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 - Maxwell's Equation

Listed below are Maxwell’s equations of electromagnetism. If magnetic monopoles exist, which of the these equations would be INCORRECT?

I. Curl H = J + ∂D/∂t
II. Curl E = −∂B/∂t
III. div D = ρ
IV. div B = 0

A. IV only
B. I and II
C. I and III
D. II and IV
E. III and IV

(GR9277 #13)

Solution:

Gauss' law of magnetism: ∇ · B = 0
→ There is no magnetic monopole
If there is one, ∇ · B ≠ 0

Faraday's law of induction: ∇ × E = −∂B/∂t
→ if there is magnetic monopole, ∇ × E = −∂B/∂t + (current of magnetic monopole)

Answer: D

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 - Gauss' Law

An isolated sphere of radius R contains a uniform volume distribution of positive charge. Which of the curves on the graph above correctly illustrated the dependence of the magnitude of the electric field of the sphere as a function of the distance r from its center?

A. A
B. B
C. C
D. D
E. E

(GR8677 #10)

Solution:

The graph shows different lines for condition inside the solid sphere, r  R

Gauss’ Law:  


Inside the sphere:

E is linearly proportional to r.

Answer: C

Electromagnetism - Gauss' Law

A cube has a constant electric potential V on its surface. If there are no charges inside the cube, the potential at the center of the cube is 

A. zero
B. V/8
C. V/6
D. V/2
E. V

(GR8677 #52)

Solution:

Gauss’ Law:  

No charges inside the cube →q_enc = 0 → E = 0

Electric field: E = ∇V
E = 0 → V = constant

Since the potential function has to remain continuous everywhere, at the center of the cube, potential is V.

Answer: E

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 - 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 - Gauss' Law

An infinite, uniformly charged sheet with surface charged density σ cuts through a spherical Gaussian surface of radius R at a distance x from its center, as shown in figure. The electric flux Φ through the Gaussian surface is:

A.  πR²σ / ε₀
B.  2πR²σ / ε₀
C.  π(R − x)²σ / ε₀
D.  π(R² −  x²)σ / ε₀
E.  2π(R² −  x²)σ / ε₀

(GR0177 #60)

Solution:

The Gaussian surface marks a circle around the plane with radius:  

r² = R² −  x²

The net charge within the Gaussian surface:

q_enc = σA = σπr² = σπ(R² −  x²)

Electric Flux:

Φ = q_enc / ε₀ = σπ(R² −  x²) / ε₀ 

Answer: D

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

Electromagnetism – Gauss' Law

A uniformly charged sphere of total charge Q expands and contract between radii R₁ and R₂  at a frequency f. The total power radiated by the sphere is

A. proportional to Q
B. proportional to f²
C. proportional to f⁴
D. proportional to (R₁ / R₂)
E. zero

(GR0177 #96)

Solution:

Power, P = VI
I = dq / dt
Q is constant → dq = 0 → I = 0  → P = 0

Answer: E