Essential Electromagnetism Solutions

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Essential Electromagnetism: Solutions

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Raymond John Protheroe

Essential Electromagnetism

Solutions

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Essential Electromagnetism: Solutions

First edition

ISBN 978-87-403-0404-6

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4

Contents

Contents

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5

Preface

Preface

This book gives the solutions to the exercises at the end of each chapter of my book Essential Electromagnetism (also published by Ventus) I recommend that you attempt a particular exercise after reading the relevant chapter, and before looking at the solutions published here Often there is more than one way to solve a problem, and obviously one should use any valid method that gets the result with the least effort Usually this means looking for symmetry in the problem for example from the information given can we say that from symmetry arguments the field we need to derive can only be pointing in a certain direction If so, we only need to calculate the component of the field in that direction, or we may be able to use Gauss law or Amp re s law to enable us to write down the result In some of these exercise solutions the simplest route to the solution is deliberately not taken in order to illustrate other methods of solving a problem, but in these cases the simpler method is pointed out

The solutions to the exercise problems for Each chapter of Essential Electromagnetism are presented here in the corresponding chapters of Essential Electromagnetism - Solutions

I hope you ﬁnd these exercises useful If you ﬁnd typos or errors I would appreciate you letting me know Suggestions for improvement are also welcome please email them to me at protheroe.essentialphysics@gmail.com

Raymond J Protheroe, January 2013

School of Chemistry & Physics, The University of Adelaide, Australia

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6

Electrostatics

1 Electrostatics

1 1 The surface of a non-conducting sphere of radius a centred on the origin has surface charge density σ(a, θ, φ) = σ0cos θ and is uniformly ﬁlled with charge of density ρ0 Find the electric ﬁeld at the origin

Solution

z

d

dS=r dS

E

O

φ

d

x

φ

θ

d θ

At the centre of the sphere the electric ﬁeld due to the volume charge will be zero because

the contribution of a volume element located at r′ will be exactly cancelled by that of an

equivalent volume element at −r′, so we only need to consider the surface charge

E(0, θ, φ) = 1

4πε0

∫ σ(a, θ, φ)

= 1 4πε0a2

∫ 2π 0

∫ π 0

σ(a, θ, φ)(−�r)[a2sin θdθdφ], (1.2)

= 1 4πε0

∫ 2π 0

dφ

∫ 1

−1

dcos θ(σ0cos θ)(−�r). (1.3)

Because of the symmetry of the problem, the electric ﬁeld at the centre can only be in the

±z direction, and so we only need to ﬁnd the z-component

E(0, θ, φ) · �z = 4πε1

0

∫ 2π 0

dφ

∫ 1

−1

dcos θ(σ0cos θ)(−�r) · �z, (1.4)

= 1 4πε02π

∫ 1

−1

dcos θ(σ0cos θ)(− cos θ), (1.5)

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7

Electrostatics

∴ E(0, θ, φ) · �z = − σ0

2ε0

∫ 1

−1

dcos θ cos2θ = − σ0

∴ E(0, θ, φ) = − σ0

1 2 A spherically symmetric charge distribution has the following charge density proﬁle

ρ(r, θ, φ) =

{

ρ0 (r < a)

where β is a constant (2 < β < 3) Find the electric ﬁeld and electrostatic potential everywhere

Solution

The charge density is spherically symmetric, with no dependence on θ or φ, so the electric ﬁeld must be in the radial direction and depend only on r This is the ideal case to exploit Gauss law in integral form,

∮

E · dS = 1

ε0

∫

For r < a

4πr2Er= 1

ε0

4

3πr

3ρ0, ∴ E(r) = ρ0r

For r > a

4πr2Er = 1

ε0

4

3πa

ε0

∫ r a

4π(r′)2ρ0aβ(r′)−βdr′, (1.11)

= 1

ε0

4

3πa

3ρ0 + 4πρ0a

β

ε0

[ (r′)3−β

3 − β

]r a

∴ 4πr2Er = 4πρ0a

3

ε0

( 1

3 +

(r/a)3−β−1

3 − β

)

∴ Er = ρ0a

3

(3 − β)ε0r2

((r a

)3−β

−β 3

)

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8

Electrostatics

This electric ﬁeld is due entirely to the charge distribution, and so must be conservative,

and we would expect that ∇ × E = 0 as E is directed radially outward and so has no

circulation It follows that:

V(r ≥ a) = − ρ0a

3

(3 − β)ε0

∫ r

∞

( (r′

)1−βaβ−3−β(r′

)− 2

3

)

dr′

= − ρ0a

3

(3 − β)ε0

[ (r′

)2−βaβ−3

2 − β +

β(r′

)− 1

3 ]r

∞

= ρ0a

3

3(3 − β)(β − 2)ε0

( 3r2−βaβ−3−β(β − 2)r− 1) (1.17)

V(r ≤ a) = V (a) − ρ0

3ε0

∫ r a

r′

dr′

= (3 + 2β − β

2)ρ0a2 3(3 − β)(β − 2)ε0 −

ρ0 3ε0

[ (r′

)2 2 ]r a

= (3 + 2β − β

2)ρ0a2 3(3 − β)(β − 2)ε0

+ ρ0 6ε0

(a2−r2) (1.20)

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Electrostatics

1 3 The electric ﬁeld is given by E(r) = E0cos(z/z0)exp(−r/r0)�r, where z0 and r0 are con-stants Find the charge density

Solution

In this problem the electric ﬁeld is given in terms of z and r We will need to write E in

terms of either Cartesian or spherical coordinates, and then use Gauss law in diﬀerential

form Choosing spherical coordinates because E is in the radial direction,

= ε0E0 1

r2

∂

∂r

[

r2cos(rcos θ

z0

) exp(− r

r0

)]

= ε0E0

r2

[ 2rcos(rcos θ

z0

) exp(− r

r0

)

−r2sin(rcos θ

z0

) cosθ

z0 exp(− r

r0

) +

r2cos(rcos θ

z0

) exp(− r

r0

) (− 1

r0

)]

= ε0E0

r cos(rcos θ

z0

) exp(− r

r0

) [

2 −tan(rcos θ

z0

)

rcos θ

z0 −

r

r0

] , (1.24)

= ε0E0

r cos(z

z0

) exp(− r

r0

) [

2 −tan( z

z0

) z

z0 −

r

r0

]

1 4 If we had a point charge q at the origin we might choose the reference point to be some point at an arbitrary distance r0 (usually inﬁnity) from the origin Then if we wish to

ﬁnd V (r, θ, φ) it would be convenient to have the reference point at r0 = (r0, θ, φ) Al-though obtaining the potential in this case is trivial, and one would usually just write it down, obtain the potential by carrying out explicitly the line integral for an appropriately parameterised curve

Solution

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Electrostatics

= r − r’

path

x

y

Γ

0

z

r

r’

r E

q

= r − r

We start by parameterising the path from r0 to r:

r′(λ) = (r0−λ)�r; dr′= −dλ�r; (0 < λ < r0−r) (1.26) Then,

V(r) = −∫ r

r0

= −

∫ r(λ=r0 −r)

= −

∫ r(λ=r0 −r)

r(λ=0)

q 4πε0(r0−λ)2�r · (−dλ �r), (1.29)

=

∫ r0−r 0

q 4πε0(r0−λ)2dλ, (1.30)

=

[

q 4πε0(r0−λ)

]r0−r 0

4πε0[r0−(r0−r)] −

q 4πε0(r0−0), (1.32)

= q 4πε0r −

q

Hence, if we set r0= ∞we get the usual potential for a point charge q at the origin

V(r) = q

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