PH 108 : Electricity & Magnetism Hints & Answers

PH 108 : Electricity & MagnetismHints & Answers

Rwitaban GoswamiKartik Gokhale

Tirthankar Mazumder

1

PH 107 : Quantum Physics Contents Rwitaban Goswami

Contents

1 Vector Calculus 3

2 Curvilinear Coordinates 9

3 Helmholtz Theorem, Delta Functions 15

4 Electrostatic Fields, Potentials, Energy 22

5 Conductors & Image Charges 27

6 Solutions to Laplace Equation 34

7 Dipoles & Multipole expansion 41

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PH 107 : Quantum Physics Tutorial Sheet: 1 Rwitaban Goswami

1 Vector Calculus

1.1. Calculate the curl and divergence of the following vector functions. If the curl turns out to bezero, construct a scalar function φ of which the vector field is the gradient:

(a) Fx = x+ y ; Fy = −x+ y ; Fz = −2z

Solution:

~F (x, y, z) = (x+ y,−x+ y,−2z) (1.1.1)

Curl: ∇× ~F = −2 k (1.1.2)

Div: ∇ · ~F = 0 (1.1.3)

(b) Gx = 2y ; Gy = 2x+ 3z ; Gz = 3y

Solution:

~G(x, y, z) = (2y, 2x+ 3z, 3y) (1.1.4)

Curl: ∇× ~G = ~0 (1.1.5)

Div: ∇ · ~G = 0 (1.1.6)

φ: Gx =∂

∂xφ(x, y, z) = 2y (1.1.7)

To find φ(x, y, z), first integrate Gx with respect to x and keep the constant in termsof y, z. Then differentiate w.r.t y and compare with Gy · · ·

=⇒ φ(x, y, z) = 2xy + 3yz + C (1.1.8)

(c) Hx = x2 − z2 ; Hy = 2 ; Hz = 2xz

Solution:

~H(x, y, z) = (x2 − z2, 2, 2xz) (1.1.9)

Curl: ∇× ~H = −4z (1.1.10)

Div: ∇ · ~H = 4x (1.1.11)

1.2. Calculate the Laplacian of the following functions:

(a) Ta = x2 + 2xy + 3z + 4

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Solution:

∇ · ∇Ta =∂2Ta∂x2

+∂2Ta∂y2

+∂2Ta∂z2

(1.2.1)

= 2 + 0 + 0 (1.2.2)

= 2 (1.2.3)

(b) Tb = sin(x) sin(y) sin(z)

Solution:

∇ · ∇Tb =∂2Tb∂x2

+∂2Tb∂y2

+∂2Tb∂z2

(1.2.4)

= −3 sin(x) sin(y) sin(z) (1.2.5)

(1.2.6)

(c) Tc = e−5x sin(4y) cos(3z)

Solution:

∇ · ∇Tc =∂2Tc∂x2

+∂2Tc∂y2

+∂2Tc∂z2

(1.2.7)

= 25e−5x sin(4y) cos(3x)− 16e−5x sin(4y) cos(3x) (1.2.8)

− 9e−5x sin(4y) cos(3x) (1.2.9)

= 0 (1.2.10)

(d) ~v = x2 ı+ 3xz2 − 2xz k

Solution:

∇2~v = ∇2vx ı+∇2vy +∇2vz k (1.2.11)

= 2 ı+ 6x (1.2.12)

1.3. Test the Stokes’ theorem for the vector field ~v = xyi + 2yzj + 3zk using a triangular areawith vertices (0, 0, 0), (0, 2, 0) and (0, 0, 2).

Solution:

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x

y

z

(0,0,0)

(0,2,0)

(0,0,2)

Γd~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)

Paremtrize the surface as

~r (y, z) = 0 ı+ y + z k; 0 ≤ y ≤ 2, 0 ≤ z ≤ 2− y (1.3.2)

(∇× ~v)ı =∂

∂y(3zk)− ∂

∂z(2yz) = −2y (1.3.3)

RHS =

ˆ y=2

y=0

ˆ z=2−y

z=0

−2y dz dy (1.3.4)

= −8

3(1.3.5)

Γ can be parametrized as

~r1(t) = 0 ı+ t + (2− t) k; 0 ≤ t ≤ 2 =⇒ d~r1(t) = dt − dt k (1.3.6)

~r2(t) = 0 ı+ 0 + t k; 0 ≤ t ≤ 2 =⇒ d~r2(t) = dt k (1.3.7)

~r3(t) = 0 ı+ t + 0 k; 0 ≤ t ≤ 2 =⇒ d~r3(t) = dt (1.3.8)

Each of ~r1, ~r2, ~r3 is a parametrization of one side of the triangle.

LHS =

ˆ 0

2

~v · (− k) dt+

ˆ 0

2

~v · k dt+

ˆ 2

0

~v · dt (1.3.9)

= −8

3= RHS (1.3.10)

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1.4. Compute the unit normal vector n to the ellipsoidal surfaces defined by constant values of

Φ(x, y, z) = V

(x2

a2+y2

b2+z2

c2

). What is n when a = b = c?

Solution: The unit normal vector to a surface specified by the equation Φ(x, y, z) = C isgiven by the gradient direction ∇Φ

=⇒ n =∇Φ

|∇Φ| =xa2ı+ y

b2+ z

c2k√

x2

a4+ y2

c4+ z2

c4

(1.4.1)

When a = b = c, n = x ı+y +z k√x2+y2+z2

= r, which is obvious because the ellipsoids become

spheres.

1.5. A force defined by ~F = A(y2i + 2x2j) is exerted on a particle which is initially at the originof the co-ordinate system. A is a positive constant. We transport the particle on a triangularpath 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 aconservative force?

Solution:

xz

y

(0,0,0)

(1,0,0)

(1,1,0)

Γ

We need the work done by the force =¸

Γ~F · d~Γ.

Γ can be parametrized as

~r1(t) = t ı+ 0 + 0 k; 0 ≤ t ≤ 1 =⇒ d~r1(t) = dt ı (1.5.1)

~r2(t) = 1 ı+ t + 0 k; 0 ≤ t ≤ 1 =⇒ d~r2(t) = dt (1.5.2)

~r3(t) = t ı+ t + 0 k; 0 ≤ t ≤ 1 =⇒ d~r3(t) = dt ı+ dt (1.5.3)

W =

ˆ 1

0

~F · ı dt+

ˆ 1

0

~F · dt+

ˆ 0

1

~F · (ı+ ) dt (1.5.4)

= A (1.5.5)

Since the work done in a closed loop is not zero, ~F cannot be conservative. The samecan be verified if we calculate ∇× ~F which will come out to be non-zero.

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(b) The particle is placed at rest right at the origin. Is this a stable situation? Give anyargument (mathematical, physical, intuitive) to justify the stability (or instability) of thissituation.

Solution: For the particle to be in a stable equilibrium, the force it experiences uponslight displacement in any direction has to be opposite to the displacement. So it canbe visualized that if we draw a very small sphere around the origin, the flux of ~F willbe negative. So ∇ · ~F should be negative at origin as well.

∇ · ~F = 0 + 0 = 0 (1.5.6)

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

sin4 θ.What is the total amount of charge?

Solution:

x

y

θ = 0

2R

θ = π4

θ = π2

θ = −π4

θ = −π2r = 0

r = 2R cos θ

The curve is a circle, as we can see, with −π2≤ θ < π

2

Charge =

¨circle

σ dA = σ0

ˆ θ=π2

θ=−π2

ˆ r=2R cos θ

0

r

Rsin4 θ r dr dθ (1.6.1)

=32σ0R

2

105(1.6.2)

1.7. Suppose that the height of a certain hill (in feet) is given by

h(x, y) = 10(2xy − 3x2 − 4y2 + 14x+ 10y + 40),

where x is the distance (in km) east, y the distance north of the closest town

(a) Where is the top of Sameer Hills located, and how high is it?

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Solution: The top of the hills will correspond to a maxima of h(x, y). And we knowthat ∇h = ~0 at the maxima. So we need to find points satisfying ∇h = ~0.

∇h = 0 =⇒ 2y − 6x+ 14 = 0; 2x− 8y + 10 = 0 (1.7.1)

=⇒ (x, y) = (3, 2) (1.7.2)

h(3, 2) = 710 (1.7.3)

So the top of Sameer hills is located 3 km east and 2 km north of the closest town, andit 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) = 100 ı+ 40 (1.7.4)

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 ı+ 2 )

1.8. The gradient operator ∇ behaves like a vector in “ some sense”. For example, divergence of acurl (∇.∇× ~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 )−∇2 ~F . 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 presentin both LHS and RHS

Don’t be confused by this. The convention only is to simply omit the∑

symbol. You canmentally insert the

∑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 = εijkAjBk ei (1.8.2)

where ei is the unit vector in i directionWhen the

∑symbol is reinserted, this simply means

∑i=x,y,z

∑j=x,y,z

∑k=x,y,z

εijkAjBkei

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Now we will represent each component of ∇, ∂∂i

by ∂i,

∇×(∇× ~F

)= εijk∂j

(∇× ~F

)kei = εijkεklm∂j∂lFm ei = εkijεklm∂j∂lFm ei (1.8.3)

= (δilδjm − δimδjl)∂j∂lFm ei (∵ εijkεilm = δjlδkm − δjmδkl)(1.8.4)

= (∂m∂iFm − ∂l∂lFi) ei (∵ δijAj = Ai) (1.8.5)

= ∂i(∂mFm) ei − ∂l∂l(Fi ei) (1.8.6)

= ∇(∇ · ~F )−∇2 ~F �(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 )−∇2 ~F . So this serves as a fake proof ofthe identity

2 Curvilinear Coordinates

2.1. Consider a planar surface S parallel to xy plane bounded by a closed curve C. Consider a vectorfield

~F = 0.5(−y ı+ x )

Prove that the value of the integral of this vector field along the curve C is exactly equal to thearea of the planar surface S. Using this find the area inside the curve x

23 + y

23 = 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 planeis another theorem called Green’s theorem).

˛C~F · d~l = Area(S) (2.1.1)

(2.1.2)

Thus we can use this to find the area enclosed inside any curve. Let C be the curve definedby x

23 + y

23 , and S be the region enclosed by this curve. Let us parametrize C using the

paremeter t

C := { (cos3 t, sin3 t) | 0 ≤ t < 2π } (2.1.3)

Area(S) =

˛C~F · d~l (2.1.4)

=1

2

ˆ 2π

t=0

(− sin3 t)(−3 cos2 t sin t) + (cos3 t)(3 sin2 t cos t) dt (2.1.5)

=3π

8(2.1.6)

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2.2. A ship sails from the southernmost point of India (6.75oN, 93.84oE) to the southernmost pointof 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 bythe path on Earth’s centre. For that we need to convert to Cartesian coordinates

~r1 = R sin θ1 cosφ1 ı+R sin θ1 sinφ1 +R cos θ1 k = R(−0.067 ı− 0.991 + 0.118 k)(2.2.1)

~r2 = R sin θ2 cosφ2 ı+R sin θ2 sinφ2 +R cos θ2 k = R(0.774 ı− 0.282 − 0.566 k)(2.2.2)

Now we can get the distance by Rθ′ = 6400 km · 1.409 = 9017 km

(b) If instead of sailing, one had travelled in an aeroplane - by what percentage would theshortest possible distance change?

Solution: The change in distance will only be due to the change in R, as R will beincreased by the cruising altitude of an aeroplane ≈ 10 km. So percentage change =

106400

= 0.15%

2.3. Compute the divergence of the vector field given by:

~v = r cos θ r + r sin θ θ + r sin θ cosφ φ

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 =1

hrhθhφ

(∂(vrhθhφ)

∂r+∂(hrvθhφ)

∂θ+∂(hrhθvφ)

∂φ

)(2.3.1)

=⇒ ∇ · ~v = 5 cos θ − sinφ (2.3.2)

Now we need to check the divergence theorem for the hemisphere

x

y

z

RS1

S2

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According to divergence theorem

˚V

∇ · ~v dV =

‹S

~v · d~S (2.3.3)

LHS =

˚V

∇ · ~v dV =5πR3

3(2.3.4)

=⇒ RHS =

‹S1

~v · d ~S1 +

‹S2

~v · d ~S2 = πR3 +2πR3

3=

5πR3

3= LHS (2.3.5)

2.4. Compute the line integral of ~F = (2xz + y, 2yz + 3x, x2 + y2 + 5). Use Stokes’ theorem tocompute ˛

C

~F · ~dr

C is any positively oriented curve on the surface of the circular cylinder of radius 1 with it’s axisalong 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 thesurface on the cylinder enclosed between C and C1, with the normal direction such that itmatches the orientation of CNow, by Stokes theorem

−˛C1~F · d~l +

˛C~F · d~l =

¨S∇× ~F · dn (2.4.1)

But,

∇× ~F = 2 k (2.4.2)

=⇒¨S∇× ~F · dn = 0 (2.4.3)

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Since dn · k = 0

=⇒˛C~F · d~l =

˛C1~F · d~l (2.4.4)

=⇒˛C1~F · d~l = 2π (2.4.5)

2.5. If ~a and ~b are constant vectors, φ(~r) = (~a× ~r) · (~b× ~r) is the potential over all space, then findthe electric field and charge density in spherical co-ordinates.[Hint: ~E = −∇φ, ρ = ε0∇ · ~E]

Solution: First 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)

= r2(~a ·~b)− (~a · ~r)(~b · ~r) (2.5.3)

Now to find ~E

~E = −∇φ (2.5.4)

= −2~r(~a ·~b) + (~b · ~r)~a+ (~a · ~r)~b (2.5.5)

Now to find ρ

ρ = ε0∇ · ~E (2.5.6)

= −4ε0(~a ·~b) (2.5.7)

2.6. A vector field is given by~v = ay i+ bx j

where a, b are constants.

(a) Find the line integral of this field over a circular path of radius R, lying in the xy planeand centered at the origin using

i. Plane polar

Solution: Let us take the anti-clockwise line integral. We first need to write ~v interms of polar coordinates.(

ı

)=

(cos θ − sin θsin θ cos θ

)(r

θ

)(2.6.1)

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~v = ay ı+ bx = (ay cos θ + bx sin θ) r + (−ay sin θ + bx cos θ) θ (2.6.2)

= r cos θ sin θ(a+ b) r + r(−a sin2 θ + b cos2 θ) θ (2.6.3)

y

x

r

θ

For a circle, r = R

=⇒ d~Γ = dr r + r dθ θ =

(r θ +

dr

dθr

)dθ = r dθ θ (2.6.4)

˛Γ

~v · d~Γ =

ˆ θ=2π

θ=0

R2(−a sin2 θ + b cos2 θ) 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

=⇒ d~Γ = dx ı+ dy =

(ı+

dy

dx

)dx =

(ı ∓ x√

R2 − x2

)dx (2.6.6)

˛Γ

~v · d~Γ =

ˆ x=−R

x=R

(a√R2 − x2 ı+ bx ) · (ı− x√

R2 − x2) dx (2.6.7)

+

ˆ x=R

x=−R(−a√R2 − x2 ı+ bx ) · (ı+

x√R2 − x2

) dx (2.6.8)

=

ˆ x=R

x=−R−2a√R2 − x2 +

2bx2

√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

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on the circle. Use cylindrical co-ordinate system to show that the Stokes theorem is validover its surface.

Solution:

x

y

z

L

According to Stokes theorem,

¨S

∇× ~v · d~S =

˛Γ

~v · d~Γ (2.6.11)

∇× ~v = (b− a) k = (b− a) z (2.6.12)

LHS =

¨S

∇× ~v · d~S =

ˆ φ=2π

φ=0

ˆ z=L2

z=L2��

��

�:0(b− a) z · ρ dφ dz (2.6.13)

+

ˆ φ=2π

φ=0

ˆ r=R

r=0

(b− a) z · z ρ dρ dφ (2.6.14)

= (b− a)πR2 = RHS (2.6.15)

2.7. Although the gradient, divergence and curl theorems are the fundamental integral theorems ofvector 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)

(b)´V

(∇ × ~v) dV = −¸S~v × d~S [Hint: Replace ~v by ~v × ~c, where ~c is a constant in the

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divergence theorem ]

Solution: ˚V

∇ · (~v × ~c) dV =

‹S

(~v × ~c) · d~S (2.7.5)

=⇒ ~c ·˚

V

T (∇× ~v) dV −��

��

��:0˚V

~v · (∇× ~c) dV = −~c ·‹S

~v × d~S (2.7.6)

=⇒˚

V

(∇× ~v) dV = −‹S

~v × d~S �

(2.7.7)

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

∇′ · ~F (~r′)

|~r − ~r′| dτ ′ +∇× 1

4π

˚V

∇′ × ~F (~r′)

|~r − ~r′| dτ ′

Solution: We have to directly apply Helmholtz Theorem

~F (~r) = −∇U(~r) +∇× ~A(~r) (3.1.1)

Where

U(~r) =1

4π

˚V

D(~r′)

|~r − ~r′| dτ′ (3.1.2)

~A(~r) =1

4π

˚V

~C(~r′)

|~r − ~r′| dτ′ (3.1.3)

D(~r) = ∇ · ~F (~r) (3.1.4)

~C(~r) = ∇× ~F (~r) (3.1.5)

Simply putting in U(~r) and ~A(~r) into 3.1.1 gives us exactly what we want

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(b) Derive the same expression for ~F (~r) using

~F (~r) =

˚Vdτ ′ ~F (~r′)δ3(~r − ~r′)

boundary of the integral is to be understood at ∞Hint: Use the following

(i) −4πδ3(~r − ~r′) = ∇2 1|~r−~r′|

(ii) ∇×∇× ~v = ∇∇ · ~f −∇2 ~f

(iii) ∇ 1|~r−~r′| = −∇′ 1

|~r−~r′|

(iv) ∇× ~F (~r′)|~r−~r′| = −~F (~r)×∇

(1

|~r−~r′|

)(v) Q2.7b

Solution:

~F (~r) =

˚Vdτ ′ ~F (~r′)δ3(~r − ~r′) (3.1.6)

= − 1

4π

˚Vdτ ′ ~F (~r′)∇2

(1

|~r − ~r′|

)Using (i) (3.1.7)

= − 1

4π∇2

˚V

~F (~r′)

|~r − ~r′| dτ′ ∵ ~r′ is constant wrt ∇ (3.1.8)

= − 1

4π∇(∇ ·(˚

V

~F (~r′)

|~r − ~r′| dτ′))

(3.1.9)

+1

4π∇×

(∇×

(˚V

~F (~r′)

|~r − ~r′| dτ′))

Using (ii) (3.1.10)

= − 1

4π∇(˚

V~F (~r′) · ∇

(1

|~r − ~r′|

)dτ ′)

(3.1.11)

+1

4π∇×

(˚V~F (~r′)×∇

(1

|~r − ~r′|

)dτ ′)∵ product rules and (iv) (3.1.12)

=1

4π∇(˚

V~F (~r′) · ∇′

(1

|~r − ~r′|

)dτ ′)

(3.1.13)

− 1

4π∇×

(˚V~F (~r′)×∇′

(1

|~r − ~r′|

)dτ ′)

Using (iii) (3.1.14)

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=1

4π∇(˚

V∇′ ·

(~F (~r′)

|~r − ~r′|

)dτ ′)

(3.1.15)

− 1

4π∇(˚

V

∇′ · ~F (~r′)

|~r − ~r′| dτ ′)

(3.1.16)

− 1

4π∇×

(˚V∇×

(~F (~r′)

|~r − ~r′|

)dτ ′)

(3.1.17)

+1

4π∇×

(˚V

∇× ~F (~r′)

|~r − ~r′| dτ ′)

∵ product rules of div and curl (3.1.18)

=1

4π∇(��

��:

0‹S

~F (~r′)

|~r − ~r′| · d~S ′

)(3.1.19)

− 1

4π∇(˚

V

∇′ · ~F (~r′)

|~r − ~r′| dτ ′)

(3.1.20)

+1

4π∇×

��

��

��:0‹S

(~F (~r′)

|~r − ~r′|

)× d~S ′

(3.1.21)

+1

4π∇×

(˚V

∇× ~F (~r′)

|~r − ~r′| dτ ′)

Using Div theorem and 2.7b (3.1.22)

= − 1

4π∇(˚

V

∇′ · ~F (~r′)

|~r − ~r′| dτ ′)

+1

4π∇×

(˚V

∇× ~F (~r′)

|~r − ~r′| dτ ′)

�

(3.1.23)

3.2. (a) Using the identity δ(ax) = δ(x)|a| , a 6= 0, prove that:

δ(g(x)) =∑

m s.t g(xm)=0, g′(xm)6=0

δ(x− xm)

|g′(xm)|

Solution: Whenever we want to prove equality of two expressions involving delta func-

17 of 48


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)

=∑

m s.t g(xm)=0

ˆ xm+ε0

x=xm−ε0f(x)δ(g(x)) dx (3.2.2)

=∑

m s.t g(xm)=0

ˆ ε0

ε=−ε0f(xm + ε)δ(g(xm) + εg′(xm) +O(ε2)) dε Taylor Expansion

(3.2.3)

=∑

m s.t g(xm)=0

ˆ ε0

ε=−ε0f(xm + ε)δ(εg′(xm) +O(ε2)) dε g(xm) = 0 (3.2.4)

=∑

m s.t g(xm)=0, g′(xm)6=0

ˆ ε0

ε=−ε0f(xm + ε)

δ(ε)

|g′(xm)| dε Using given identity

(3.2.5)

=∑

m s.t g(xm)=0, g′(xm)6=0

ˆ xm+ε0

x=xm−ε0f(x)

δ(x− xm)

|g′(xm)| dx (3.2.6)

=

ˆ ∞−∞

f(x)∑

m s.t g(xm)=0, g′(xm)6=0

δ(x− xm)

|g′(xm)| dx (3.2.7)

Since this works for any arbitrary ordinary function f(x), the expressions δ(g(x)) and∑m s.t g(xm)=0, g′(xm)6=0

δ(x−xm)|g′(xm)| are equal

(b) Confirm that I =´∞x=0

δ(cosx)e−x dx = 1

2 sinh(π2 )

Solution: We need xm s.t. cos(xm) = 0, − sin(xm) 6= 0Thus xm = (2n+ 1)π

2, n ∈ I

We have |g′(xm)| = 1 ∀ n ∈ IThus,

I =

ˆ ∞x=0

δ(cosx)e−x dx =∑n∈N

e−(2n+1)π2 (3.2.8)

=1

2 sinh(π2

) (3.2.9)

(c) Show that,

limm→∞

sin mx

πx

is a representation of δ(x) by showing that´∞−∞ dxf(x)D(x) = f(0)

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Solution:ˆRf(x)D(x) dx =

ˆ ∞x=−∞

f(x) limm→∞

sin(mx)

πxdx (3.2.10)

=

ˆ ∞y=−∞

limm→∞

f( ym

) sin y

πydy (3.2.11)

=

ˆ ∞y=−∞

f(0)sin y

πydy (3.2.12)

=f(0)

π

ˆ ∞y=−∞

sin y

ydy (3.2.13)

= f(0) (3.2.14)

Hence D(x) is equivalent to δ(x)

3.3. Show that D(~r,ε) demonstrates the peak-character & goes to δ3(~r) as ε→ 0

D(~r, ε) = − 1

4π∇2 1√

r2 + ε2

Hint:

(i) Show that D(~r, ε) = 3ε2

4π((r2+ε2)5/2)

(ii) Check that D(0, ε)→∞ as ε→ 0

(iii) Check that D(~r, ε)→ 0 as ε→ 0, for all r 6= 0

(iv) Check that the integral of D(~r, ε) over all the space is 1

Solution: First, we will evaluate D(r, ε) in terms of ~r and ε

D(~r, ε) = − 1

4π∇2 1√

r2 + ε2(3.3.1)

=⇒ D(~r, ε) = − 1

4π∇ · ∇

(1√

r2 + ε2

)(3.3.2)

=⇒ D(~r, ε) =3ε2

4π(r2 + ε2)52

(3.3.3)

Now, we need to show that as ε → 0 , D(r, ε) behaves like δ3(r). For an ordinary functionf(~r)

limε→0

˚all space

f(~r)D(r, ε)dτ ′ (3.3.4)

= limε→0

˚all space

3ε2f(~r)

4π(r2 + ε2)52

dτ ′ (3.3.5)

= f(0) =

˚all space

f(~r)δ3(r)dτ ′ (3.3.6)

Hence limε→0D(~r, ε) is equivalent to δ3(~r)

19 of 48


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 .

˚Vr · (d− r)δ3(e− r)dτ ′ = −620 (3.4.1)

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(~r′) = δ(x′)δ(y′), ~C(~r′) = ~0. Let us find U(~r) and ~A(~r) ofHelmholtz theorem

~A(~r) =

˚ ~C(~r′)

|~r − ~r′| dτ′ = 0 (3.5.1)

U(~r) =

ˆ ∞z′=−∞

1√x2 + y2 + (z − z′)2

dz′ (3.5.2)

This integral is divergent. (You can guess so, since δ(x)δ(y) represents the charge distributionof a charged wire along z axis, whose potential we know does indeed diverge when referenceis taken as infinity). Thus we directly attempt to find ~F by calculating the grad of thispotential.

=⇒ ~F (~r) = −∇U (3.5.3)

=2x

x2 + y2ı+

2y

x2 + y2 (3.5.4)

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 arean infinite number of points where a third ball with charge +2q may be positioned so that thefirst ball continues to remain vertical when released. Find the equation of the curve describingthese points.

Solution:

20 of 48


+q

+2q

(x, y)

a

+q

We basically want to balance all the forces on the hanging +q.We balance forces in the x direction

1

4πε0

q2

a2=

1

4πε0

2q2x√x2 + y2

3 (3.6.1)

=⇒ y = +

√(2a2x)

23 − x2 (3.6.2)

3.7. After an extremely precise measurement, it was revealed that the actual force between two pointcharges is given by

~F =1

4πε0

q1q2

r2

(1 +

r

λ

)e−

rλ r

Where λ is a constant with dimensions of length, such that λ� 1, hence the correction factoris negligible, but non zeroDoes this electric field results from a scalar potential? Justify.And if yes, find the potential due to a point charge q placed at the origin using infinity as yourreference.

Solution: Yes. The field of a point charge at the origin is radial and symmetric, so ∇ ×~E(~r) = ~0

U(~r) = −ˆ r

r′=∞~E(~r′) · d~l′ = − 1

4πε0q

ˆ r

r′=∞

1

r′2

(1 +

r′

λ

)e−

r′λ dr′ (3.7.1)

=q

4πε0

e−rλ

r(3.7.2)

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 ı+ 2xz − 3yz k

)Solution: We know that the curl of the electric field has to be ~0, so if the curl of thispurported electric field is nonzero, then it cannot be the expression of an electrostaticfield.

21 of 48


In this case,

∇× ~E 6= ~0 (3.8.1)

(b) ~E = A([

3xz2 + y2]ı+ 2xy + 3x2z k

)Solution: The curl of this ~E is ~0 (verify), so therefore, this is a possible expression foran electrostatic field.

Φ(x, y, z) =3

2x2z2 + xy2 (3.8.2)

3.9. Find the electric field a distance z above center of a square loop of side l carrying uniform linecharge 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 thetotal electric field.

Hence, to find the total electric field, we simply integrate over the z-component of the electricfield due to one side of the square loop and then multiply by 4:

Ez =4

4πε0

ˆdq sin θ

r2=

2zλ

πε0

l/2ˆ

0

dy((l2

)2+ z2 + y2

)3/2(3.9.1)

Therefore,

~F =zlλ

πε0

(l2

4+ z2

)√l2

2+ z2

k (3.9.2)

4 Electrostatic Fields, Potentials, Energy

4.1. Consider a conducting sphere A which is initially uncharged. Another conducting sphere B isgiven a charge +Q, brought into contact with A and then moved far away. The charge on B

22 of 48


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 limitQq

Q− q , where q is

the charge acquired by A after its first contact with B.

Solution: It is obvious that after each transfer of charge from B to A, A has k fraction oftotal charge of A and BSo qA → (. . .+ k3 + k2 + k)Q = Qk

1−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 ofthe hemisphere.

Solution: We know that the field at ~r due to a 3D charge distribution is given by

~E(r) =1

4πε0

˚V

ρ(~r′)

| ~ |2 ˆ dV ′ =1

4πε0

˚V

ρ(~r′)

| ~ |3 ~ dV ′ (4.2.1)

We will use spherical coordinates for convenience

Here ~r = ~0 (origin), so ~ = −r′ r = −r′(

sin θ cosφ ı+ sin θ sinφ + cos θ k)

=⇒ ~E(~0) =1

4πε0

˚V

ρ(~r′)

| ~ |3 ~ =1

4πε0

ˆ φ=2π

φ=0

ˆ θ=π2

θ=0

ˆ r′=R

r′=0

ρ

r′3(−r′ r) r′2 sin θ dr dθ dφ

(4.2.2)

= − ρ

4πε0��

��

��

��:0ˆ φ=2π

φ=0

ˆ θ=π2

θ=0

ˆ r′=R

r′=0

sin2 θ cosφ dr dθ dφ ı (4.2.3)

− ρ

4πε0��

��

��

��:0ˆ φ=2π

φ=0

ˆ θ=π2

θ=0

ˆ r′=R

r′=0

sin2 θ sinφ dr dθ dφ (4.2.4)

(4.2.5)

− ρ

4πε0

ˆ φ=2π

φ=0

ˆ θ=π2

θ=0

ˆ r′=R

r′=0

sin θ cos θ dr dθ dφ k (4.2.6)

= −ρR4ε0

k (4.2.7)

4.3. The potential takes the constant value φ0 on the closed surface S which bounds a volume V .The total charge inside V is Q. There is no charge anywhere else. Show that the electrostaticenergy contained in the space outside of S is UE(out) = Qφ0

2

23 of 48


Solution:

U =ε02

˚V(∇ · ~E)V dτ (4.3.1)

Use integration by parts to transfer the derivative from ~E to V , and apply Gauss law at theend

=⇒ U =ε02

(−˚V~E · (∇V ) dτ +

‹SV ~E · d~S

)(4.3.2)

U − UE(in) = UE(out) =φ0ε0

2

‹S~E · d~S (4.3.3)

=Qφ0

2� (4.3.4)

4.4. The inside of a grounded spherical metallic shell (inner radius R1 and outer radius R2) is filledwith 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.Hence, by symmetrical arguments, the potential and magnitude of electric field will beindependent of θ and φ coordinates.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 asphere of radius r. Use Gauss Law to get

=⇒ ~E(r) =1

ε0

(ar

3+br2

4

)

24 of 48


Taking a line integral to find potential

=⇒ V (0) =1

ε0

(aR2

1

6+bR3

1

12

)(4.4.1)

(b) Calculate the electrostatic energy of the system

Solution:

Energy =1

2

˚Vρ(~r)V (~r)dτ

=2πR5

1

ε0

(a2

45+abR1

36+b2R2

1

112

)

(c) Show that the system attains minimum energy configuration when b = −14a9R1

Solution:

For minimum energy, we differentiate wrt b

∂E

∂b=

2πR51

ε0

(aR1

36+bR2

1

56

)= 0

=⇒ b =−14a

9R1

(4.4.2)

4.5. Show that the field is uniquely determined when the charge density ρ is specified within abounded region and either the potential φ or its normal derivative ∂φ

∂nis 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|2 dτ (4.5.4)

=⇒‹

boundary

V3~E3 · n|dS| = −

˚| ~E3|2 dτ (4.5.5)

25 of 48


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 · n = ∂φ∂n− ∂φ

∂n= 0. Either way V3

~E3 · n = 0

=⇒˚| ~E3|2 dτ = 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

4πε0R

Field at centre = 0

(b) Calculate field at a height z above the center. Compare this result with Coulomb’s law inlarge z limit.

Solution: At height z above centre, field is

Qzz

4πε0(R2 + z2)32

As z →∞, E → Qz4πε0z2

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 chargewill execute simple harmonic motion for small displacements perpendicular to the plane ofthe ring. Calculate time period.

Solution: Force acting on the charged particle

F = − Q2z

4πε0(R2 + z2)32

Compare with force equation of SHM to get time period as

2π

√k

m=

2πR

Q

√4πε0Rm

26 of 48


4.7. Assume that an electron is a small sphere of radius R in which the charge −|e| is distributeduniformly over its volume. Calculate the total energy of the system as a function of R. Nowsuppose you equate this energy to m0c

2 where m0 is the rest mass of the electron. What valueof R do you get? How does it compare with the size of a hydrogen atom?

Solution: Energy equals3q2

20πε0R

Equating this to m0c2, we get

R =3e2

20πε0m0c2(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 zerocharge. Now connect the spheres together using a fine conducting wire. Assume that the spheresare separated by a distance R which is large enough that the charge distribution on each ballremains 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 wholeconductor system will become an equipotential

Vleft end = Vright end (5.1.1)

Q1

R1

+Q−Q1

R=Q−Q1

R2

+Q1

R(5.1.2)

=⇒ Q1 =(R−R2)R1

RR1 +RR2 − 2R1R2

(5.1.3)

R� R1, R2

=⇒ Q1 ≈ QR1

R1 +R2

(1 +

R2(R1 −R2)

R(R1 +R2)

)(5.1.4)

27 of 48


5.2. (a) Show that the capacitance, C, of a conducting sphere of radius a is given by C = 4πε0a

Solution: Assume a conducting sphere of radius a carrying a total charge Q, whichwill be distributed uniformly. Potential on surface of the sphere is

Q

4πε0a

Since our reference for potential measurement is with respect to infinity, potential atinfinity is 0 Hence, Capacitance is

C =Q

V= 4πε0a (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 untilequilibrium is reached, whereupon the wire is removed. Show that the total electrostaticenergy stored in the spheres decreases by an amount U , given by

∆U =1

16πε0a(q1 − q2)2

What happens to this energy?

Solution: Initial Charges on the spheres are q1 and q2. When the spheres are connectedby the wire, they become equipotential. Calculate the final charges.

Calculate energies to get

∆U =1

16πε0a(q1 − q2)2 (5.2.2)

As the charges move from one sphere to another under the influence of an electricfield(higher to lower potential), they gain kinetic energy. When the reach the secondsphere, this energy gets dissipated as heat. In other words, this energy gets dissipateddue to resistive heating of the wire

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: Let S represent the surface of the ”northern” hemisphere. Electrostatic PressureP = σ2

2ε0= Q2

32π2ε0R4

F =

‹SP (r) d ~A (5.3.1)

=⇒ F =Q2

32πε0R2(5.3.2)

28 of 48


5.4. A solid spherical conductor encloses 3 cavities, a cross-section of which are as shown in thefigure. A net charge +q resides on the outer surface of the conductor. Cavities A and C containpoint 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 thebulk 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 Gaussiansurface is 0. Hence,

Qinside cavity +Qon surface = 0

Hence charge on surface of cavity A = −qSimilarly, charge on surface of cavity C = +q

5.5. A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell(inner radius a, outer radius b). The shell carries no net charge.

(a) Find the surface charge density σ at R, at a, and at b.

Solution: σR = q4πR2 , as any leftover charge on conductor will distribute itself uni-

formly on the surfaceThis +q in the cavity of the shell induces a total charge of −q on the inner surface ofshell. Hence σa = −q

4πa2

The leftover charge on shell distributes itself uniformly on the outer surface. Henceσb = q

4πb2

(b) Find the potential at the center, using infinity as the reference point.

Solution: V (b) = q4πε0b

(The interior of shell is shielded)V (a) = V (b)V (R) = V (a) + q

4πε0

(1R− 1

a

)V (a) = V (R) = q

4πε0

(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?

29 of 48


Solution: The only difference is there will be no leftover charge +q on the outer surfaceof shell. So σb = 0Also, V (a) = V (b) = 0So V (0) = q

4πε0

(1R− 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 theplane, by energy conservation. Then integrate velocity inverse to get time

T =πd

q(2πε0md)

12

5.7. Two infinite parallel grounded conducting plates are held a distance a apart. A point charge qis placed between them, at a distance x from one plate. Find the force on q. Check that youranswer is correct for the special cases a→∞ and x = a

2.

Solution:

F = − q2

16πε0

{1

x2− 4ax

∞∑n=1

n

(n2a2 − x2)2

}(5.7.1)

Now when a→∞, the sum term tends to 0, and we are left with

F =q2

4πε0

1

(2x)2(5.7.2)

which is the same force as the case for a single plane.If x = a

2then we expect no force, and:

F =q2

16πε0

{4

a2− 2

a2

∞∑n=1

n(n2 − 1

4

)2

}= 0 (5.7.3)

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

30 of 48


Solution: Already calculated in 5.3

(b) Now suppose one has a solid sphere of radius R with charge Q distributed uniformly overits volume.

Solution: Let V represent the volume of the sphere.

~F =

˚Vρ ~E(r)dτ (5.8.1)

=⇒ F =3Q2

64πε0R2(5.8.2)

(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 theconductor, 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 qlocated inside. Show that, in general,

Vave = Vcenter +Qenc

4πε0R

where Vcenter is the potential at the center due to all the external charges and Qenc is thetotal enclosed charge.

Solution: The average value of V at a point r over a spherical surface of radius R isgiven as (assuming S represents the sphere surface)

Vave(r) =1

4πR2

‹SV 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 tothe externally located charges.The potential Vint due to the point charge q located inside the sphere at ~r′ = (0, 0, z)can be calculated in polar coordinates to be q

4πε0R

Thus from superposition principle, Vint = Qenc

4πε0R

For Vext, the potential is same as given at the center of the sphere Vcenter. So finally, wehave

Vave = Vcenter +Qenc

4πε0R

(b) Find the general solution to Laplace’s equation in spherical coordinates for the case whereV depends only on r. Do the same for cylindrical coordinates assuming V depends onlyon s.

31 of 48


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 =−cr

+ 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√

d+Rd−R .

Solution:

− qRd q

R2

d

d

III α

We need the ratio of charge on I and II

˜Iσ dS˜

IIσ dS

(5.10.1)

But remember, σ = ε0 ~En, because this is a conductor. Replacing this in the above ratio,we get

=

˜I~E · d~S˜

II~E · d~S

=FluxI

FluxII

(5.10.2)

Here we are talking about flux of the electric field just outside. We can calculate these fluxes

32 of 48


using solid angles to get

−q

(1−√d2−R2

d

)2ε0

− qR2ε0d

q

(1−√d2−R2

d

)2ε0

− qR2ε0d

=R + d−

√d2 −R2

R− d+√d2 −R2

=

√d+R

d−R �

(5.10.3)

5.11. Two infinite conducting plates (both grounded and perpendicular to the x− y plane) meet atan angle of 60°. A point charge +q in the xy plane has plane polar coordinates (a, 20°). Findall 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 metalsides at y = 0, y = a and x = 0. The fourth side at x = b is maintained at a specified potentialV0(y).

(a) Develop a general formula for the potential within the pipe.

Solution: We need to solve∂2V

∂x2+∂2V

∂y2= 0

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 = −Ak =

nπ

a

33 of 48


Using the above we can write the most general form of the solution as

V (x, y) =∞∑n=1

Cn sinh(nπx/a) sin(nπy/a)

wherein Cn can be determined using the boundary condition V (b, y) = V0(y) and isgiven 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 2anπ

when n is odd. So, finally wewill get

V (x, y) =4V0

π

∑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 bymetal strips maintained at a constant potential V0, (a thin layer of insulation at each cornerprevents them from shorting out). Find the potential inside the resulting rectangular pipe.

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) =∑k

(Akekx +Bke

−kx)(Ck cos ky +Dk sin ky) (6.1.6)

34 of 48


Apply the boundary conditions to get,

V (x, y) =∞∑n=1

2DnAn cosh(nπ

x

a

)sin(nπ

y

a

)(6.1.7)

Find the coefficients using Fourier’s trick

V0 =∞∑n=1

Kn cosh

(nπ

b

a

)sin(nπ

y

a

)(6.1.8)

Using Fourier’s trick

Kn =

{0 n even4V0nπ

1

cosh(nπ ba)n odd

(6.1.9)

Putting this back into (6.1.7),

V (x, y) =4V0

π

∑n=1,3,5...

cosh(nπ x

a

)n cosh

(nπ b

a

) sin(nπ

y

a

)(6.1.10)

We can indeed verify that this does satisfy the boundary conditions and the Laplace’sequation, 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. Thesixth plate at the top is insulated from the rest and maintained at V0.

(a) Argue that the potential at the centre should be V06

Solution: Let us consider 6 seperate cases, where in the ith case the ith face is atpotential Vi and all others are at 0 potential (the charges on the faces arrange in sucha 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 ofthe charges on each face. Thus the potential at the centre must be proportional to Viin the ith case. Hence if we take a superposition of all cases, we can easily see that theresultant scenario in which ith face is at Vi must have the potential at centre ∝∑i Vi.If we take the case of all Vi = V0, we can see that the proportionality constant is 1

6

Hence the potential at centre is V06

when one face is held at V0 and all others at 0

35 of 48


(b) Find the potential inside the box.


∂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 ≤ aUsing separation of variables we know that the general solution can be written as

V (x, y, z) =∑k,l

(Ak cos kx+Bk sin kx)(Cl cos ly +Dl sin ly)(Ekle√k2+l2z +Kkle

−√k2+l2z)

(6.2.8)

Simplify (indexing the coefficients with m,n instead of k, l)

V (x, y, z) =∞∑n=1

∞∑m=1

2BnDmEnm sin(nπ

x

a

)sin(mπ

y

a

)sinh

(√n2 +m2π

z

a

)(6.2.9)

In order to solve this, we use Fo

V0 =∞∑n=1

∞∑m=1

Knm sin(nπ

x

a

)sin(mπ

y

a

)sinh

(√n2 +m2π

)(6.2.10)

Applying Fourier’s trick,

=⇒ Knm =

{0 n even or m even16V0π2nm

1

sinh(√n2+m2π)

n odd and m odd(6.2.11)

Thus we can write the solution as

V (x, y, z) =16V0

π2

∑n=1,3,5...

∑m=1,3,5...

sinh(√

n2 +m2π za

)nm sinh

(√n2 +m2π

) sin(nπ

x

a

)sin(mπ

y

a

)(6.2.12)

Summing it up numerically upto 49 terms gives us 0.166666666676V0 ≈ V06

, confirmingour “guess”

6.3. A rectangular pipe, running parallel to the z-axis (from −∞ to +∞), has three grounded metalsides, at y = 0, y = a, and x = 0. The fourth side, at x = b, is maintained at a specifiedpotential V0(y). Develop a general formula for the potential inside the pipe.

36 of 48



∂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 (and simplify using boundaryconditions)

V (x, y) =∞∑n=1

Kn sinh(nπ

x

a

)sin(nπ

y

a

)(6.3.6)


Kn =2

a

1

sinh(nπ b

a

) ˆ a

y=0

V0(y) sin(nπ

y

a

)dy (6.3.7)

We can finally write the solution as

V (x, y) =2

a

∞∑n=1

´ ay=0

V0(y) sin(nπ y

a

)dy

sinh(nπ b

a

) sinh(nπ

x

a

)sin(nπ

y

a

)(6.3.8)

6.4. Two infinitely long metal plates at y = 0 and y = a are connected at x = ±b by metal stripsmaintained at a constant potential V0. The potential on the bottom (y = 0) is zero, howeverthe potential on the top (y = a) is a nonzero constant V1. A thin layer of insulation at eachcorner prevents the plates from shorting out. Find the potential inside the resulting rectangularpipe.

Solution: To solve this we need to break this up into 2 situations

x

y

−b

a

−b

V0V0

0

0x

y

−b

a

−b

00

V1

0

Case I Case II

37 of 48


We already know the solution to Case I (from Q6.1),

VI(x, y) =4V0

π

∑n=1,3,5...

cosh(nπ x

a

)n cosh

(nπ b

a

) sin(nπ

y

a

)(6.4.1)

Solve for Case II (Hint: Use a shift of coordinates to make the calculations easier):

VII(x, y) =4V1

π

∑n=1,3,5...

sinh(nπ y

2b

)n sinh

(nπ a

2b

) sin

(nπ

x+ b

2b

)(6.4.2)

We finally have

V (x, y) = VI + VII (6.4.3)

=4V0

π

∑n=1,3,5...

cosh(nπ x

a

)n cosh

(nπ b

a

) sin(nπ

y

a

)(6.4.4)

+4V1

π

∑n=1,3,5...

sinh(nπ y

2b

)n sinh

(nπ a

2b

) sin

(nπ

x+ b

2b

)(6.4.5)

6.5. Consider a spherical surface of a large radius R, the potential on the spherical surface is givenbelow. 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 π2≤ 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 π2≤ θ ≤ π

Boundary Conditions (6.5.2)

limr→∞

V (r, θ) = 0 (6.5.3)

in the region r ≥ R

Using separation of variables we can write the general solution as

V (r, θ) =∞∑l=0

(Alr

l +Bl

rl+1

)Pl(cos θ) (6.5.4)

38 of 48


Using (6.5.3), Al = 0 ∀ l


ˆ π

θ=0

V (R, θ)Pl′(cos θ) sin θ dθ =∞∑l=0

Bl

Rl+1

ˆ π

θ=0

Pl(cos θ)Pl′(cos θ) sin θ dθ (6.5.5)

=⇒ Bl =

{0 l even

(2l + 1)V Rl+1´ 1

x=0Pl(x) dx l odd

(6.5.6)

We only need to calculate upto l = 5 (O( 1r6

))

V (r, θ) = V

(3R2

2r2P1(cos θ)− 7R4

8r4P3(cos θ) +

11R6

16r6P5(cos θ)

)+O(

1

r8) (6.5.7)

6.6. Let a sphere of radius R have potential V (R, θ, φ) = V0 cos2 θ. Find the potential everywhereinside 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


Vout(R, θ) = V0 cos2 θ Boundary Conditions (6.6.2)

limr→∞

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) =∞∑l=0

(Bl

rl+1

)Pl(cos θ) (6.6.4)

Solve for coefficients by applying boundary conditions

Comparing the coefficients of like degrees of cos θ,

B2 =2V0R

3

3(6.6.5)

B1 = 0 (6.6.6)

B0 =V0R

3(6.6.7)

=⇒ Vout(r, θ) =V0R

3r

(1 +

R2

r2

(3 cos2 θ − 1

))(6.6.8)

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We can similarly find Vin(r, θ),

Vin(r, θ) =V0

3

(1 +

r2

R2

(3 cos2 θ − 1

))(6.6.9)

6.7. An uncharged, conducting sphere of radius R is placed in a region where the electric field isuniform i.e. ~E = ~E0. Can you guess whether l value will be odd or even in potential outsidethe sphere? Find the electric field in the region after the sphere is put in place.

Solution: We can begin by observing that charges will distribute themeselves such that thecharge density function will be odd in cos θ, and hence l will also be oddWLOG, Let the external electric field be along k, thus ~E0 = E0k. In spherical polarcoordinates, ~E0 = E0 cos θ r − E0 sin θ θThis time we must be careful in setting the reference of potential, since we cannot set it toits 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 intuitionsregarding absolute potentials go out the window. Any potentials we must specify must alsobe 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 thewhole xy plane will also be at 0 potential wrt the sphere (∵ due to symmetry, ~E · r = 0 onthe 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


V (R, θ) = 0 Boundary Conditions (6.7.2)

limr→∞

V (r, θ) = −E0r cos θ (6.7.3)

in the region r ≥ R Solve using separation of variables

V (r, θ) = −E0

(r − R3

r2

)cos θ (6.7.4)

Of course, remember that this potential is in reference to the sphereTo find electric field,

~E = −∇V (6.7.5)

=(E0 cos θ r − E0 sin θ θ

)+ E0

R3

r3

(2 cos θ r + sin θ θ

)(6.7.6)

We can see that in addition to ~E0, a dipole field is induced by the sphere

6.8. Suppose the potential V0(θ) at the surface of the sphere is specified, and there is no charge

40 of 48


outside or inside the sphere. Show that charge density on the surface of the sphere is given by

σ(θ) =ε02R

∞∑l=0

(2l + 1)2ClPl(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 byFourier’s trick)


´ πθ=0


rl+1

(2l + 1)Rl+1

2Pl(cos θ) (6.8.1)

=∞∑l=0

Clrl+1

(2l + 1)Rl+1

2Pl(cos θ) (6.8.2)

Vin(r, θ) =∞∑l=0

´ πθ=0


Rl

(2l + 1)rl

2Pl(cos θ) (6.8.3)

=∞∑l=0

ClRl

(2l + 1)rl

2Pl(cos θ) (6.8.4)

We also know that

σ(θ)

ε0= −∂Vout

∂r

∣∣∣∣r=R

+∂Vin

∂r

∣∣∣∣r=R

(6.8.5)

=∞∑l=0

(2l + 1)ClPl(cos θ)

2

(l

R+l + 1

R

)(6.8.6)

=1

2R

∞∑l=0

(2l + 1)2ClPl(cos θ) (6.8.7)

=⇒ σ(θ) =ε02R

∞∑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 + cos2 θ)

Using solutions of Laplace’s equation, find the potential V (r, θ) everywhere, both for r > R andr < R.

41 of 48


Solution: As before, we know that,


Bl

rl+1Pl(cos θ) (7.1.1)

Vin(r, θ) =∞∑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) = ε0∂Vin(r, θ)

∂r

∣∣∣∣r=R

− ε0∂Vout(r, θ)

∂r

∣∣∣∣r=R

(7.1.3)

Comparing coefficients,

A2 =2σ0

15Rε0(7.1.4)

A1 =σ0

3ε0(7.1.5)

A0 =σ0R

3ε0(7.1.6)

Thus we can write the potentials

Vout =σ0R

2

3ε0r+σ0R

3

3ε0r2cos θ +

σ0R4

15ε0r3

(3 cos2 θ − 1

)(7.1.7)

Vin =σ0R

3ε0+σ0r

3ε0cos θ +

σ0r2

15ε0R

(3 cos2 θ − 1

)(7.1.8)

7.2. In the following system (see figure), the inner conducting sphere of radius a carries charge Qand the outer sphere of radius b is grounded. The distance between the centres is c which is asmall quantity.

c

r b

θ

(a) Show that to the first order in c, the equation describing the outer sphere, using the centreof inner sphere as origin, is r(θ) = b+ c cos θ.

42 of 48


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 +√b2 − c2 sin2 θ (7.2.2)

= c cos θ + b

√1− c2 sin2 θ

b2(7.2.3)

= c cos θ + b

(1− c2 sin2 θ

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.

Solution: As before, we have

V (r, θ) =∞∑l=0

(Alr

l +Bl

rl+1

)Pl(cos θ) (7.2.6)

∂V (a, θ)

∂θ= 0 Inner sphere equipotential (7.2.7)

−ε0ˆ

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)

4πε0B0 +∞∑l=1

al+1(2l + 1)Al

ˆ 1

x=−1

Pl(x) dx = Q (7.2.11)

=⇒ B0 =Q

4πε0(7.2.12)

43 of 48


Applying (7.2.9)

0 = A0 +B0

(b+ c cos θ +O(c2)

)−1(7.2.13)

+(A1

(b+ c cos θ +O(c2)

)+B1

(b+ c cos θ +O(c2)

)−2)

cos θ +O(cos2 θ) (7.2.14)

=⇒ A0 +B0

b+

(A1b+

B1

b2− B0c

b2

)cos θ +O(c2) +O(cos2 θ) = 0 (7.2.15)

Solving upto first order in c and equating coefficients of 1 and cos θ,

A0 = − Q

4πε0b(7.2.16)

A1 = − Q

4πε0(b3 − a3)(7.2.17)

Putting it all together,

V (r, θ) =Q

4πε0

(1

r− 1

b− c

b3 − a3

(r − a3

r2

)cos θ

)+O(c2) +O(cos2 θ) (7.2.18)

7.3. Static charges are distributed along the x axis (one-dimensional) in the interval −a ≤ x′ ≤ a.The charge density is : {

ρ(x′) −a ≤ x′ ≤ a

0 |x′| > a

(a) Write down the multipole expansion for the electrostatic potential φ(x) at a point x on theaxis in terms of ρ(x′), valid for x > a

Solution:

φ(x) =1

4πε0

ˆ a

x=−a

ρ(x′)

|x− x′|dx′ (7.3.1)

=∞∑l=0

1

4πε0

ˆ a

x=−a

ρ(x′)x′l

xl+1dx′ (7.3.2)

(b) For each charge configuration given in below figure, find (a) total charge Q =´ρ(x′) dx′,

(b) dipole moment P =´x′ρ(x′) dx′

(I)

q

x′ = 0 x

(II)

−q

x′ = −a2

+q

x′ = a2

x

44 of 48


(III)

+q

x′ = −a2

−2q

x′ = 0

+q

x′ = a2

x

Solution: (I) Q = q, P = 0(II) Q = 0, P = qa(III) Q = 0, P = 0

7.4. A circular disc of radius R lies in the z = 0 plane, centred at the origin. It has the followingcharge density frozen on

σ(r′, φ) = σ0r′ cos(φ)

(a) What is the monopole moment of the configuration?

Solution:

Q =

ˆall space

ρ dτ (7.4.1)

= σ0

ˆ R

r′=0

ˆ 2π

φ=0

r′2 cosφ dr′ dφ (7.4.2)

= 0 (7.4.3)

(b) Calculate the dipole contribution to the potential due to the configuration at (0, 0, z) usingthe expression in polar form

Solution: We know that θ′ = π2, so cos θ′ = 0, therefore dipole contribution is also 0

(c) Now calculate the the cartesian components of the dipole moment of the con

guration. Use this to calculate the dipole contribution at (0, 0, z). Verify your answer withthe expression obtained in (b)

Solution:

~P =

ˆdisc

~rρ(r′) dτ ′ (7.4.4)

= σ0

ˆ R

r′=0

ˆ 2π

φ=0

r′3(cosφ ı+ sinφ ) cosφ dr′ dφ (7.4.5)

=πσ0R

4

4ı (7.4.6)

V (0, 0, z) = 14πε0z2

πσ0R4

4ı · k = 0

7.5. If the total amount of charge (monopole) contained in a distribution is zero, show that thedipole moment is independent of the choice of the origin.

45 of 48


Solution:

~P =

ˆ~r′ρ(~r′) dτ ′ (7.5.1)

~r′ → ~r′ − ~a

~P1 =

ˆ~r′ρ(~r′) dτ ′ − ~a

ˆρ(~r′) dτ ′ (7.5.2)

=

ˆ~r′ρ(~r′) dτ ′ = ~P (7.5.3)

7.6. Find the dipole moment of:

(a) A ring with charge per unit length λ = λ0 cosφ where φ is the angular variable in cylindricalcoordinates.

Solution:

~P = λ0R2

ˆ 2π

φ=0

(cosφ ı+ sinφ ) cosφ dφ (7.6.1)

= λ0πR2 ı (7.6.2)

(b) A sphere with charge per unit area σ = σ0 cos θ where θ is the polar angle measured fromthe positive z axis.

Solution:

~P = σ0R3

ˆ π

θ=0

ˆ 2π

φ=0

(sin θ cosφ ı+ sin θ sinφ + cos θk) cos θ sin θ dθ dφ (7.6.3)

=4π

3σ0R

3 k (7.6.4)

7.7. An electric dipole of moment ~P = (Px, 0, 0) is located at the point (x0, y0, 0) where x0 > 0 andy0 > 0. The planes x = 0 and y = 0 are conducting plates with a tiny gap at the origin. Thepotential of the plate at x = 0 is maintained at V0 and the plate at y = 0 is grounded. Thedipole is sufficiently weak so that you can ignore the charges induced on the plates.

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φ = V0

φ = 0

(x0, y0)

~P

φ =constant

θ

(a) Based on the figure, deduce a simple expression for the electrostatic potential φ(x, y).


1

ρ

∂

∂ρρ∂φ(ρ, θ)

∂ρ+

1

ρ2

∂2φ(θ)

∂θ2= 0 Laplace’s Equation (7.7.1)

φ(ρ, 0) = 0 Boundary Conditions (7.7.2)

φ(ρ,π

2) = V0 (7.7.3)

in the region 0 ≤ θ ≤ π2

We can write the general solution using separation of variables (the linear solution inθ no longer needs to be periodic since our domain is no longer the full range of θ)

φ(ρ, θ) = (A0 +B0 ln ρ)(C0 +D0θ) +∞∑m=1

(Amρm +Bmρ

−m)(Cm cosmθ +Dm sinmθ)

(7.7.4)

Applying (7.7.2) and (7.7.3)

B0 = 0 (7.7.5)

Cm = 0∀m (7.7.6)

Dm = 0∀m > 0 (7.7.7)

(7.7.8)

We are left with

φ(ρ, θ) = A0D0θ (7.7.9)

Applying (7.7.3) gives us A0D0 = 2V0π

=⇒ φ(θ) =2V0

πθ (7.7.10)

=⇒ φ(x, y) =2V0

πtan−1

(yx

)(7.7.11)

(b) Calculate the force on the dipole

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Solution:

~E = −∇φ (7.7.12)

= −1

ρ

2V0

πθ (7.7.13)

=2V0y

(x2 + y2) πı− 2V0x

(x2 + y2) π (7.7.14)

~F = (~P · ∇) ~E (7.7.15)

= − 4V0x0y0Px

(x20 + y2

0)2πı+

(− 2V0Px

(x20 + y2

0)π+

4V0x20Px

(x20 + y2

0)2π

) (7.7.16)

=2V0Px

(x20 + y2

0)2π

(2x0y0 ı+ (x2

0 − y20) )

(7.7.17)

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PH 108 : Electricity & Magnetism Hints & Answers

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