ECE 3209 — Electromagnetic Fields

University of Virginia

Fall 2012

Midterm 1

(print name above)

This exam is closed book and closed note.

Answer any two out of three problems

Problem 1(25%): Problem 2(25%):

Problem 3(25%):

TOTAL(50%):

On my honor as a student, I have neither given nor received unauthorized assistanceon this exam.

(sign name above)

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Problem 1

1. Consider an R-C circuit with an AC voltage source V = V0 cosωt. Using phasor notation,write down Kirchhoff’s equation for current, and take the real part to find I(t). Showthat this current can be written as I(t) = V0 cos (ωt+ φ)/Z0 and identify φ and Z0 [6points].

What the question asks of you:

Write down everything in phasor notation like we did in class, find the phaseand amplitude for the impedance, and voila!

I = V0ejωt/Z, where Z = R+ 1/jωC (Caution! Using phasors means you should

write exp jωt instead of cosωt)

All we need now to do is to write Z in phasor notation (You should prac-tice this! Go back to HW1 if needed)

Z = R + 1/jωC = Z0e−jφ in phasor notation, with Z0 =

√R2 + 1/ω2C2 and

φ = tan−1(1/ωRC).

(Note that I used a negative phase, because the imaginary part of Z, 1/jωC =−j/ωC, was negative).

This gives I = V0ej(ωt+φ)/Z0, which upon taking real part gives the expression

above

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2. Find the power dissipated in the resistor through Joule loss, i.e., I2R, averaged over oneAC period, T = 2π/ω (Hint: What’s the average of the cosine squared over one AC pe-riod?) [4 points].

P = V 20 R/2Z

2, since average of cos2 (ωt+ φ) is 1/2.

Potential mis-steps: Do not need to do any complex integrals! Read thehint on the question – it is already given! Just calculate I2R, and replace thecosine squared with its average, i.e., 1/2

3. Show that the average power dissipated can be reduced to zero if we run the circuit at veryslow frequencies, ω � 1/RC (In other words, we can trade energy cost for speed) [4 points].

Don’t jump into the maths rightaway - just think! We have in the fig-ure two resistors in series - one is R and the impedance of the capacitoris 1/jωC. At low frequency, the impedance of the capacitor becomes large,effectively creating an open circuit. This makes current I go to zero, so thatdissipated energy I2R goes to zero.

We can recover this from the maths too - P =V 20 R

2(R2 + 1/ω2C2)=V 20

2R

(ω2R2C2

1 + ω2R2C2

),

and this power vanishes when ωRC � 1, as the factor in the brackets vanishes.

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4. If instead we ran the circuit very fast, ω � 1/RC, how much is the average power dissi-pation? [4 points]

Again, this is easy to see without any serious maths. The impedance of thecapacitor now goes to zero at high frequency, effectively shorting it out. Thecurrent now is I = V/R as we only have R in action, so that the power isV 2/R = V 2

0 cos2 (ωt+ φ)/R. The average of the cosine square is once again 1/2,so that the time-averaged power is V 2

0 /2R.

From the maths, P =V 20

2R

(ω2R2C2

1 + ω2R2C2

), and for ωRC � 1 this reaches V 2

0 /2R.

5. Let us now move on to a transmission line. A matched filter is a transmission line thatsatisfies the identity, R′/G′ = L′/C ′. Let us call this ratio φ0. Show that the characteris-tic impedance Z0 for a matched filter (see equations at the back) is real and a function ofφ0 alone (3 points).

R′

G′=

L′

C ′= φ0. In other words, R′ = φ0G

′ and L′ = φ0C′. This gives Z0 =√

R′ + jωL′

G′ + jωC ′=

√φ0G

′ + jωφ0C′

G′ + jωC ′=

√φ0(G

′ + jωC ′)

G′ + jωC ′=√φ0

Just brute force, you can twist yourself into a pretzel over the maths. What’sthe logic behind the matched filter? Someone like you dreamt it up!

It’s similar to the impedance matching we did in class. When we want aratio of two complex numbers to be real (as we do here), the real and imagi-nary parts of the numerator and denonimator must be separately proportional(e.g. if the real above is twice the real below, AND the imaginary above istwice that below, then the overall ratio is simply two!). So you see here thatthe ratio of the complex numbers inside the square root must be φ0

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6. Write down the propagation constant γ for the matched filter, and show that its imag-inary part β = ω/v corresponds to a speed v that is frequency independent. In otherwords, the matched filter gives a distortionless (but not a lossless) line (4 points).

γ =√

(R′ + jωL′)(G′ + jωC ′) =√φ0(G′ + jωC ′)(G′ + jωC ′) = (G′+ jωC ′)

√φ0. Taking

the imaginary part we get β = ωC ′√φ0.

Using φ0 = L′/C ′, we get β = ω√L′C ′. Since by definition β = ω/v, we find that

the speed v = 1/√L′C ′. v is frequency independent, so the line is distortionless.

But there is a loss, since there is a real part of γ, given by α = G′√φ0 =

√R′G′,

using φ0 = R′/G′

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Problem 2

The picture shows two metallic spheres which have equal and opposite charges +Q and -Qsmeared homogeneously on their surfaces. The sphere radii are a and b (a < b).

1. Using a suitable Gaussian surface in each case, find the electric field ~E at all points Rfrom inside the innermost sphere to outside the outermost sphere. In other words, thethree regions (i) R < a, (ii) a < R < b, and (iii) R > b (6 points).

E = 0 for R < a and R > b because the net charge enclosed is zero in both cases. For aGaussian surface with radius between the two spheres, E = Q/4πε0R

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This makes sense! A capacitor should only store electric field lines between the plates.

It maybe tempting to say that the field could well be non-zero inside oroutside, but Gauss’ law guarantees that the total fluxes there are zero as theenclosed charge in each case is zero, and since the fields are the same all overeach Gaussian sphere, so must be the fields!

Caution! When calculating the field at a point R in between the two spheres,you must consider a sphere of radius R going through that point. Some ofyou are using R = a or R = b but you want it at some variable radius R lyingbetween the plates, and your Gaussian surface must go through that point.

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2. Plot E(R) vs R all the way from R=0 to R=∞ schematically (5 points).

3. Let us try to rationalize the plot using superposition. Plot E(R) arising from the innersphere alone, and also plot E(R) due to the outer sphere alone, both on the same graphwith the same axes (label them!), and show how their sum yields the graph you plottedearlier (3 + 3 = 6 points).

Some of you drew a larger negative field for the negative sphere than thepositive sphere. It may be tempting to think that the negative field must belarger than the positive, as we’re closer to the negative sphere! But for thenegative sphere, we also have points on the opposite side that are further thanthe positive sphere. So it’s a toss-up!

The way to think is that the field outside any sphere looks the same as that dueto a point charge at the center of the sphere. And since both spheres have thesame center, any point outside is effectively equidistant from both spheres!

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4. The potential difference across the two spheres is given by V = −∫ R=bR=a

~E · d~R. Find thepotential difference, and thus the capacitance C = |Q/V | (5 + 3 = 8 points).

V = −∫ b

a

Q

4πε0R2dR =

Q

4πε0

(1

a− 1

b

). Thus C = Q/V =

4πε01/a− 1/b

Once again, remember units!1

1/a− 1/bhas units of length (meter) while the

numerator 4πε0 has units of Farads/meter, so that C has the unit of F/m×m,i.e., Farads

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Problem 3

Consider an electric field ~E = xx - yy and a magnetic field ~H = yx + xy. (Careful! Thesigns have changed relative to the HW problem, so you expect entirely different plots!

1. Find ~∇ · ~E, ~∇ · ~H, ~∇× ~E and ~∇× ~H. (2 x 4 = 8 points).

~∇ · ~E =∂Ex∂x

+∂Ey∂y

=∂x

∂x+∂(−y)

∂y= 0

~∇ · ~H =∂Hx

∂x+∂Hy

∂y=∂y

∂x+∂x

∂y= 0

~∇× ~E =

( x y z∂

∂x

∂

∂y

∂

∂zx −y 0

)= 0

~∇× ~H =

( x y z∂

∂x

∂

∂y

∂

∂zy x 0

)= 0

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2. Convert ~E and ~H into cylindrical coordinates, using the formulae at the back of the ex-amination sheet, simplify them, and show that (i) at the angle φ = π/4 the electric fieldbecomes tangential and the magnetic field becomes radial, and (ii) at the angle φ = 0 theelectric field is radial and the magnetic field is tangential. (2.5 x 2 = 5 points)

~E = (r cosφ)(r cosφ− φ sinφ)− (r sinφ)(r sinφ+ φ cosφ) = r

(r cos 2φ− φ sin 2φ

)~H = (r sinφ)(r cosφ− φ sinφ) + (r cosφ)(r sinφ+ φ cosφ) = r

(r sin 2φ+ φ cos 2φ

).

At φ = π/4, ~E = −rφ (tangential) and ~H = rr (radial), while at φ = 0, ~E = rrand ~H = rφ

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3. Find ~∇ · ~E and ~∇ × ~H in cylindrical coordinates using the formulae at the back of thesheet. Show relevant algebraic steps – don’t just guess the result! (3 x 2 = 6 points)

~∇· ~E =1

r

∂

∂r

(rEr

)+

1

r

∂Eφ∂φ

=1

r

∂

∂r

(r2 cos 2φ

)+

1

r

∂(−r sin 2φ)

∂φ= 2 cos 2φ−2 cos 2φ = 0

~∇× ~H =1

r

( r rφ z∂

∂r

∂

∂φ

∂

∂zHr rHφ Hz

)=

1

r

( r rφ z∂

∂r

∂

∂φ

∂

∂zr sin 2φ r2 cos 2φ 0

)

=z

r

(2r cos 2φ− r(2 cos 2φ)

)= 0

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4. Plot ~E and ~H (3 x 2 = 6 points)

Hint1: Keep in mind all the clues from the previous parts, such as fields which becometangential or radial along the x-y axes, and along the axes oriented at π/4 to them

Hint2: While not necessary, you may choose to solve the differential equations for thelines, as was worked out in class. In that case, a list of 2D graphs with equations is shownon the last page

Note that the fields don’t diverge or curl consistently, and if you draw anylittle box, as many lines go in as they come out (zero divergence), and asmany lines go clockwise as anticlockwise around it (zero curl). The fields havea net shear (∂Hx/∂y+∂Hy/∂x 6= 0). The fields are orthogonal to each other.

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Equations that you may or may not need:

Z0 =

√Z ′

Y ′

γ = α+ jβ =√Z ′Y ′

Z ′ = R′ + jωL′

Y ′ = G′ + jωC ′

dV = dx dy dz

d~S = dx dy z

= dx dz y

= dy dz x

d~l = dx x+ dy y + dz z

dV = r dr dφ dz

d~S = r dr dφ z

= r dφ dz r

= dr dz φ

d~l = dr r + r dφ φ+ dz z

dV = R2 sin θ dR dθ dφ

d~S = R2 sin θ dθ dφ R

= R sin θ dR dφ θ

= RdRdθ φ

d~l = dR R+Rdθ θ +R sin θ dφ φ

∇V =∂V

∂rr +

1

r

∂V

∂φφ+

∂V

∂zz

∇ · ~A =1

r

∂

∂r(rAr) +

1

r

∂Aφ∂φ

+∂Az∂z

∇× ~A =

(1

r

∂Az∂φ− ∂Aφ

∂z

)r +

(∂Ar∂z− ∂Az

∂r

)φ+

1

r

[∂

∂r(rAφ)− ∂Ar

∂φ

]z

∇2V =1

r

∂

∂r

(r∂V

∂r

)+

1

r2∂2V

∂φ2+∂2V

∂z2

x = r cosφ, y = r sinφ

r =√x2 + y2, φ = tan−1(y/x)

x = r cosφ− φ sinφ, y = r sinφ+ φ cosφ

r = x cosφ+ y sinφ, φ = −x sinφ+ y cosφ

cos 2φ = cos2 φ− sin2 φ, sin 2φ = 2 sinφ cosφ

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