Special Relativity - Electric Field

The infinity xy-plane is a nonconducting surface, with surface charge density σ, as measured by an observer at rest on the surface. A second observer moves with velocity v relative to the surface, at height h above it. Which of the following expressions gives the electric measured by this second observer?

A.

B.

C.

D.

E. 

(GR9677 #49)

Solution:

E_z = γE_0z 
with γ = 1/(1 − ^v2/_c²)^½ 

E_z = E₀/(1 − ^v2/_c²)^½

For infinity plane: E = σ/2ɛ₀ (show)

For infinity xy-plane: E₀  = σ / 2ɛ₀

E_z =  σ / 2ɛ₀(1 − ^v2/_c²)^½

Answer: C

Electromagnetism - Superposition

Questions 54-55 concern a plane electromagnetic wave that is a superposition of two independent orthogonal plane waves and can be written as the real part of 

 E = x̂ E₁ exp [i(kz −ωt)] +  ŷ E₂ exp [i(kz − ωt + π)]
where k, ω, E₁ and E₂ are real

If E₂ = E₁, the tip of the electric field vector will describe a trajectory that, as viewed along the z-axis from positive z and looking toward the origin, is a

A. Line at 45^o to the + x-axis
B. Line at 135^o to the + x-axis
C. Clockwise circle
D. Counterclockwise circle
E. Random path

(GR9677 #54)

Solution:

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂ e^{i(kz − ωt +π)} 
E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂ e^{i(kz − ωt)} · e^iπ 

with 
E₂ = E₁ = E
e^iπ = −1

E = E e^{i(kz − ωt)} x̂ − E e^{i(kz − ωt)} ŷ 
E = a x̂ − a ŷ 

tan θ  = a / (−a) = −1

tan 45  = 1
tan 135 = tan 315 = −1

Answer: B


Note: 
e^iϕ = cos ϕ + isin ϕ
e^iπ = cos π + isin π =  −1 + 0 = −1

Electromagnetism - Polarization

Questions 54-55 concern a plane electromagnetic wave that is a superposition of two independent orthogonal plane waves and can be written as the real part of 

E = x̂ E₁ exp [i(kz −ωt)] +  ŷ E₂ exp [i(kz − ωt + π)]

where k, ω, E₁ and E₂ are real

If the plane wave is split and recombined on a screen after the two portions, which are polarized in the x- and y- directions, have traveled an optical path difference of 2π/k, the observed average intensity will be proportional to

A. E₁² + E₂² 
B. E₁² − E₂² 
C. (E₁+ E₂)² 
D. (E₁− E₂)² 
E. 0

(GR9677 #55)

Solution:

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i(kz − ωt +π )} 

Path difference of 2π/k
E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[k(z + 2π/k) − ωt + π)} 

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[kz − ωt)} · e^i3π 
e^i3π = −1

E = x̂ E₁ e^{i(kz − ωt)}  −  ŷ E₂  e^{i(kz − ωt)} 

Intensity in the x- directions, I_x = |E₁|²  
Intensity in the x- directions, I_y = |−E₂|² 

I_total = I_x + I_y = E₁² + E₂²

Answer: A

Electromagnetism - Gauss’ Law, Electric Field

A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar², where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric field is:

A. A/4πɛ₀
B. AR³/40ɛ₀
C. AR³/24ɛ₀
D. AR³/5ɛ₀
E. AR³/3ɛ₀

(GR9677 #61)

Solution:

The net charge within the Gaussian surface with ρ as a function of r (non-uniformly charged sphere):

dq_enclose = ρ dV = Ar² d(⁴⁄₃ πr³) = Ar² ⁴⁄₃ π3r² dr = A4πr⁴ dr





Answer: B

Electromagnetism - Particle Trajectory

The figure above shows the trajectory of a particle that is deflected as it moves through the uniform electric field between parallel plates. There is potential difference V and distance d between the plates, and they have length L. The particle (mass m, charge q) has non relativistic speed v before it enters the field, and its direction at this time is perpendicular to the field. For small deflections, which of the following expressions is the best approximation to the deflection angle θ?

A. Arctan ( (L/d)(Vq/mv²) )
B. Arctan ( (L/d)(Vq/mv²)² )
C. Arctan ( (L/d)² (Vq/mv²) )
D. Arctan ( (L/d)(2Vq/mv²)^½ )
E. Arctan ( (L/d)^½(2Vq/mv²) )

(GR9677 #71)

Solution:

tan θ = v_y / v_x = at / v 
v  = L/t → t = L/v
F  =  ma = qE = qV / d → a = qV / md

tan θ = at / v =  (qV / md) (L/v) / v = (qVL /mdv²)
→  θ = arctan (qVL /mdv²) = arctan ( (L/d)(Vq/mv²) )

Answer: A

Electromagnetism - Electric and Magnetic Force

A positively charged particle is moving in the xy-plane in a region where there is a non-zero uniform electric magnetic field B in the +z –direction and a non-zero uniform electric field in the +y-direction. Which of the following is a posible trajectory for the particle?

(GR9677 #86)

Solution:

v is in the xy-plane
E is in +y-direction


→ F is in +y-direction
→ particle will be deflected by E in +y-direction

B is in +z-direction


→ F,v, B orthogonal to each other
→ E and B orthogonal to each other

Particle moving in an orthogonal direction with B will exhibit cyclotron (helix shaped motion).                    
Answer:  B

Electromagnetism - Electric Potential

Two large conducting plates form a wedge of angle α as shown-in the diagram above. The plates are insulated from each other; one has a potential V₀ and the other is grounded. Assuming that the plates are large enough so that the potential difference between them is independent of the cylindrical coordinates z and ρ, the potential anywhere between the plates as s function of the angle φ is

A. V₀/α
B. V₀φ/α
C. V₀α/φ
D. V₀φ²/α
E. V₀α/φ²

(GR9277 #12)

Solution:

Boundary conditions:
V(φ = 0) = 0 
V(φ = α) = V₀ 

(A) FALSE
V₀/α does not depend on φ 

(B) TRUE
V₀φ/α = 0 for φ = 0
V₀φ/α = V₀ for φ = α

(C) FALSE. V₀α/φ = ∞ for φ = 0

(D) FALSE
For φ = α, V₀φ²/α  = V₀α 

(E) FALSE
For φ = α, V₀φ²/α  = V₀α

Answer: B

Electromagnetism - Gauss' Law

If an electric field is given in a certain region by E_x = 0, E_y = 0, E_z = kz, where k is a nonzero constant, which of the following is true? 

A. There is a time-varying magnetic field.
B. There is charge density in the region.
C. The electric field cannot be constant in time.
D. The electric field is impossible under any circumstances.
E. None of the above.

(GR9277 #64)

Solution:

Gauss’ Law: ∇ ∙ E = ρ/ε₀

For E_x = 0, E_y = 0, E_z = kz

∇ ∙ E = ∂Ex/∂x + ∂Ey/∂y + ∂Ez/∂z = 0 + 0 + k = k

Since ∇ ∙ E ≠ 0 → There is charge density in the region.

Answer: B 

Electromagnetism - EM Radiation

An electron oscillates back and forth along the + and − x-axes, consequently emitting electromagnetic radiation. Which of the following statements concerning the radiation is NOT true?

1. The total rate of radiation of energy into all directions is proportional to the square of the electron's acceleration.
2. The total rate of radiation of energy into all directions is proportional to the square of the electron's charge.
3. Far from the electron, the rate at which radiated energy crosses a perpendicular unit area decreases as the inverse square of the distance from the electron
4. Far from the electron, the rate at which radiated energy crosses a perpendicular unit area is a maximum when the unit area is located on the + or − x-axes
5. Far from the electron, the radiated energy is carried equally by the transverse electric and the transverse magnetic fields.

(GR9277 #84)

Solution:

Power radiated, P = ∮ S ∙ dA
S is poynting vector, where S = (1/μ₀) E × B  = (1/cμ₀) |E|²
Electric field, E = − (kq/r)(a_⊥/c²)

(A) TRUE
P ∝ a²  since |E|² ∝  a_⊥²

(B) TRUE,
P ∝ q² since |E|² ∝ q²

(C) TRUE
P ∝ 1/r² since |E|² ∝ 1/r²

(D) FALSE
The electron oscillates back and forth along the + and − x-axes. Thus, the direction of acceleration is along the + and − x axes. P_max is when the area of electric flux perpendicular, ⊥ to the acceleration of the particle (not located on the + or − x-axes).
P_max ∝ a_⊥² since |E|² ∝  a_⊥²

(E) TRUE, 
Power radiated, P = ∮ S ∙ dA
S is poynting vector, where S = (1/μ₀) E × B
P is carried equally by E and B field.

Answer: D

Electromagnetism - Drift Velocity

A wire of diameter 0.02 meter contains 10²⁸ free electrons per cubic meter. For an electric current of 100 amperes, the drift velocity for free electrons in the wire is most nearly

A. 0.6 × 10⁻²⁹ m/s
B. 1 × 10⁻¹⁹ m/s
C. 5 × 10⁻¹⁰ m/s
D. 2 × 10⁻⁴ m/s
E. 8 × 10³ m/s

(GR8677 #09)

Solution:

Drift velocity is the average velocity of a carrier that is moving under the influence of an electric field.

Velocity: v =  L/t

In a wire with length L and cross sectional area A, there are n electrons with charge q_e per cubic meter.

Total number of mobile electrons in the wire, Q = nq_eLA

Current: I = Q/t = nq_eLA/t = nq_evA
v = I/nq_eA

q_e = charge of an electron = 1.6 × 10⁻¹⁹ C
n = 10²⁸ electrons/m³
A = πr² =  0.5π × 10⁻⁴
I =100 Ampere

v = 10² / (10²⁸× 1.6 × 10⁻¹⁹× 0.5π × 10⁻⁴)
= 10^2−28+19+4 / (1.6 × 0.5π)
≈ 10⁻⁴

Answer: D

Special Relativity - Electromagnetic Field

Which of the following statements most accurately describes how an electromagnetic field behaves under a Lorentz transformation?

1. The electric field transforms completely into a magnetic field.
2. If initially there is only an electric field, after the transformation there maybe both an electric and magnetic field.
3. The electric field is unaltered.
4. The magnetic field is unaltered.
5. It cannot be determined unless a gauge transformation is also specified.

(GR8677 #22)

Solution:

see analysis: Electromagnetism and Relativity

Answer: A

Electromagnetism - Electric and Magnetic Forces

A charge particle is released from rest in a region where there is a constant electric field and a constant magnetic field. If the two fields are parallel to each other, the path of the particle is a

A. circle
B. parabola
C. helix
D. cycloid
E. straight line

(GR8677 #25)

Solution:



Electric Force:
→ the path of the particle is parallel to the electric field.
→ the velocity of particle has the same direction with the electric field.

Magnetic Force/Lorentz Force:  

If
→ no magnetic field contribution
→ the path of the particle is a straight line, parallel to the electric field.

Answer: E

Electromagnetism - EM Radiation

A charge particle oscillates harmonically along the x-axis as shown above. The radiation from the particle is detected at a distant point P which lies in the xy-plane. The electric field at P is in the

A. ±z direction and has a maximum amplitude at θ = 90^o
B. ±z direction and has a minimum amplitude at θ = 90^o
C. xy-plane and has a maximum amplitude at θ = 90^o
D. xy-plane and has a minimum amplitude at θ = 90^o
E. xy-plane and has a maximum amplitude at θ = 45^o

(GR8677 #53)

Solution:

Since point P lies in the xy-plane, the electric field at P also lies in the xy-plane, and its maximum amplitude is at θ = 90^o (perpendicular to the acceleration of particle).

Answer: C

Electromagnetism - Dielectric

A dielectric of dielectric constant K is placed in contact with a conductor having surface charge density σ, as shown above. What is the polarization (bound) charge density σ_p on the surface of the dielectric at the interface between the two materials? 

A. σK / (1−K)
B. K / (1 + K)
C. σK
D. σ(1 + K) / K
E. σ(1 − K) / K

(GR8677 #54)

Solution:

Polarization, P = ɛ₀χ_eE = ɛ₀(K − 1)E
Thus, K  1.
If K = 1 → P = 0 and σ_p = 0 → no polarized dielectric

Check answers for K = 1→ σ_p = 0

(A) σ_p = σK/(1 − K) = ∞ → FALSE
(B) σ_p = K/(1 + K) = 1/2 → FALSE
(C) σ_p = σK = σ → FALSE
(D) σ_p = σ(1 + K)/K = 2σ → FALSE
(E) σ_p = σ(1 − K)/K = 0 → TRUE

Answer: E 


Alternative Solution:

In dielectric, E = σ/ɛ = σ/ɛ₀K
Polarization, P = ɛ₀(K − 1)E
σ_p = ɛ₀(K − 1)(σ/ɛ₀K)
= (K − 1)(σ/K)
= σ(1 − K)/K

Electromagnetism - Gauss' Law

Which of the following electric fields could exist in a finite region of space that contains no charges? (in these expressions, A is a constant, and i, j, and k are unit vectors pointing in the x, y, and z directions, respectively.)

A. A(2xyi − xzk)
B. A(−xyj + xzk)
C. A(xzi + xzj)
D. Axyz(i + j)
E. Axyzi

(GR8677 #80)

Solution:

Gauss’ Law for no charges:



Use method of elimination:

(A). FALSE
A(2xyi − xzk)
→

(B). TRUE
A(−xyj + xzk)
→

Answer: B

Electromagnetism - Coulomb's Law

The exponent in Coulomb’s inverse square law has been found to differ from two by less than one part in a billion by measuring which of the following?

A. The charge on an oil drop in the Millikan experiment
B. The deflection of an electron beam in an electric field
C. The neutrality of charge of an atom
D. The electric force between two charged objects
E. The electric field inside a charged conducting shell

(GR8677 #86)

Solution:

The Coulomb force can be expressed by .

The smallest value for n = (2.7 ± 3.1) × 10⁻¹⁶ has been obtained by Williams, Faller, and Hill experiment (Phys. Rev. Lett. 26, 721, 1971).

The experiment was done by measuring the electric field inside a charged conducting shell.

See: Improved result for the accuracy of Coulombs law: A Review of the Williams, Faller, and Hill experiment.

Answer: E

Electromagnetism - Gauss' Law

Five positive charges of magnitude q are arranged symmetrically around the circumference of a circle of radius r. What is the magnitude of the electric field at the center of the circle? (k = 1/4πε₀)

A. 0
B. kq/r²
C. 5kq/r²
D. (kq/r²) cos (2π/5)
E. (5kq/r²) cos (2π/5)

(GR0177 #09)

Solution:

Gauss’ Law:  

There is no q_enc inside the Gaussian surface → E = 0  

Answer: A

Electromagnetism - Lorentz Force

A proton moves in the +z direction after being accelerated from rest through a potential difference V. The proton then passes through a region with a uniform electric field E in the +x direction and a uniform magnetic field B in the +y direction, but the proton’s trajectory is not affected. If the experiment were repeated using a potential difference of 2V, the proton would then be

A. Deflected in the +x-direction
B. Deflected in the -x-direction
C. Deflected in the +y-direction
D. Deflected in the -y-direction
E. Undeflected

(GR0177 #58)

Solution:

Lorentz Force: 

Since it is not deflected, 

Potential energy transforms into kinetic energy, qV = ½ mv²

Thus, velocity is  

For 2V, velocity is  

  

Therefore,

Particle is deflected in −x direction

Answer: B

Electromagnetism - Conductor

An electromagnetic plane wave, propagating in vacuum, has an electric field given by E = E₀ cos (kx−ωt) and is normally incident on a perfect conductor at x = 0, as shown in figure. Immediately to the left of the conductor, the total electric field E and the total magnetic field B are given by which of the following?

	E	B
A.	0	0
B.	2E0 cos ωt	0
C.	0	2(E0/c) cos ωt
D.	2E0 cos ωt	2(E0/c) cos ωt
E.	2E0 cos ωt	2(E0/c) sin ωt

(GR0177 #61)

Solution:

Perfect conductor: 

Its charge and current are distributed on the surface so that external fields can not penetrate it.

Boundary Condition (BC) on the surface of perfect conductor:

• E_tangential = 0
• E_normal ≠ 0 
• B_tangential ≠ 0
• B_normal = 0 

EM wave: E ⊥ B (perpendicular to each other).

If E normal to the surface of conductor, B is tangential.

BC: E_normal ≠ 0

E_initial is reflected by the conductor with the same magnitude but opposite direction.

Total E = E_initial − E_reflected = 0

(B), (D), (E) FALSE

BC: B_tangential ≠ 0

B is not reflected, it's parallel to the surface.

Total B ≠ 0

(A) FALSE

Answer: C

Electromagnetism - Electric Force

Two spherical, nonconducting, and very thin shells of uniformly distributed positive charge Q and radius d are located a distance 10d from each other. A positive point charge q is placed inside one of the shells at a distance d/2 from the center, on the line connecting the centers of the two shells, as shown in figure. What is the net force on the charge q? 

A.     to the left

B.     to the right

C.     to the left

D.     to the right

E.     to the left

(GR0177 #87)

Solution:

Inside thin shell, no charge → E = 0

Outside thin shell,  E = ^kQ/_r² where k = ¹/_4πɛ0

Thus, we just have to consider the force exerted by the opposite sphere, to the left.

r = 10d − ^d/₂ = ^19d/₂

E = ^kQ/_(19d/2)² = ^4kQ/_361d² = ^Q/_361πɛ0d²

F = qE = ^qQ/_361πɛ0d²

Answer: A