EEC130A: Practice Problems for Midterm 2 PDF

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EEC130A: Practice Problems for Midterm 2∗ Instructor: Xiaoguang “Leo” Liu ([email protected]) Updated: Mar. 6th 2012 P-1. Charge is distributed along the z-axis from z = 5 m to ∞ and from z = −5 m to −∞ with charge density ρl = 20 nC/m. Find E at (2, 0, 0) m. z ρl 5 y (2,0,0) 5 x Figure 1: Problem ...

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EEC130A: Practice Problems for Midterm 2∗ Instructor: Xiaoguang “Leo” Liu ([email protected]) Updated: Mar. 6th 2012 P-1. Charge is distributed along the z-axis from z = 5 m to ∞ and from z = −5 m to −∞ with charge density ρl = 20 nC/m. Find E at (2, 0, 0) m. z ρl

5

y (2,0,0)

5

x

Figure 1: Problem 1. The z components of the electric field vanishes due to symmetry. The differential electric field is 2 ρl dz p dE = xˆ 2 2 4πǫ0 (2 + z ) 22 + z2

The total electric field is ρl E= 4πǫ0

"Z

∞ 5

2dz

(4 + z2 )3/2

+

Z −5 −∞

2dz

(4 + z2 )3/2

#

xˆ = 13ˆx

V/m. 

P-2. Charge lies on the circular disk r ≤ a, z = 0 with density ρs = ρ0 sin2 φ. Determine E at (0, φ, h) ∗ Some

problems are adapted from “The Schaum’s Outlines on Electromagnetics” and “2008+ Solved Problems in Electromagnetics”.

1

z ρl

dl

5

dE2

y

(2,0,0)

x

5

dE1

Figure 2: Solution to Problem 1. Choose a differential surface element on the disk dSz . Due to symmetry, the rˆ component of the electric field vanishes. The differential electric field is ! h ρ0 (sin2 φ)rdrdφ p zˆ dEz = 4πǫ0 (r2 + h2 ) r 2 + h2 It follows that

ρ0 h E= 4πǫ0

Z 2π Z a (sin2 φ)rdrdφ 0

0

(r 2 + h2 )

3/2

ρ0 h zˆ = 4ǫ0



 −1 1 √ zˆ + a2 + h2 h 

P-3. A point charge Q is at the origin of a spherical coordinate system. Find the flux which crosses the portion of a spherical shell described by α ≤ θ ≤ β. What is the result if α = 0 and β = π/2. The total flux that crosses a full spherical shell (total area of 4πr2 ) is Ψt = Q (Gauss’s Law). The area of the strip is A=

Z 2π Z β 0

α

r2 sin θdθdφ = 2πr2 (cos α − cos β)

Then the flux through the strip is Ψ=

Q A Q = (cos α − cos β) 2 4πr2

For α = 0 and β = π/2 (a hemisphere), this becomes Ψ = Q/2.



P-4. Two identical uniform line charges lie along the x and y axes with charge densities ρl = 20 µC/m. Obtain D at (3, 3, 3) m. 2

Figure 3: Problem 3.

√ The distance from the observation point to either line charge is 3 2 m. Consider first the line charge on the x-axis, ρl D1 = 2πr1



yˆ + zˆ √ 2



.

ρl D2 = 2πr2



xˆ + zˆ √ 2



.

Then the y-axis line charge,

The total flux density is the vector sum of the two,   10 xˆ + yˆ + 2ˆz √ D = D1 + D2 = √ 3 2π 2

µC/m2 . 

P-5. A charge configuration in cylindrical coordinate system is given by ρ = 5re−2r C/m3 . Use Gauss’s Law to find D. Since ρ is not a function of φ or z, the flux Ψ is completely radial. It is also true that, for r constant, the flux density D must be of constant magnitude due to symmetry. Then a proper Gaussian surface is a closed cylinder. The integrals over the end surfaces vanish (because D is parallel to the end surfaces), so that Gauss’s law becomes Z

sidewall

D · dS = Qenclosed 3

z

D2 ρl

D1 (3,3,3) y

ρl x

Figure 4: Problem 4.

Z L Z 2π Z r 0



5πL e Therefore,

0

0

−2r



5re−2r rdrdφdz = D2πrL

1 −r − r − 2 2



 1 + = D2πrL 2

   2.5 1 1 −2r 2 D= rˆ. −e r +r+ r 2 2 

P-6. The volume in cylindrical coordinate system between r = 2 m and r = 4 m contains a uniform charge density ρ (C/m3 ). Use Gauss’s law to find D in all regions. z 4m 2m

Figure 5: Problem 6.

4

Due to the symmetry of the problem, the electric flux density has only rˆ component and is independent of φ and z. Define a cylindrical Gaussian surface as illustrated in Fig. 6. The flux out of the top and bottom surfaces is 0 because the electric flux density is parallel to these surfaces. For 0 < r < 2 m, Gauss’s Law gives D (2πrL) = Qenclosed = 0 For 2 m ≤ r ≤ 4 m,

D (2πrL) = Qenclosed = πρL(r2 − 22 ) D=

For r > 4 m,

ρ 2 (r − 4)rˆ 2r

C/m2

(42 − 22 )πρL = D (2πrL) D=

6ρ rˆ r

C/m2

z 4m 2m

r

Gaussian surface

Figure 6: Solution to Problem 6. 

P-7. Given that D = 10r3 /4ˆr in cylindrical coordinate system, evaluate both sides of the divergence theorem for the volume enclosed by r = 1 m, r = 2 m, z = 0, and z = 10 m. I

D · dS =

Z

(∇ · D)dv

Since D has no z component, D · dS is zero for the top and bottom surfaces.

On the outer cylindrical surface, dSr is in the rˆ direction; on the inner cylindrical surface, dSr is in the −rˆ direction. 5

z 2m 1m

Figure 7: Problem 7.

I

Z 10 Z 2π Z 10 Z 2π 10 3 10 3 r · rdφdz |r=1 + r · rdφdz |r=2 = 750π D · dS = − 0

0

4

0

0

4

(C)

The right hand side of the divergence theorem is   1 ∂ 10r4 ∇·D = = 10r2 r ∂r 4 Z

∇ · Ddv =

Z 10 Z 2π Z 2 0

0

1

(10r2 )rdrdφdz = 750π

(C) 

P-8. Determine the value of E in a material for which the electric susceptibility is 3.5 and P = 2.3 × 10−7 C/m2 . E=

1 P = 7.42 × 103 χ e ǫ0

V/m 

P-9. Region 1, defined by x < 0, is free space, while region 2, x > 0, is a dielectric material for which ǫr2 = 2.4. Given D1 = 3ˆx − 4yˆ + 6ˆz C/m2 , find E2 and the angles θ1 and θ2 . The x components are normal to the interface: Dn and Et are continuous. D1 = 3ˆx − 4yˆ + 6ˆz

E1 = 6

6 4 3 xˆ − yˆ + zˆ ǫ0 ǫ0 ǫ0

z

1

2 D2 θ2

O θ1

x

D1

Figure 8: Problem 9.

E2 = Ex2 xˆ −

D2 = 3ˆx + Dy2 yˆ + Dz2 zˆ

6 4 yˆ + zˆ ǫ0 ǫ0

Then D2 = ǫ0 ǫr2 E2 gives 3ˆx + Dy2 yˆ + Dz2 zˆ = ǫ0 ǫr2 Ex2 xˆ − 4ǫr2 yˆ + 6ǫr2 zˆ Therefore, Ex2 =

1.25 3 = ǫ0 ǫr2 ǫ0

To find the angles:

Dy2 = −4ǫr2 = −9.6

Dz2 = 6ǫr2 = 14.4

D1 · xˆ = |D1 | cos(90◦ − θ1 ) θ1 = 22.6◦

Similarly, θ2 = 9.83◦ .



P-10. A free-space parallel-plate capacitor is charged by momentary connection to a voltage source V, which is then removed. Determine how WE , D, E, C, and V change as the plates are moved apart to a separation distance d2 = 2d1 without disturbing the charge. Since the capacitor is disconnected from the source, the total charge Q remains the same as the distance between the plates changes. 

7

Relationship

Explanation

D2 = D

D = Q/A

E2 = E1

E = D/ǫ0

WE2 = 2WE1

WE2 =

C2 = 21 C1

C = ǫA/d

V2 = 2V1

V = Q/C

1 2

R

ǫ0 E2 dv, and the volume is doubled

P-11. A spherical conducting shell of radius a, centered at the origin, has a potential of V = V0 , for r ≤ a

,and V0 a/r, for r > a

with the zero reference at infinity. Find an expression for the stored energy that this potential represents. ,and (V0 a/r2 )rˆ, for r > a

E = −∇V = 0, for r ≤ a, The total electric energy 1 WE = 2

Z

ǫ0 ǫ0 E dv = 0 + 2 2

 Z 2π Z π Z ∞  V0 a 2 2 r sin θdrdθdφ = 2πǫ0 V02 a 2 0

r

a

0



P-12. Find H at the center of a square current loop of side L. y

L/2

I

L/2

-L/2

x

-L/2

Figure 9: Problem 12. 8

Choose the Cartesian coordinate system because of the square shape. By symmetry, each half side contributes the same amount to H at the center. Consider an increment source on the half side of 0 ≤ x ≤ L/2, y = − L/2, the Biot-Savart law gives the increment H y

L/2

I

L/2

-L/2

x R

-L/2

Figure 10: Problem 12.

− xxˆ + ( L/2)yˆ Idxxˆ × p 1 x2 + ( L/2)2 dH = 4π x2 + ( L/2)2 Idx ( L/2)zˆ = 3/2 4π [ x2 + ( L/2)2 ] Therefore, H = 8

Z L/2 0

√ 2 2I = zˆ πL

Idx ( L/2)zˆ 4π [ x2 + ( L/2)2 ]

3/2

(1) 

P-13. In the region 0 < r < 0.5 m, in cylindrical coordinate system, the current density is J = 4.5e−2r zˆ (A/m2 ) and J = 0 elsewhere. Find H. Because the current density is symmetrical around the z-axis, a circular path on the constant z plane is chosen as the Ampere contour. 9

For r < 0.5 m, 2πrH =

H=

Z 2π Z r 0

0

4.5e−2r rdrdφ

 1.125  1 − e−2r − 2re−2r φˆ r

(A/m)

For r ≥ 0.5 m, the enclosed current is the same, 0.594πA. Then, 2πrH =

Z 2π Z 0.5 0

Therefore, H=

0

4.5e−2r rdrdφ = 0.594π.

0.297 φˆ r

(A/m) 

P-14. In region 1 of Fig. 11, B1 = 1.2ˆx + 0.8yˆ + 0.4ˆz (T). Find H2 and the angles θ1 and θ2 .

z μr2=1

2

x μr1=15

1

Figure 11: Problem 14. Magnetic boundary conditions dictates that a) the normal components of B are continuous; and b) the tangential components of H are contiuous. Therefore,

H1 =

B1 = 1.2ˆx + 0.8yˆ + 0.4ˆz

(T)

B2 = Bx2 xˆ + By2 yˆ + 0.4ˆz

(T)

1 (1.2ˆx + 0.8yˆ + 0.4ˆz) µ0 µr1

10

(A/m)

H2 = It follows that,

1 (1.2ˆx + 0.8yˆ + µ0 µr1 Hz2 zˆ ) µ0 µr1 Bx2 = µ0 µr2 Hx2 = 8 × 10−2 By2 = 5.33 × 10−2

Hz2 =

Bz2 = 3.18 × 105 µ0 µr2

(A/m)

(T )

(T ) (A/m)

Angle θ1 can be found by cos(90◦ − θ1 ) =

B1 · zˆ = 0.27 | B1 |

Therefore, θ1 = 15.5◦ . Similarly, θ2 = 76.5◦ . 

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