0) |~r − ~r0| ! × d ~S0 (3.1.21) + 1 4π∇ × ˚ V ∇ × ~F (~r0) |~r − ~r0| dτ0 !
Using Div theorem and 2.7b (3.1.22)
=− 1 4π∇ ˚ V ∇0· ~F (~r0) |~r − ~r0| dτ0 ! + 1 4π∇ × ˚ V ∇ × ~F (~r0) |~r − ~r0| dτ0 ! (3.1.23)
3.2. (a) Using the identity δ(ax) = δ(x)|a| , a6= 0, prove that: δ(g(x)) = X
m s.t g(xm)=0, g0(xm)6=0
δ(x− xm)
|g0(xm)|
func-tions, we need to prove them under the integral sign with an arbitrary test function ˆ ∞ x=−∞ f (x)δ(g(x)) dx (3.2.1) = X m s.t g(xm)=0 ˆ xm+0 x=xm−0 f (x)δ(g(x)) dx (3.2.2) = X m s.t g(xm)=0 ˆ 0 =−0
f (xm+ )δ(g(xm) + g0(xm) +O(2)) d Taylor Expansion
(3.2.3) = X m s.t g(xm)=0 ˆ 0 =−0 f (xm+ )δ(g0(xm) +O(2)) d g(xm) = 0 (3.2.4) = X m s.t g(xm)=0, g0(xm)6=0 ˆ 0 =−0 f (xm+ ) δ() |g0(x m)|
d Using given identity (3.2.5) = X m s.t g(xm)=0, g0(xm)6=0 ˆ xm+0 x=xm−0 f (x)δ(x− xm) |g0(xm)| dx (3.2.6) = ˆ ∞ −∞ f (x) X m s.t g(xm)=0, g0(xm)6=0 δ(x− xm) |g0(xm)| dx (3.2.7)
Since this works for any arbitrary ordinary function f (x), the expressions δ(g(x)) and P
m s.t g(xm)=0, g0(xm)6=0
δ(x−xm)
|g0(x
m)| are equal
(b) Confirm that I =´x=0∞ δ(cos x)e−xdx = 1 2 sinh(π
2)
Solution: We need xm s.t. cos(xm) = 0, − sin(xm)6= 0
(c) Show that,
lim
m→∞
sin mx πx
is a representation of δ(x) by showing that ´−∞∞ dxf (x)D(x) = f (0) Solution: ˆ R f (x)D(x) dx = ˆ ∞ x=−∞ f (x) lim m→∞ sin(mx) πx dx (3.2.13) = ˆ ∞ y=−∞ lim m→∞f y m sin y πy dy (3.2.14) = ˆ ∞ y=−∞ f (0)sin y πy dy (3.2.15) = f (0) π ˆ ∞ y=−∞ sin y y dy (3.2.16) = f (0) (3.2.17) Hence D(x) is equivalent to δ(x)
3.3. Show that D(~r,) demonstrates the peak-character & goes to δ3(~r) as → 0
D(~r, ) =− 1 4π∇
2√ 1
r2+ 2
Hint:
(i) Show that D(~r, ) = 32
4π((r2+2)5/2)
(ii) Check that D(0, ) → ∞ as → 0
(iii) Check that D(~r, )→ 0 as → 0, for all r 6= 0
(iv) Check that the integral of D(~r, ) over all the space is 1
Now, we need to show that as → 0 , D(r, ) behaves like δ3(r). For an ordinary function f (~r) lim→0 ˚ all space f (~r)D(r, )dτ0 (3.3.6) = lim →0 ˚ all space 32f (~r) 4π(r2+ 2)52 dτ0 (3.3.7) = lim →0 ˆ 2π φ=0 ˆ π θ=0 ˆ ∞ r=0 32f (~r) 4π(r2+ 2)52 · r 2dr sin θ dθ dφ (3.3.8) = lim →0 ˆ ∞ r=0 32r2f (~r) (r2+ 2)52 dr (3.3.9) = lim →0 f (~r)· ˆ 32r2 (r2+ 2)52dr ∞ r=0 − ˆ ∞ r=0 ˆ 32r2 (r2+ 2)52 ! · f0(~r) ! (3.3.10) = lim →0 f (~r)r3 (r2+ 2)32 r =∞0 ! − ˆ ∞ r=0 lim →0 f0(~r)r3 (r2+ 2)32 ! (3.3.11) = f (0) = ˚ all space f (~r)δ3(r)dτ0 (3.3.12) Hence lim→0D(~r, ) is equivalent to δ3(~r)
3.4. Evaluate the following integral
˚
V
~r· (~d − ~r)δ3(~e− ~r) dτ
where ~d = (5, 5, 5), ~e = (15, 19, 17), and V is a sphere of radius 7 centered at (10, 15, 19).
Solution: First, we verify if the vector ~e∈ V .
~e = (15, 19, 17) and centre of sphere is ~r1 = (10, 15, 19)
Since k~e − ~r1k = √ 45 < 7 =⇒ ~e ∈ V Using property, ˚ V f (r0)δ3(r0− r0)dτ0 = f (r0), r0 ∈ V (3.4.1)
the required integral ˚
V
r· (d − r)δ3(e− r)dτ0 = e· (d − e) (3.4.2) = (15, 19, 17)· (−10, −14, 12) (3.4.3)
3.5. Let ~F be a vector field whose divergence and curl are given as ∇ · ~F = δ(x)δ(y) and ∇ × ~F = ~0 Using the Helmholtz theorem, determine ~F (x, y, z).
Solution: We are given that D(~r0) = δ(x0)δ(y0), ~C(~r0) = ~0. Let us find U (~r) and ~A(~r) of
Helmholtz theorem ~ A(~r) =˚ ~C(~r0) |~r − ~r0|dτ0 = 0 (3.5.1) U (~r) = ˚ D(~r0) |~r − ~r0|dτ 0 (3.5.2) = ˆ ∞ x0=−∞ ˆ ∞ y0=−∞ ˆ ∞ z0=−∞ 1 p(x − x0)2+ (y− y0)2+ (z− z0)2 dz 0δ(y0) dy0δ(x0) dx0 (3.5.3) = ˆ ∞ z0=−∞ 1 px2+ y2+ (z− z0)2 dz 0 (3.5.4)
Now, this integral is divergent. (You can guess so, since δ(x)δ(y) represents the charge distribution of a charged wire along z axis, whose potential we know does indeed diverge when reference is taken as infinity). Thus we directly attempt to find ~F by calculating the grad of this potential.
3.6. A small ball with a positive charge +q hangs by an insulating thread.Holding this ball vertical, a second ball having charge +q is kept at a distance a along the horizontal direction. There are an infinite number of points where a third ball with charge +2q may be positioned so that the first ball continues to remain vertical when released. Find the equation of the curve describing these points. Solution: +q +2q (x, y) a +q
We basically want to balance all the forces on the hanging +q. Now, ~ Fthread = f ˆ (3.6.1) ~ F+q= q ~E+q = 1 4π0 q2 a2ˆı (3.6.2) ~ F+2q = q ~E+2q =− 1 4π0 2q2x px2+ y23ˆı− 1 4π0 2q2y px2+ y23 ˆ (3.6.3)
Now we know that the thread can only apply a pulling force, so f > 0. So to balance the forces in the y direction, y > 0.
Next we balance forces in the x direction 1 4π0 q2 a2 = 1 4π0 2q2x px2+ y23 (3.6.4) =⇒ (x2+ y2)3 = (2a2x)2 (3.6.5) =⇒ y2 = (2a2x)23 − x2 (3.6.6) =⇒ y = + q (2a2x)23 − x2 (3.6.7)
3.7. After an extremely precise measurement, it was revealed that the actual force between two point charges is given by ~ F = 1 4π0 q1q2 r2 1 + r λ e−rλrˆ
Where λ is a constant with dimensions of length, such that λ 1, hence the correction factor is negligible, but non zero
Does this electric field results from a scalar potential? Justify.
Solution: Yes. The field of a point charge at the origin is radial and symmetric, so ∇ × ~ E(~r) = ~0 U (~r) =− ˆ r r0=∞ ~ E(~r0)· d~l0 =− 1 4π0 q ˆ r r0=∞ 1 r02 1 + r 0 λ e−r0λ dr0 (3.7.1) = 1 4π0 q ˆ ∞ r0=r 1 r02 1 + r 0 λ e−r0λ dr0 = q 4π0 ˆ ∞ r0=r 1 r02e −r0 λ dr0+ 1 λ ˆ ∞ r0=r 1 r0e −r0 λ dr0 (3.7.2) Now ´ 1 r02e− r0 λ dr0 =−e − r0 λ r0 − 1λ ´ e− r0 λ
r0 dr0 ←− exactly right to kill the last term. Therefore
V (~r) = q 4π0 − e−r0λ r0 ∞ r0=r ! = q 4π0 e−rλ r (3.7.3)
3.8. Which one of the following is possible expression for an electrostatic field? For the right expres-sion, find a potential which determines this field with the origin as the reference.
(a) E = A x2yz ˆı + 2xz ˆ− 3yz ˆk
Solution: We know that the curl of the electric field has to be 0, so if the curl of this purported electric field is nonzero, then it cannot be the expression of an electrostatic field. In this case, ∇ × E = ˆı ˆ kˆ ∂x ∂y ∂z x2yz 2xz −3yz (3.8.1) = − 3z − 2x ˆı+ x2yz ˆ + 2z− x2zˆ k (3.8.2) 6= 0 (3.8.3)
(∂x denotes the partial derivative w.r.t. x.)
(b) E = A 3xz2+ y2 ˆı+ 2xy ˆ+ 3x2z ˆk
Solution: The curl of this E is 0 (verify), so therefore, this is a possible expression for an electrostatic field.
∂Φ ∂x = 3xz 2 + y2 (3.8.4) ∂Φ ∂y = 2xy (3.8.5) ∂Φ ∂z = 3x 2z (3.8.6)
Solving them, we get that
Φ = 3 2x 2z2+ xy2+ g(y, z) (3.8.7) Φ = xy2 + h(x, z) (3.8.8) Φ = 3 2x 2z2+ f (x, y) (3.8.9) (3.8.10) where f (x, y) is a function purely of x and y, g(y, z) is a function purely of y and z, and h(x, z) is a function purely of x and z.
Putting it all together, we get that
Φ(x, y, z) = 3 2x
2z2+ xy2 (3.8.11)
3.9. Find the electric field a distance z above center of a square loop of side l carrying uniform line charge density λ. Solution: x y z l •(0, 0, z)
By symmetry, we know that only the z component of the electric field contributes to the total electric field.
field due to one side of the square loop and then multiply by 4: Ez = 4 4πε0 ˆ dq sin θ r2 = 4 4πε0 ˆ z dq r3 (3.9.1) Ez = 1 πε0 l/2 ˆ −l/2 zλ dy l 2 2 + z2+ y23/2 (3.9.2) Ez = 2zλ πε0 l/2 ˆ 0 dy l 2 2 + z2+ y23/2 (3.9.3)
Now, to solve the integral, we use the trig. substitution y = a tan θ, where a = l 2 2 + z2: Ez = 2zλ πε0 ˆ y = l/2 y = 0 a sec2θ dθ a3sec3θ (3.9.4) Ez = 2zλ πε0 ˆ y = l/2 y = 0 cos θ dθ a2 (3.9.5) Ez = 2zλ πε0 · sin θ a2 y = l/2 y = 0 (3.9.6) Ez = 2zλ πε0 · 1 l 2 2 + z2 · y q y2+l2 4 + z2 y = l/2 y = 0 (3.9.7) Ez = zlλ πε0 l 2 4 + z2 ql2 2 + z2 (3.9.8) Therefore, ~ F = zlλ πε0 l 2 4 + z2 ql2 2 + z2 ˆ k (3.9.9)
4
Electrostatic Fields, Potentials, Energy
4.1. Consider a conducting sphere A which is initially uncharged. Another conducting sphere B is given a charge +Q, brought into contact with A and then moved far away. The charge on B is then increased to its original value +Q and again brought into contact with A. Show that if this process is repeated many times, the charge on A will tend to the limit Qq
Solution: It is obvious that after each transfer of charge from B to A, A has k fraction of total charge of A and B
Initially after one transfer qA= q = kQ
After another transfer qA= (k2+ k)Q
After 2 transfers qA= (k3+ k2+ k)Q
So qA→ (. . . + k3+ k2+ k)Q = 1Qk−k = QQq−q
4.2. A hemisphere of radius R has z = 0 as its equatorial plane and lies entirely in the region z ≥ 0. The hemisphere has a uniform volume charge density ρ. Determine the field at the center of the hemisphere.
Solution: We know that the field at ~r due to a 3D charge distribution is given by ~ E(r) = 1 4π0 ˚ V ρ(~r0) | ~ |2 ˆ dV 0 = 1 4π0 ˚ V ρ(~r0) | ~ |3 ~ dV 0 (4.2.1)
We will use spherical coordinates for convenience
Here ~r = ~0 (origin), so ~ =−r0r =ˆ −r0sin θ cos φ ˆı + sin θ sin φ ˆ + cos θ ˆk
=⇒ ~E(~0) = 1 4π0 ˚ V ρ(~r0) | ~ |3 ~ = 1 4π0 ˆ φ=2π φ=0 ˆ θ=π 2 θ=0 ˆ r0=R r0=0 ρ r03(−r 0r) rˆ 02sin θ dr dθ dφ (4.2.2) =− ρ 4π0 :0 ˆ φ=2π φ=0 ˆ θ=π 2 θ=0 ˆ r0=R r0=0 sin2θ cos φ dr dθ dφ ˆı (4.2.3) − 4πρ 0 :0 ˆ φ=2π φ=0 ˆ θ=π 2 θ=0 ˆ r0=R r0=0 sin2θ sin φ dr dθ dφ ˆ (4.2.4) (4.2.5) − 4πρ 0 ˆ φ=2π φ=0 ˆ θ=π 2 θ=0 ˆ r0=R r0=0 sin θ cos θ dr dθ dφ ˆk (4.2.6) =−ρR 40 ˆ k (4.2.7)
4.3. The potential takes the constant value φ0 on the closed surface S which bounds a volume V .
Solution: We know, for a volume charge density ρ, U = 1 2 ˝ VρV dτ ρ = 0∇ · ~E (4.3.1) =⇒ U = 0 2 ˚ V (∇ · ~E)V dτ (4.3.2) Use integration by parts to transfer the derivative from ~E to V
=⇒ U = 0 2 − ˚ V ~ E· (∇V ) dτ + ‹ S V ~E· d~S (4.3.3) =⇒ U = 0 2 ˚ V ~ E· ~E dτ + ‹ S V ~E· d~S (4.3.4) =⇒ U = UE(in) + 0 2 ‹ S V ~E· d~S (4.3.5) U − UE(in) = UE(out) = 0 2 ‹ S V ~E· d~S = φ00 2 ‹ S ~ E· d~S (4.3.6) = Qφ0 2 (4.3.7)
Notice that the last step is due to Gauss’s law
4.4. The inside of a grounded spherical metallic shell (inner radius R1 and outer radius R2) is filled
with space charge of charge density ρ(~r) = a + br. (a) Find the potential at the center.
Solution:
R2
R1
ρ(~r)
We know ρ(~r) = a + br.
Additionally, the charge in the bulk of the conductor and outside the conductor is 0. So is the potential and the electric field. Consider a Gaussian surface in the form of a sphere of radius r. Using Gauss Law,
˚ V ρ(~r)dτ = 0 ‹ ~ E(r)· d ~A =⇒ ˆ r 0 4πr2(a + br)dr = 4π0r2E(r)~ =⇒ ~E(r) = 1 0 ar 3 + br2 4
Taking a line integral
ˆ ~b ~a ~ E(r) = V (~a)− V (~b) ˆ R1 r ~ E(r)dr = V (r)− V (R1) =⇒ 1 0 ˆ R1 r ar 3 + br2 4 dr = V (r)− V (R1) =⇒ V (r) = 1 0 a 6(R 2 1− r2) + b 12(R 3 1− r3) (4.4.1) =⇒ V (0) = 1 0 aR2 1 6 + bR3 1 12 (4.4.2)
(b) Calculate the electrostatic energy of the system Solution:
Now to find energy
Energy = 1 2 ˚ V ρ(~r)V (~r)dτ = ˆ R1 0 (a + br)4πr2 0 a 6(R 2 1− r2) + b 12(R 3 1− r3) = 2πR 5 1 0 a2 45+ abR1 36 + b2R2 1 112
(c) Show that the system attains minimum energy configuration when b = −14a9R
1
For minimum energy, we differentiate wrt b ∂E ∂b = 2πR5 1 0 aR1 36 + bR2 1 56 = 0 =⇒ b = −14a 9R1 (4.4.3)
4.5. Show that the field is uniquely determined when the charge density ρ is specified within a bounded region and either the potential φ or its normal derivative ∂φ∂n is specified on each boundary
Solution: We let there be 2 solutions to the ~E field, ~E1 and ~E2, with potentials V1 and V2
such that ~E1 =−∇V1, ~E2 =−∇V2 Let ~E3 = ~E2− ~E1, V3 = V2− V1 ∇ · ( ~E3V3) = V3∇ · ~E3+ ~E3· ∇V3 (4.5.1) = V3(∇ · ~E2− ∇ · ~E1)− | ~E3|2 (4.5.2) = :0 V3(ρ− ρ) − | ~E3|2 (4.5.3) =⇒ ˚ ∇ · ( ~E3V3) dτ = − ˚ | ~E3|2dτ (4.5.4) =⇒ ‹ boundary V3E~3 · ˆn|dS| = − ˚ | ~E3|2dτ (4.5.5)
Now if φ is specified at boundary, then V3 = V2 − V1 = φ− φ = 0 at boundary. If ∂φ∂n is
specified at boundary, then ~E3· ˆn = ∂φ∂n− ∂φ∂n = 0. Either way V3E~3· ˆn = 0
=⇒ ˚
| ~E3|2dτ = 0 (4.5.6)
=⇒ ~E3 = 0 (4.5.7)
=⇒ ~E1 = ~E2 (4.5.8)
Hence no two distinct solutions can exist
4.6. A ring of radius R has a total charge +Q uniformly distributed on it. (a) Calculate the electric field and potential at the center of the ring.
Solution: Potential at centre =
Q 4π0R
(b) Calculate field at a height z above the center. Compare this result with Coulomb’s law in large z limit.
Solution: At height z above centre, field is Qz ˆz 4π0(R2+ z2)
3 2
As z → ∞, E → 4πQˆ0zz2
which resembles the field from a point charge.
(c) Consider a charge−Q constrained to slide along the axis of the ring. Show that the charge will execute simple harmonic motion for small displacements perpendicular to the plane of the ring. Calculate time period.
Solution: We know the equation for simple harmonic motion is md
2x
dt2 =−kx
Hence,Force acting on the charged particle
F = −Q ~E(z)ˆz =− Q 2z 4π0(R2+ z2) 3 2 Thus, k =−F0(0) = Q2 4π0(R2+ z2) 3 2 Time period is 2πr k m = 2πR Q p4π0Rm
4.7. Assume that an electron is a small sphere of radius R in which the charge −|e| is distributed uniformly over its volume. Calculate the total energy of the system as a function of R. Now suppose you equate this energy to m0c2 where m0 is the rest mass of the electron. What value
of R do you get? How does it compare with the size of a hydrogen atom?
Solution: We can first calculate the energy of a spherical charge distribution with constant volume charge density.
This energy equals
3q2
20π0R
Equating this to m0c2, we get R = 3e 2 20π0m0c2 (4.7.1) =⇒ R ≈ 10−7m ≈ 1000ra (4.7.2)
where ra is Bohr radius of Hydrogen atom
5
Conductors & Image Charges
5.1. A metal sphere with radius R1 has charge Q. A second metal sphere with radius R2 has zero
charge. Now connect the spheres together using a fine conducting wire. Assume that the spheres are separated by a distance R which is large enough that the charge distribution on each ball remains uniform. Derive an expression for the final charge on the sphere with radius R1.
Solution:
Q1 Q− Q1
We need to equate the potentials at each end of the wire, because that is when the whole conductor system will become an equipotential
Vleft end= Vright end (5.1.1)
5.2. (a) Show that the capacitance, C, of a conducting sphere of radius a is given by C = 4π0a
Solution: Assume a conducting sphere of radius a carrying a total charge Q, which will be distributed uniformly. Potential on surface of the sphere is
Q 4π0a
Since our reference for potential measurement is with respect to infinity, potential at infinity is 0 Hence, Capacitance is
C = Q
V = 4π0a (5.2.1) (b) Two isolated conducting spheres, both of radius a, initially carry charges of q1 and q2
and are held far apart. The spheres are connected together by a conducting wire until equilibrium is reached, whereupon the wire is removed. Show that the total electrostatic energy stored in the spheres decreases by an amount U , given by
∆U = 1 16π0a
(q1− q2)2
What happens to this energy?
Solution: Initial Charges on the spheres are q1and q2. When the spheres are connected
by the wire, they become equipotential. Assume the final charges are q10 and q20 q1+ q2 = q10 + q02 q10 4π0a = q 0 2 4π0a =⇒ q10 = q02 = q1+ q2 2 Initial Energy is E1 = q2 1 + q22 2C Final Energy is E2 = q02 1 + q022 2C Substituting ∆U = 1 16π0a (q1− q2)2 (5.2.2)
5.3. A metal sphere of radius R carries a total charge Q. What is the force of repulsion between the ”northern” hemisphere and the ”southern” hemisphere?
Solution: Since the sphere is made of metal(conductor), all the charge will be uniformly distributed on the surface(with σ = 4πRQ2). Let S represent the surface of the ”northern”
hemisphere. Electrostatic Pressure P = 2σ20 = 32πQ220R4
F = ‹
S
P (r) d ~A (5.3.1)
By symmetry, we know that force along z-axis only will survive. Hence we take the compo-nent of each elemental contribution to force along with z-axis
F = ‹ S Q2 32π2 0R4 cos θ d ~A (5.3.2) = Q 2 32π2 0R4 ˆ 2π φ=0 ˆ π 2 θ=0 R2cos θ sin θ dθ dφ (5.3.3) =⇒ F = Q 2 32π0R2 (5.3.4)
5.4. A solid spherical conductor encloses 3 cavities, a cross-section of which are as shown in the figure. A net charge +q resides on the outer surface of the conductor. Cavities A and C contain point charges +q and−q, respectively. What are the net charges on the surfaces of the cavities?
B • +q A • −q C +q
Solution: Consider a Gaussian surface just outside the cavity A inside the interior of the bulk of the conductor. Since electric field is 0 everywhere inside the bulk of a conductor, flux through the Gaussian surface is 0. By Gauss’ Law, the net charge inside the Gaussian surface is 0. Hence,
Qinside cavity+ Qon surf ace = 0
Hence charge on surface of cavity A = −q Similarly, charge on surface of cavity C = +q
(a) Find the surface charge density σ at R, at a, and at b.
Solution: σR = 4πRq 2, as any leftover charge on conductor will distribute itself
uni-formly on the surface
This +q in the cavity of the shell induces a total charge of −q on the inner surface of shell. Hence σa = 4πa−q2
The leftover charge on shell distributes itself uniformly on the outer surface. Hence σb = 4πbq2
(b) Find the potential at the center, using infinity as the reference point. Solution: V (b) = 4πq0b (The interior of shell is shielded)
V (a) = V (b) V (R) = V (a) +4πq 0 1 R− 1 a V (a) = V (R) = 4πq0 1b − 1 a + 1 R
(c) Now the outer surface is touched to a grounding wire, which drains off charge and lowers its potential to zero (same as at infinity). How do your answers to (a) and (b) change?
Solution: The only difference is there will be no leftover charge +q on the outer surface of shell. So σb = 0 Also, V (a) = V (b) = 0 So V (0) = 4πq 0 1 R− 1 a
5.6. A point charge q of mass m is released from rest at a distance d from an infinite grounded conducting plane. How long will it take for the charge to hit the plane?
Solution: We will first find the velocity of the charge after falls to a distance of x from the plane, by energy conservation
1 2mv 2 = q2 16π0 1 x − 1 d =⇒ v = s q2 8π0m 1 x − 1 d =⇒ T =r 8π0m q2 ˆ x=0 x=d 1 x− 1 d −12 = πd q (2π0md) 1 2
5.7. Two infinite parallel grounded conducting plates are held a distance a apart. A point charge q is placed between them, at a distance x from one plate. Find the force on q. Check that your answer is correct for the special cases a→ ∞ and x = a
Solution:
The image configuration is shown in the figure, the positive image charges cancel in pairs. The net force of the negative image charges is:
F = 1 4π0 q2 ( 1 [2(a− x)]2 + 1 [2a + 2(a− x)]2 + 1 [4a + 2(a− x)]2 + . . . (5.7.1) − 1 (2x)2 − 1 (2a + 2x)2 − 1 (4a + 2x)2 − . . . ) (5.7.2) =− q 2 16π0 ( ∞ X n=0 1 (na + x)2 − ∞ X n=1 1 (na− x)2 ) (5.7.3) =− q 2 16π0 ( 1 x2 − 4ax ∞ X n=1 n (n2a2− x2)2 ) (5.7.4) Now when a→ ∞, the sum term tends to 0, and we are left with
F = q
2
4π0
1
(2x)2 (5.7.5)
which is the same force as the case for a single plane. If x = a
2 then we expect no force, and:
F = q 2 16π0 ( 4 a2 − 2 a2 ∞ X n=1 n n2 −1 4 2 ) = 0 (5.7.6)
5.8. A conducting sphere (or a shell) of radius R has a charge Q. (a) Find the force of repulsion between the two hemispheres
Solution: Already calculated in 5.3
(b) Now suppose one has a solid sphere of radius R with charge Q distributed uniformly over its volume.
Since we know field is in z-direction by symmetry = ˚ V 3Q 4πR3 Qr 4π0R3 cos θdτ (5.8.3) = 3Q 2 16π2 0R6 ˆ R r=0 ˆ 2π φ=0 ˆ π 2 θ=0 r3cos θ sin θ dr dθ dφ (5.8.4) =⇒ F = 3Q 2 64π0R2 (5.8.5)
(c) Which case (a) vs (b) has the larger force of repulsion?
Solution: Clearly, the force of repulsion is greater in (b) than (a) as in the case of the conductor, the charges redistribute to minimise repulsion.
5.9. (a) Find the average potential over a spherical surface of radius R due to a point charge q located inside. Show that, in general,
Vave= Vcenter+
Qenc
4π0R
where Vcenter is the potential at the center due to all the external charges and Qenc is the
total enclosed charge.
Solution: The average value of V at a point r over a spherical surface of radius R is given as (assuming S represents the sphere surface)
Vave(r) = 1 4πR2 ‹ S V dS (5.9.1)
In this case since the point charge is located inside so we can break Vave = Vint+ Vext
where Vint is the average due to the internally located charges and Vext is the due to
the externally located charges.
The potential Vint due to the point charge q located inside the sphere at ~r0 = (0, 0, z)
can be calculated as follows Vint = 1 4πR2 ‹ S 1 4π0 q |~r − ~r0| dS (5.9.2) = q 16π2 0R2 ˆ π θ=0 ˆ 2π φ=0 1 √ R2+ z2− 2zR cos θR 2sin θ dθ dφ (5.9.3) = q 8π0 ˆ π θ=0 sin θ √ R2+ z2− 2zR cos θdθ (5.9.4) = q 4π0R (5.9.5) Thus from superposition principle, Vint = 4πQenc0R
So finally, we have
Vave= Vcenter+
Qenc
4π0R
(b) Find the general solution to Laplace’s equation in spherical coordinates for the case where V depends only on r. Do the same for cylindrical coordinates assuming V depends only on s.
Solution: In case of the spherical coordinates we will have ∇2V = 1 r2 d dr r2dV dr = 0 for which the solution will be
V = −c r + k where c, k is some constant.
In case of the cylindrical coordinates we will have ∇2V = 1 s d ds sdV ds = 0 and the solution for the above case is
V = c ln s + k where c, k is some constant.
5.10. A point charge +q is placed at a distance d from the centre of a conducting sphere of radius R (d > R). Show that if the sphere is grounded, the ratio of the charge on the part of the sphere visible from +q to that on the rest is qd+R
d−R. Solution: −qR d q R2 d d I II α
We need the ratio of charge on I and II ˜
Iσ dS
˜
IIσ dS
But remember, σ = 0E ˆ~n, because this is a conductor. Replacing this in the above ratio, we get = ˜ IE~ · d~S ˜ IIE~ · d~S = FluxI FluxII (5.10.2) Here we are talking about flux of the electric field. Now, as we can see contribution of q towards flux through I and II is the same (except for the sign), and is 4πqΩ
0, where Ω is the
solid angle subtended by the sphere on q. But Ω = 2π(1− cos α) = 2π1− √d2−R2
d
. So the flux due to q is
q 1− √ d2−R2 d 20 through II and −q 1− √ d2−R2 d 20 through I
Now, the flux due to the image charge−qRd is− qR
20d through both I and II
So ultimately, −q 1− √ d2−R2 d 20 − qR 20d q 1− √ d2−R2 d 20 − qR 20d = R + d− √ d2− R2 R− d +√d2− R2 = r d + R d− R (5.10.3)
5.11. Two infinite conducting plates (both grounded and perpendicular to the x− y plane) meet at an angle of 60°. A point charge +q in the xy plane has plane polar coordinates (a, 20°). Find all the image charges and their positions in polar coordinates.
Solution: (a,−20°) (a, 100°) (a,−100°) (a, 140°) (a,−140°)
5.12. A rectangular pipe running parallel to the z-axis (from−∞ to +∞) has three grounded metal sides at y = 0, y = a and x = 0. The fourth side at x = b is maintained at a specified potential V0(y).
(a) Develop a general formula for the potential within the pipe. Solution: We need to solve
∂2V
∂x2 +
∂2V
with the boundary conditions given as
V (x, 0) = 0 V (x, a) = 0 V (0, y) = 0 V (b, y) = V0(y)
Using separation of variables, the solution to the above equation can be written as V (x, y) = Aekx+ Be−kx (C sin ky + D cos ky)
Using the boundary conditions we can deduce that D = 0 B =−A k = nπ
a
Using the above we can write the most general form of the solution as V (x, y) =
∞
X
n=1
Cnsinh(nπx/a) sin(nπy/a)
wherein Cn can be determined using the boundary condition V (b, y) = V0(y) and is
given as Cn = 2 a sinh(nπb/a) ˆ a 0 V0(y) sin(nπy/a)dy
(b) Find the potential explicitly, for the case V0(y) = V0 (a constant).
Solution: In this case Cn turns out to be
Cn = 2 a sinh(nπb/a)V0 ˆ a 0 sin(nπy/a)dy The integral in the above eq. is 0 when n is even and 2a
nπ when n is odd. So, finally we
will get V (x, y) = 4V0 π X n=1,3,5 sinh(nπx/a) sin(nπy/a) n sinh(nπb/a)
6
Solutions to Laplace Equation
6.1. Two infinitely long grounded metal plates, at y = 0 and y = a, are connected at x = ±b by metal strips maintained at a constant potential V0, (a thin layer of insulation at each corner
Solution: We need to solve the equations ∂2V (x, y)
∂x2 +
∂2V (x, y)
∂y2 = 0 Laplace’s Equation (6.1.1)
V (−b, y) = V0 Boundary Conditions (6.1.2)
V (b, y) = V0 (6.1.3)
V (x, 0) = 0 (6.1.4)
V (x, a) = 0 (6.1.5)
in the region|x| ≤ b, 0 ≤ y ≤ a
Using separation of variables, we know that the general solution can be written as V (x, y) = X
k
(Akekx+ Bke−kx)(Ckcos ky + Dksin ky) (6.1.6)
Using (6.1.4) and (6.1.5) 0 =X k Ck(Akekx+ Bke−kx) (6.1.7) 0 =X k
(Akekx+ Bke−kx)(Ckcos ka + Dksin ka) (6.1.8)
The trivial way to satisfy (6.1.7) is Ck = 0 ∀ k
Putting this back into (6.1.8) we can see that sin ka = 0 will satisfy it, hence only those Dk
terms are non zero, which satisfy k = nπ a
We can thus index our coefficients by n instead of k We get the general solution as
V (x, y) =X n (DnAnenπ x a + D nBne−nπ x a) sin nπy a (6.1.9) Using (6.1.2) and (6.1.3) we can see that V (b, y) = V (−b, y)
=⇒ X n Anenπ b a + B ne−nπ b a = X n Ane−nπ b a + B nenπ b a (6.1.10) =⇒ X n (An− Bn)(enπ b a + e−nπ b a) = 0 (6.1.11) =⇒ An = Bn ∀ n (6.1.12)
We can finally simplify our general solution as V (x, y) = ∞ X n=1 2DnAncosh nπx a sinnπy a (6.1.13) In order to solve this, we use (6.1.3) (and write our coefficient as Kn= 2DnAn)
Using Fourier’s trick ˆ a y=0 V0sin n0πy a dy = ∞ X n=1 Kncosh nπb a ˆ a y=0 sinnπy a sinn0πy a dy (6.1.15) =⇒ ∞ X n=1 aKn 2 cosh nπb a δnn0 = ( 0 n0 even 2aV0 n0π n0 odd (6.1.16) =⇒ Kn = (0 n even 4V0 nπ 1 cosh(nπb a) n odd (6.1.17) Putting this back into (6.1.13),
V (x, y) = 4V0 π X n=1,3,5... cosh nπx a n cosh nπb a sin nπy a (6.1.18)
We can indeed verify that this does satisfy the boundary conditions and the Laplace’s equation, hence by Uniqueness theorem it must be the only solution
6.2. A cubical box of side length a consists of five metal plates welded together and grounded. The sixth plate at the top is insulated from the rest and maintained at V0.
(a) Argue that the potential at the centre should be V0
6
Solution: Let us consider 6 seperate cases, where in the ith case the ith face is at
potential Vi and all others are at 0 potential (the charges on the faces arrange in such
a way to facilitate this scenario in each case)
It is not very difficult to see that each case is identical upto a rotation and scaling of the charges on each face. Thus the potential at the centre must be proportional to Vi
in the ith case. Hence if we take a superposition of all cases, we can easily see that the
resultant scenario in which ith face is at V
i must have the potential at centre ∝PiVi.
If we take the case of all Vi = V0, we can see that the proportionality constant is 16
Hence the potential at centre is V0
(b) Find the potential inside the box.
Solution: We need to solve the equations ∂2V (x, y, z) ∂x2 + ∂2V (x, y, z) ∂y2 + ∂2V (x, y, z) ∂z2 = 0 Laplace’s Equation (6.2.1) V (x, y, a) = V0 Boundary Conditions (6.2.2) V (x, y, 0) = 0 (6.2.3) V (0, y, z) = 0 (6.2.4) V (a, y, z) = 0 (6.2.5) V (x, 0, z) = 0 (6.2.6) V (x, a, z) = 0 (6.2.7)
For the region 0≤ x, y, z ≤ a
Using separation of variables we know that the general solution can be written as V (x, y, z) =X
k,l
(Akcos kx + Bksin kx)(Clcos ly + Dlsin ly)(Ekle √ k2+l2z + Kkle− √ k2+l2z ) (6.2.8) Applying (6.2.4), (6.2.5), (6.2.6), (6.2.7), we can see that
Ak = 0 ∀ k (6.2.9) Bk = 0 ∀ k 6= nπ a (6.2.10) Cl = 0 ∀ l (6.2.11) Dl = 0 ∀ l 6= mπ a (6.2.12) Applying (6.2.3), Ekl+ Kkl = 0 ∀ k, l (6.2.13)
Thus we can write the general solution as (indexing the coefficients with m, n instead of k, l) V (x, y, z) = ∞ X n=1 ∞ X m=1 2BnDmEnmsin nπx a sinmπy a sinh√n2+ m2πz a (6.2.14) In order to solve this, we can use (6.2.2) (and write the coefficient Knm = 2BnDmEnm)
V0 = ∞ X n=1 ∞ X m=1 Knmsin nπx a sinmπy a sinh√n2+ m2π (6.2.15)
=⇒ ∞ X n=1 ∞ X m=1 a2K nm 4 sinh √ n2+ m2πδ nn0δmm0 = ( 0 n0 even or m0 even 4a2V 0
π2n0m0 n0 odd and m0 odd
(6.2.18)
=⇒ Knm =
(0 n even or m even
16V0
π2nmsinh(√n12+m2π) n odd and m odd
(6.2.19) Thus we can write the solution as
V (x, y, z) = 16V0 π2 X n=1,3,5... X m=1,3,5... sinh √n2+ m2πz a nm sinh √n2+ m2π sin nπx a sinmπy a (6.2.20) Summing it up numerically upto 49 terms gives us 0.166666666676V0 ≈ V60, confirming
our “guess”
6.3. A rectangular pipe, running parallel to the z-axis (from−∞ to +∞), has three grounded metal sides, at y = 0, y = a, and x = 0. The fourth side, at x = b, is maintained at a specified potential V0(y). Develop a general formula for the potential inside the pipe.
Solution: We need to solve the equations ∂2V (x, y)
∂x2 +
∂2V (x, y)
∂y2 = 0 Laplace’s Equation (6.3.1)
V (b, y) = V0(y) Boundary Conditions (6.3.2)
V (0, y) = 0 (6.3.3)
V (x, 0) = 0 (6.3.4)
V (x, a) = 0 (6.3.5)
We can use separation of variables to write the general solution V (x, y) = X
k
(Akekx+ Bke−kx)(Ckcos ky + Dksin ky) (6.3.6)
We can apply (6.3.3), (6.3.4), (6.3.5) to get
Ak+ Bk = 0∀ k (6.3.7)
Ck = 0∀ k (6.3.8)
Dk = 0∀ k 6=
nπ
a (6.3.9)
We use (6.3.2) (Writing the coefficient as Kn= 2AkDk) V0(y) = ∞ X n=1 Knsinh nπb a sinnπy a (6.3.11) Applying Fourier’s trick,
Kn= 2 a 1 sinh nπab ˆ a y=0 V0(y) sin nπy a dy (6.3.12) We can finally write the solution as
V (x, y) = 2 a ∞ X n=1 ´a
y=0V0(y) sin nπ y a dy sinh nπba sinh nπx a sinnπy a (6.3.13)
6.4. Two infinitely long metal plates at y = 0 and y = a are connected at x = ±b by metal strips maintained at a constant potential V0. The potential on the bottom (y = 0) is zero, however
the potential on the top (y = a) is a nonzero constant V1. A thin layer of insulation at each
corner prevents the plates from shorting out. Find the potential inside the resulting rectangular pipe.
Solution: To solve this we need to break this up into 2 situations
x y −b a −b V0 V0 0 0 x y −b a −b 0 0 V1 0 Case I Case II We already know the solution to Case I (from Q6.1),
VI(x, y) = 4V0 π X n=1,3,5... cosh nπxa n cosh nπab sin nπy a (6.4.1) Let us solve for Case II:
We need to solve the equations ∂2V
II(x, y)
∂x2 +
∂2V
II(x, y)
∂y2 = 0 Laplace’s Equation (6.4.2)
VII(x, a) = V1 Boundary Conditions (6.4.3)
VII(x, 0) = 0 (6.4.4)
VII(y,−b) = 0 (6.4.5)
in the region|x| ≤ b, 0 ≤ y ≤ a
Using separation of variables we can write the general solution as VII(x, y) =
X
k
(Akcos kx + Bksin kx)(Ckeky+ Dke−ky) (6.4.7)
Using (6.4.4) we get
Ck+ Dk= 0 (6.4.8)
Thus our general solution simplifies to VII(x, y) =
X
k
2Ck(Akcos kx + Bksin kx) sinh(ky) (6.4.9)
Putting (6.4.5) and (6.4.6) we get
Akcos kb + Bksin kb = 0 (6.4.10) Akcos kb− Bksin kb = 0 (6.4.11) =⇒ Akcos kb = Bksin kb = 0 (6.4.12) =⇒ Ak = 0∀ k = 2nπ 2b (6.4.13) Bk = 0∀ k = (2n + 1)π 2b (6.4.14) Ak= Bk = 0∀ k 6= nπ 2b (6.4.15)
Applying Fourier’s trick, Kn= (0 n even 4V1 nπ 1 sinh(nπa 2b) n odd (6.4.22) Thus VII(x, y) = 4V1 π X n=1,3,5... sinh nπ2by n sinh nπ2ba sin nπx + b 2b (6.4.23) We finally have V (x, y) = VI+ VII (6.4.24) = 4V0 π X n=1,3,5... cosh nπx a n cosh nπb a sin nπy a (6.4.25) + 4V1 π X n=1,3,5... sinh nπ2by n sinh nπa 2b sin nπx + b 2b (6.4.26)
6.5. Consider a spherical surface of a large radius R, the potential on the spherical surface is given below. Using the separation of variables find the potential V (r, θ) for r≥ R upto order O(1
r6) V (R, θ) = ( +V 0≤ θ < π 2 −V π 2 ≤ θ ≤ π
Solution: We need to solve the equations (remembering that we have azimuthal symmetry) ∂ ∂r r2∂V (r, θ) ∂r + 1 sin θ ∂ ∂θ sin θ∂V (r, θ) ∂θ = 0 Laplace’s Equation (6.5.1) V (R, θ) = ( +V 0≤ θ < π 2 −V π 2 ≤ θ ≤ π Boundary Conditions (6.5.2) lim r→∞V (r, θ) = 0 (6.5.3) in the region r ≥ R
Using (6.5.3), Al = 0 ∀ l
We can solve for the coefficients using (6.5.3) V (R, θ) = ∞ X l=0 Bl Rl+1Pl(cos θ) (6.5.5)
Applying Fourier’s trick, ˆ π θ=0 V (R, θ)Pl0(cos θ) sin θ dθ = ∞ X l=0 Bl Rl+1 ˆ π θ=0
Pl(cos θ)Pl0(cos θ) sin θ dθ (6.5.6)
=⇒ V ˆ π 2 θ=0 Pl0(cos θ) sin θ dθ− V ˆ π θ=π 2 Pl0(cos θ) sin θ dθ (6.5.7) = ∞ X l=0 Bl Rl+1 ˆ π θ=0
Pl(cos θ)Pl0(cos θ) sin θ dθ (6.5.8)
=⇒ Bl = ( 0 l even (2l + 1)V Rl+1´1 x=0Pl(x) dx l odd (6.5.9) Thus our solution is
V (r, θ) = V X l=1,3,5... (2l + 1)R l+1 rl+1Pl(cos θ) ˆ 1 x=0 Pl(x) dx (6.5.10)
6.6. Let a sphere of radius R have potential V (R, θ, φ) = V0cos2θ. Find the potential everywhere
inside and outside the sphere.
Solution: Let us begin by calculating potential outside. As always, we have to solve ∂ ∂r r2∂Vout(r, θ) ∂r + 1 sin θ ∂ ∂θ sin θ∂Vout(r, θ) ∂θ = 0 Laplace’s Equation (6.6.1) Vout(R, θ) = V0cos2θ Boundary Conditions (6.6.2)
lim
r→∞Vout(r, θ) = 0 (6.6.3)
in the region r ≥ R We can write the general solution using separation of variables Vout(r, θ) = ∞ X l=0 Alrl+ Bl rl+1 Pl(cos θ) (6.6.4) Using (6.6.3), Al = 0
We can solve for the coefficients using (6.6.2) V0cos2θ = ∞ X l=0 Bl Rl+1Pl(cos θ) (6.6.5)
We can see that Bl = 0 ∀ l > 2 by observing the degree of cos θ in the LHS
=⇒ V0cos2θ = B0 R + B1 R2 cos θ + B2 2R3 3 cos 2θ− 1 (6.6.6) Comparing the coefficients of like degrees of cos θ,
B2 = 2V0R3 3 (6.6.7) B1 = 0 (6.6.8) B0 = V0R 3 (6.6.9) =⇒ Vout(r, θ) = V0R 3r 1 + R 2 r2 3 cos 2θ − 1 (6.6.10) We can similarly find Vin(r, θ),
Vin(r, θ) = V0 3 1 + r 2 R2 3 cos 2θ − 1 (6.6.11)
6.7. An uncharged, conducting sphere of radius R is placed in a region where the electric field is uniform i.e. ~E = ~E0. Can you guess whether l value will be odd or even in potential outside
Solution: We can begin by observing that charges will distribute themeselves such that the charge density function will be odd in cos θ, and hence l will also be odd
WLOG, Let the external electric field be along ˆk, thus ~E0 = E0ˆk. In spherical polar
coordinates, ~E0 = E0cos θ ˆr− E0sin θ ˆθ
This time we must be careful in setting the reference of potential, since we cannot set it to its usual at infinity (since everything near the sphere will have an infinite potential difference wrt infinity due to the constant ~E0 field).
Remember that since our potential reference is now not at infinity, all of our intuitions regarding absolute potentials go out the window. Any potentials we must specify must also be accompanied by what they are in respect to.
We can simply choose the reference to be the sphere surface itself. We can see that then the whole xy plane will also be at 0 potential wrt the sphere (∵ due to symmetry, ~E· ˆr = 0 on the xy plane).
Thus we observe that far away from sphere (where charges on sphere won’t have any effect), V =−E0z =−E0r cos θ Thus we need to solve the equations
∂ ∂r r2∂V (r, θ) ∂r + 1 sin θ ∂ ∂θ sin θ∂V (r, θ) ∂θ = 0 Laplace’s Equation (6.7.1) V (R, θ) = 0 Boundary Conditions (6.7.2) lim r→∞V (r, θ) =−E0r cos θ (6.7.3)
in the region r≥ R As always, we can write the general solution using separation of variables V (r, θ) = ∞ X l=0 Alrl+ Bl rl+1 Pl(cos θ) (6.7.4) Applying (6.7.2), Bl =−AlR2l+1∀ l (6.7.5)
Applying (6.7.3), we can see that Al = 0 ∀ l > 1. Comparing coefficients,
−E0r cos θ = A0 + A1r cos θ (6.7.6)
=⇒ A0 = 0 (6.7.7)
A1 =−E0 (6.7.8)
Putting it all together,
V (r, θ) =−E0 r−R 3 r2 cos θ (6.7.9) Of course, remember that this potential is in reference to the sphere
6.8. Suppose the potential V0(θ) at the surface of the sphere is specified, and there is no charge
outside or inside the sphere. Show that charge density on the surface of the sphere is given by σ(θ) = 0 2R ∞ X l=0 (2l + 1)2C lPl(cos θ) where Cl is Cl= ˆ π θ=0 V0(θ)Pl(cos θ) sin θ dθ
Solution: We know that (after a simple application of separation of variables followed by Fourier’s trick) Vout(r, θ) = ∞ X l=0 ´π θ=0V0(θ)Pl(cos θ) sin θ dθ rl+1 (2l + 1)Rl+1 2 Pl(cos θ) (6.8.1) = ∞ X l=0 Cl rl+1 (2l + 1)Rl+1 2 Pl(cos θ) (6.8.2) Vin(r, θ) = ∞ X l=0 ´π θ=0V0(θ)Pl(cos θ) sin θ dθ Rl (2l + 1)rl 2 Pl(cos θ) (6.8.3) = ∞ X l=0 Cl Rl (2l + 1)rl 2 Pl(cos θ) (6.8.4) We also know that
σ(θ) 0 = −∂Vout ∂r r=R + ∂Vin ∂r r=R (6.8.5) = ∞ X l=0 (2l + 1)ClPl(cos θ) 2 l R + l + 1 R (6.8.6) = 1 2R ∞ X l=0 (2l + 1)2ClPl(cos θ) (6.8.7) =⇒ σ(θ) = 0 2R ∞ X l=0 (2l + 1)2ClPl(cos θ) (6.8.8)
7
Dipoles & Multipole expansion
7.1. Consider a thin spherical shell (thickness→ 0) of radius R with a surface charge density σ(θ) = σ0(cos θ + cos2θ)
Solution: As before, we know that, Vout(r, θ) = ∞ X l=0 Bl rl+1Pl(cos θ) (7.1.1) Vin(r, θ) = ∞ X l=0 AlrlPl(cos θ) (7.1.2)
We can apply Vin(R, θ) = Vout(R, θ) to get Bl= AlR2l+1
We also know that
σ(θ) = 0 ∂Vin(r, θ) ∂r r=R − 0 ∂Vout(r, θ) ∂r r=R (7.1.3) = 0 ∞ X l=0 (2l + 1)AlRl−1Pl(cos θ) (7.1.4)
We can immediately see that Al = 0 ∀ l > 2. Comparing coefficients,
σ0cos θ + σ0cos2θ = A00 R + 3A10cos θ + 5A2R0 2 3 cos 2θ − 1 (7.1.5) =⇒ A2 = 2σ0 15R0 (7.1.6) A1 = σ0 30 (7.1.7) A0 = σ0R 30 (7.1.8) Thus we can write the potentials
Vout = σ0R2 30r + σ0R 3 30r2 cos θ + σ0R 4 150r3 3 cos2θ− 1 (7.1.9) Vin= σ0R 30 + σ0r 30 cos θ + σ0r 2 150R 3 cos2θ− 1 (7.1.10)
c r b θ
(a) Show that to the first order in c, the equation describing the outer sphere, using the centre of inner sphere as origin, is r(θ) = b + c cos θ.
Solution: We can use trigonometry and observe
b2 = c2+ r2− 2cr cos θ (7.2.1) Solving using the Quadratic Formula (and taking the positive root)
r = c cos θ +pb2− c2sin2θ (7.2.2) = c cos θ + b r 1− c 2sin2θ b2 (7.2.3) = c cos θ + b 1− c 2sin2θ 2b2 +O(c4) (7.2.4) = b + c cos θ +O(c2) (7.2.5)
We can also see this geometrically
c cos θ b
(b) If the potential between two spheres contains only l = 0 and l = 1 angular components, determine it to first order in c.
∂V (a, θ)
∂θ = 0 Inner sphere equipotential (7.2.7) −0 ˆ inner sphere ∂V (r, θ) ∂r r=a
= Q Total charge on inner sphere (7.2.8) V (b + c cos θ +O(c2), θ) = 0 Outer sphere grounded (7.2.9) Using (7.2.7), Bl =−Ala2l+1∀ l > 0 (7.2.10) Using, (7.2.8) −2π0a2 ˆ π θ=0 − B0 a2 + X l=1 (2l + 1)Alal−1Pl(cos θ) ! sin θ dθ = Q (7.2.11) =⇒ 4π0B0+ ∞ X l=1 al+1(2l + 1)A l ˆ 1 x=−1 Pl(x) dx = Q (7.2.12) =⇒ 4π0B0+ ∞ X l=1 al+1(2l + 1)Al ˆ 1 x=−1 Pl(x)P0(x) dx = Q (7.2.13) =⇒ B0 = Q 4π0 (7.2.14) Applying (7.2.9) 0 = A0+ B0 b + c cos θ +O(c2) −1 (7.2.15) +A1 b + c cos θ +O(c2) + B1 b + c cos θ +O(c2)
−2 cos θ +O(cos2θ) (7.2.16) =⇒ A0+ B0 b + A1b + B1 b2 − B0c b2
cos θ +O(c2) +O(cos2θ) = 0 (7.2.17) Solving upto first order in c and equating coefficients of 1 and cos θ,
A0 =− Q 4π0b (7.2.18) A1 =− Q 4π0(b3− a3) (7.2.19) Putting it all together,
V (r, θ) = Q 4π0 1 r − 1 b − c b3 − a3 r− a 3 r2 cos θ +O(c2) +O(cos2θ) (7.2.20)