ECE221

ECE221: Electric and Magnetic FieldsMidterm- Tuesday, February 23 2016

Instructors: Tome Kosteski (LEC01, LEC02)Costas Sarris (LEC03)

Last name: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

First name: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Student number: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Tutorial section number: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Instructions

• Duration: 1 hour 30 minutes (6:10 pm to 7:40 pm)

• Exam Paper Type: A. Closed book. Only the aid sheet provided at the end of this booklet is permitted.

• Calculator Type: 2. All non-programmable electronic calculators are allowed.

• Only answers that are fully justified will be given full credit!

Marks: Q1: /20 Q2: /20 Q3: /20 TOTAL: /60

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Question 1

Consider, in spherical coordinates, the volume charge density:

𝜌𝑣(𝑅) =

{𝑘0𝑅, 𝑅 ≤ 𝛼

0, 𝑅 > 𝛼

Moreover, a constant surface charge density 𝜌𝑠 = 𝜎0 (C/m2) exists on the surface of the sphere 𝑅 = 𝛼. Theconstants 𝑘0 and 𝜎0 are both positive (> 0). Throughout space 𝜖 = 𝜖0.

1. Using Gauss’ law determine the electric field E everywhere. State and justify your assumptions on thedirection of the electric field and its dependence on the spherical coordinates 𝑅, 𝜃, 𝜑.

You may need the integral∫𝑅𝑝𝑑𝑅 =

𝑅𝑝+1

(𝑝 + 1). (8 pts)

2. Using the expression for E that you derived, find the electric potential 𝑉 everywhere (8 pts). Sketch theequipotential surfaces (2 pts) and confirm that the electric field E points in the direction of decreasingpotential (1 pt) and that it is perpendicular to the equipotential surfaces (1 pt).

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Summary of your results for 1.1:

E(𝑅 < 𝛼) =

E(𝑅 > 𝛼) =

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Summary of your results for 1.2:

𝑉 (𝑅 < 𝛼) =

𝑉 (𝑅 > 𝛼) =

Equipotential surfaces:

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Question 2

Consider the parallel plate capacitor shown in the figure below. The capacitor is half-filled with a dielectric ofrelative dielectric permittivity 𝜖𝑟,1; its other half is filled with a dielectric of relative dielectric permittivity 𝜖𝑟,2.The electric field throughout the capacitor is E = −𝐸0z, where 𝐸0 is a positive constant.

1. Confirm that the given field satisfies the electric field boundary conditions at the interface between thetwo dielectrics. (4 pts)

2. Show that the voltage between the plates is 𝑉 (𝑧 = ℎ) − 𝑉 (𝑧 = 0) = 𝐸0 · ℎ. (4 pts)

3. Use the boundary conditions at 𝑧 = 0 to show that the total charge on the upper plate of the capacitor is𝑄 = 𝜖0 · 𝐸0 · (𝜖𝑟,1 + 𝜖𝑟,2) · 𝑑 · 𝑤. (6 pts)

4. Using the results above, find the capacitance 𝐶 (3 pts) and the total electrostatic energy stored in thecapacitor (3 pts).

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Summary of your results for 2d:

𝐶 =

𝑊𝑒 =

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Question 3

The following questions are equally weighted (4 pts each) and independent from each other.

1. A fundamental component of printed circuit boards is the microstrip line.

Approximating the microstrip as a surface (−𝑤/2 ≤ 𝑥 ≤ +𝑤/2,−𝐿/2 ≤ 𝑦 ≤ 𝐿/2, 𝑧 = 0), charged withsurface charge density 𝜌𝑠(𝑥, 𝑦), the electric field it creates at an observation point 𝑃 (0, 0, 𝑧) along the𝑧−axis is:

E =1

4𝜋𝜖

∫ 𝑤/2

−𝑤/2

∫ 𝐿/2

−𝐿/2

𝑑𝑄(𝑥′, 𝑦′)|R−R′|3

(R−R′)

Fill out the details of this formula and briefly explain:

𝑑𝑄 =

R−R′ =

|R−R′| =

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2. Gauss’ law in dielectrics can be written as:∮

𝑆𝜖0E · dS = 𝑄

The charge on the right hand-side is (choose the correct answer and briefly explain):

a) Bound (polarization) charge enclosed by the surface 𝑆.

b) Free charge enclosed by the surface 𝑆.

c) Total (free and bound) charge enclosed by the surface 𝑆.

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3. Consider the electric current density J in a medium of uniform conductivity 𝜎. Then, the closed path lineintegral

∮𝐶 J · dl is (choose the correct answer and briefly explain):

a) Non-zero.

b) Zero.

c) We need to know the dielectric permittivity of the medium to answer this question.

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4. An empty box with metallic (perfectly conducting) walls is used to shield (protect) the circuit Σ from theelectrostatic field of a charged sphere. Two configurations are proposed. Shielding is achieved (choosethe correct answer and briefly explain):

a) In case (a).

b) In case (b).

c) In both cases.

d) Neither in (a), nor in (b).

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5. The dielectric breakdown voltage 𝑉𝐵,𝑑 in a capacitor with a dielectric of 𝜖𝑟 > 2 between its electrodes,compared to the breakdown voltage 𝑉𝐵,𝑎𝑖𝑟 of the same capacitor with air between its electrodes (choosethe correct answer and briefly explain):

a) is higher (𝑉𝐵,𝑑 > 𝑉𝐵,𝑎𝑖𝑟).

b) is lower (𝑉𝐵,𝑑 < 𝑉𝐵,𝑎𝑖𝑟).

c) is the same (𝑉𝐵,𝑑 = 𝑉𝐵,𝑎𝑖𝑟).

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1 Coordinate Systems

1.1 Cartesian coordinates

Position vector: R = 𝑥x + 𝑦y + 𝑧zDifferential length elements: dl𝑥 = x𝑑𝑥, dl𝑦 = y𝑑𝑦, dl𝑧 = z𝑑𝑧Differential surface elements: dS𝑥 = x𝑑𝑦𝑑𝑧, dS𝑦 = y𝑑𝑥𝑑𝑧, dS𝑧 = z𝑑𝑥𝑑𝑦Differential volume element: 𝑑𝑉 = 𝑑𝑥𝑑𝑦𝑑𝑧

1.2 Cylindrical coordinates

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Position vector: R = 𝑟r + 𝑧zDifferential length elements: dl𝑟 = r𝑑𝑟, dl𝜑 = ��𝑟𝑑𝜑, dl𝑧 = z𝑑𝑧

Differential surface elements: dS𝑟 = r𝑟𝑑𝜑𝑑𝑧, dS𝜑 = ��𝑑𝑟𝑑𝑧, dS𝑧 = z𝑟𝑑𝜑𝑑𝑟Differential volume element: 𝑑𝑉 = 𝑟𝑑𝑟𝑑𝜑𝑑𝑧

1.3 Spherical coordinates

Position vector: R = 𝑅rDifferential length elements: dl𝑅 = r𝑑𝑅, dl𝜃 = 𝜃𝑅𝑑𝜃, dl𝜑 = ��𝑅 sin 𝜃𝑑𝜑

Differential surface elements: dS𝑅 = r𝑅2 sin 𝜃𝑑𝜃𝑑𝜑, dS𝜃 = 𝜃𝑅 sin 𝜃𝑑𝑅𝑑𝜑, dS𝜑 = ��𝑅𝑑𝑅𝑑𝜃Differential volume element: 𝑑𝑉 = 𝑅2 sin 𝜃𝑑𝑅𝑑𝜃𝑑𝜑

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2 Relationship between coordinate variables

- Cartesian Cylindrical Spherical

𝑥 𝑥 𝑟 cos𝜑 𝑅 sin 𝜃 cos𝜑

𝑦 𝑦 𝑟 sin𝜑 𝑅 sin 𝜃 sin𝜑

𝑧 𝑧 𝑧 𝑅 cos 𝜃

𝑟√

𝑥2 + 𝑦2 𝑟 𝑅 sin 𝜃

𝜑 tan−1 𝑦

𝑥𝜑 𝜑

𝑧 𝑧 𝑧 𝑅 cos 𝜃

𝑅√

𝑥2 + 𝑦2 + 𝑧2𝑟

sin 𝜃𝑅

𝜃 cos−1 𝑧√𝑥2 + 𝑦2 + 𝑧2

tan−1 𝑟

𝑧𝜃

𝜑 tan−1 𝑦

𝑥𝜑 𝜑

3 Dot products of unit vectors

· x y z r �� z R 𝜃 ��

x 1 0 0 cos𝜑 − sin𝜑 0 sin 𝜃 cos𝜑 cos 𝜃 cos𝜑 − sin𝜑

y 0 1 0 sin𝜑 cos𝜑 0 sin 𝜃 sin𝜑 cos 𝜃 sin𝜑 cos𝜑

z 0 0 1 0 0 1 cos 𝜃 − sin 𝜃 0

r cos𝜑 sin𝜑 0 1 0 0 sin 𝜃 cos 𝜃 0

�� - sin𝜑 cos𝜑 0 0 1 0 0 0 1

z 0 0 1 0 0 1 cos 𝜃 − sin 𝜃 0

r sin 𝜃 cos𝜑 sin 𝜃 sin𝜑 cos 𝜃 sin 𝜃 0 cos 𝜃 1 0 0

𝜃 cos 𝜃 cos𝜑 cos 𝜃 sin𝜑 − sin 𝜃 cos 𝜃 0 − sin 𝜃 0 1 0

�� - sin𝜑 cos𝜑 0 0 1 0 0 0 1

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4 Relationship between vector components

= Cartesian Cylindrical Spherical𝐴𝑥 𝐴𝑥 𝐴𝑟 cos𝜑−𝐴𝜑 sin𝜑 𝐴𝑅 sin 𝜃 cos𝜑+𝐴𝜃 cos 𝜃 cos𝜑−

𝐴𝜑 sin𝜑

𝐴𝑦 𝐴𝑦 𝐴𝑟 sin𝜑+𝐴𝜑 cos𝜑 𝐴𝑅 sin 𝜃 sin𝜑+𝐴𝜃 cos 𝜃 sin𝜑−𝐴𝜑 cos𝜑

𝐴𝑧 𝐴𝑧 𝐴𝑧 𝐴𝑅 cos 𝜃 −𝐴𝜃 sin 𝜃

𝐴𝑟 𝐴𝑥 cos𝜑 + 𝐴𝑦 sin𝜑 𝐴𝑟 𝐴𝑅 sin 𝜃 + 𝐴𝜃 cos 𝜃

𝐴𝜑 -𝐴𝑥 sin𝜑 + 𝐴𝑦 cos𝜑 𝐴𝜑 𝐴𝜑

𝐴𝑧 𝐴𝑧 𝐴𝑧 𝐴𝑅 cos 𝜃 −𝐴𝜃 sin 𝜃

𝐴𝑅 𝐴𝑥 sin 𝜃 cos𝜑+𝐴𝑦 sin 𝜃 sin𝜑+𝐴𝑧 cos 𝜃

𝐴𝑟 sin 𝜃 + 𝐴𝑧 cos 𝜃 𝐴𝑅

𝐴𝜃 𝐴𝑥 cos 𝜃 cos𝜑+𝐴𝑦 cos 𝜃 sin𝜑−𝐴𝑧 sin 𝜃

𝐴𝑟 cos 𝜃 −𝐴𝑧 sin 𝜃 𝐴𝜃

𝐴𝜑 −𝐴𝑥 sin𝜑 + 𝐴𝑦 cos𝜑 𝐴𝜑 𝐴𝜑

5 Differential operators

5.1 Gradient

∇𝑉 =𝜕𝑉

𝜕𝑥x +

𝜕𝑉

𝜕𝑦y +

𝜕𝑉

𝜕𝑧z =

𝜕𝑉

𝜕𝑟r +

1

𝑟

𝜕𝑉

𝜕𝜑�� +

𝜕𝑉

𝜕𝑧z =

𝜕𝑉

𝜕𝑅r +

1

𝑅

𝜕𝑉

𝜕𝜃𝜃 +

1

𝑅 sin 𝜃

𝜕𝑉

𝜕𝜑��

5.2 Divergence

∇·A =𝜕𝐴𝑥

𝜕𝑥+𝜕𝐴𝑦

𝜕𝑦+𝜕𝐴𝑧

𝜕𝑧=

1

𝑟

𝜕(𝑟𝐴𝑟)

𝜕𝑟+

1

𝑟

𝜕𝐴𝜑

𝜕𝜑+𝜕𝐴𝑧

𝜕𝑧=

1

𝑅2

𝜕(𝑅2𝐴𝑅)

𝜕𝑅+

1

𝑅 sin 𝜃

𝜕(sin 𝜃𝐴𝜃)

𝜕𝜃+

1

𝑅 sin 𝜃

𝜕𝐴𝜑

𝜕𝜑

5.3 Laplacian

∇2𝑉 =𝜕2𝑉

𝜕𝑥2+𝜕2𝑉

𝜕𝑦2+𝜕2𝑉

𝜕𝑧2=

1

𝑟

𝜕

𝜕𝑟

(𝑟𝜕𝑉

𝜕𝑟

)+

1

𝑟2𝜕2𝑉

𝜕𝜑2+𝜕2𝑉

𝜕𝑧2=

1

𝑅2

𝜕

𝜕𝑅

(𝑅2𝜕𝑉

𝜕𝑅

)+

1

𝑅2 sin 𝜃

𝜕

𝜕𝜃

(sin 𝜃

𝜕𝑉

𝜕𝜃

)+

1

𝑅2 sin2 𝜃

𝜕2𝑉

𝜕𝜑2

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5.4 Curl

∇×A =(𝜕𝐴𝑧

𝜕𝑦− 𝜕𝐴𝑦

𝜕𝑧

)x +

(𝜕𝐴𝑥

𝜕𝑧− 𝜕𝐴𝑧

𝜕𝑥

)x +

(𝜕𝐴𝑦

𝜕𝑥− 𝜕𝐴𝑥

𝜕𝑦

)x

=(

1

𝑟

𝜕𝐴𝑧

𝜕𝜑− 𝜕𝐴𝜑

𝜕𝑧

)r +

(𝜕𝐴𝑟

𝜕𝑧− 𝜕𝐴𝑧

𝜕𝑟

)�� +

1

𝑟

(𝜕(𝑟𝐴𝜑)

𝜕𝑟− 𝜕𝐴𝑟

𝜕𝜑

)z

=1

𝑅 sin 𝜃

(𝜕(sin 𝜃𝐴𝜑)

𝜕𝜃− 𝜕𝐴𝜃

𝜕𝜑

)r +

1

𝑅

(1

sin 𝜃

𝜕𝐴𝑅

𝜕𝜑− 𝜕(𝑅𝐴𝜑)

𝜕𝑅

)𝜃

+1

𝑅

(𝜕(𝑅𝐴𝜃)

𝜕𝑅− 𝜕𝐴𝑅

𝜕𝜃

)��

6 Electromagnetic formulas

Table 1 Electrostatics

F =1

4𝜋𝜀

𝑄1𝑄2

|R2 −R1|3(R2 −R1) E =

1

4𝜋𝜀

∫

𝑣

(R−R′)|R−R′|3𝑑𝑄

′

∇×E = 0 ∇ ·D = 𝜌𝑣∮

𝐶E · dl = 0

∮

𝑆D · dS = 𝑄

E = −∇𝑉 𝑉 =1

4𝜋𝜀

∫

𝑣

𝑑𝑄′

|R−R′|

𝑉 = 𝑉 (𝑃2) − 𝑉 (𝑃1) =𝑊

𝑄= −

∫ 𝑃2

𝑃1

E · dl

D = 𝜀0E + P = 𝜀E P = 𝜒𝑒𝜀0E

n

1

2

n · (D2 −D1) = 𝜌𝑠

𝐸1,𝑡 = 𝐸2,𝑡

𝑄 = 𝐶𝑉 𝑊𝑒 =1

2𝑄𝑉

𝑊𝑒 =1

2

∫

𝑣𝑜𝑙𝜌𝑣𝑉 𝑑𝑣 =

1

2

∫

𝑣𝑜𝑙D ·E𝑑𝑣

∇ · (𝜀∇𝑉 ) = −𝜌𝑣 ∇ · (𝜀∇𝑉 ) = 0

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Table 2 Magnetostatics

F𝑚 = 𝑞u×B F𝑚 = 𝐼l×B

∇×B = 𝜇J ∇ ·B = 0∮

𝐶B · dl = 𝜇𝐼

∮

𝑆B · dS = 0

B =𝜇𝐼

4𝜋

∮

𝐶

dl′ × (R−R′)|R−R′|3 Φ =

∫

𝑆B · dS

B = 𝜇0 (H + M) = 𝜇H M = 𝜒𝑚H

J𝑚,𝑠 = M× a𝑛 J𝑚 = ∇×M

𝐵1,𝑛 = 𝐵2,𝑛 a𝑛 × (H2 −H1) = J𝑠

∇×H = J

∮H · dl = 𝐼

𝑊𝑚 =1

2

∫

𝑣𝑜𝑙B ·H𝑑𝑣

𝐿 =𝑁Φ

𝐼=

2𝑊𝑚

𝐼2𝐿12 =

𝑁2Φ12

𝐼1=

𝑁2

𝐼1

∫

𝑆2

B1 · dS2

ℛ =𝑙

𝜇𝑆𝑉𝑚𝑚𝑓 = 𝑁𝐼

Table 3 Faraday’s law, Ampere-Maxwell law

𝑉𝑒𝑚𝑓 =

∮

𝐶(u×B) · dl−

∫

𝑆

𝜕B

𝜕𝑡· dS 𝑉𝑒𝑚𝑓 =

∮

𝐶E · dl = −𝑑Φ

𝑑𝑡∮

𝐶H · dl =

∫

𝑆J · dS +

𝑑

𝑑𝑡

∫

𝑆D · dS ∇×H = J +

𝜕D

𝜕𝑡

Table 4 Currents

𝐼 =

∫

𝑆J · dS J = 𝜌u = 𝜎E

∇ · J = −𝜕𝜌𝑣𝜕𝑡

P =

∫

𝑣𝑜𝑙(E · J) 𝑑𝑣

𝐽1,𝑛 = 𝐽2,𝑛 𝜎2𝐽1,𝑡 = 𝜎1𝐽2,𝑡

𝑅 =𝑙

𝜎𝑆𝜎 = −𝜌𝑒𝜇𝑒 =

1

𝜌

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