Electromagnetism - Oscillation

Question 3-4: refer to a thin, nonconducting ring of radius R, as shown below, which has a charge Q uniformly spread out on it.

A small particle of mass m and charge –q is placed at point P and released. If R ≫ x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to:

(GR9677 #04)

Solution:

F_electric = kqQ/r²
F_centripetal = mv²/r = mω²r

F_e = F_c
kqQ/r² = mω²r
ω² = kqQ/mr³

with
k = 1/4πɛ₀
r²  = R² + x²
R ≫ x
r² ∼ R²  → r³ ∼ R³

ω = √(^qQ/_4πɛ0mR³)

Answer: A

Notes: see problem GR9277 #65

Electromagnetism - Electric Force

Two identical conducting spheres, A and B, carry equal charge. They are initially separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B is equal to.

A. 0
B. F/16
C. F/4
D. 3F/8
E. F/2

(GR9677 #24)

Solution:

Coulomb's Law, F_i = kq_Aq_B/r² 
q_A = q_B = Q
F_i = F = kQ² /r² 

When uncharged sphere C touches A, charge A distributes itself evenly:
q_C = q_A' = ¹/₂ q_A = ¹/₂Q

When sphere C touches B:
q_C'  = q_B'  = ¹/₂(q_C + q_B) = ¹/₂(¹/₂Q + Q) = ³/₄Q

Therefore,
F_f  = kq_A' q_B'/r² 
= k (¹/₂Q) (³/₄Q)/r²
= ³/₈ kQ²/r² 
= ³/₈ F

Answer: D

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 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 - 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