Physics Problems & Solutions: #61

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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:

$\textup{Gauss Law:} \oint \bar{E} \cdot \hat{n} dA=\frac{q_{\textup{enc}}}{\varepsilon_0}$

$E\cdot 4\pi \bigg(\frac{R}{2}\bigg)^2=\frac{q_{\textup{enc}}}{\varepsilon_0} \rightarrow E=\frac{q_{\textup{enc}}}{\pi R^2 \varepsilon_0}$

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 π3r²dr = A4πr⁴ dr

$\\q_{\textup{enc}}=\int_0^{R/2} A4\pi r^4 dr\\=A4 \pi \frac{1}{5} \bigg(\frac{R}{2}\bigg)^5\\=\frac{A\pi R^5}{40}$

$E=\frac{1}{\pi R^2 \varepsilon_0}\cdot \frac{A \pi R^5}{40}=\frac{AR^3}{40\varepsilon_0}$

Answer: B

Classical Mechanics - Physical Pendulum

A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, shown above, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of configuration I?

A. (6/5)^½
B. (3/2)^½
C. 6/5
D. 3/2
E. 5/3

(GR9277 #61)

Solution:

Angular frequency for physical pendulum:

ω = (MgL/I)^1/2

Pendulum I
M = 2m
I = mr² + mr² = 2mr²
L = r + r = 2r

ω_I= (2mg2r/2mr²)^1/2= (2g/r)^1/2

Pendulum II
M = 2m
I = mr² + m(r/2)² = (⁵/₄)mr²

L = ½ r + r = (³/₂)r

ω_II=[ 2mg (³/₂)r / (⁵/₄)mr²]^1/2= (12g/5r)^1/2

The ratio:

ω_II/ ω_I= (12g/5r)^1/2/ (2g/r)^1/2= 6/5

Answer: A

Classical Mechanics - Equation of Motion

Questions 61-62

The Equation of motion of a rocket in free space can be written where m is the rocket’s mass, v is its velocity, t is time, and u is a constant.

$m \frac{dv}{dt}+u \frac{dm}{dt}=0$

The constant u represents the speed of the

A. Rocket at t = 0
B. Rocket after its fuel is spent
C. Rocket in its instantaneous rest frame
D. Rocket’s exhaust in a stationary frame
E. Rocket’s exhaust relative to the rocket

(GR8677 #61)

Solution:

In free space → without gravity → no external force:

$\\F= \frac{dp}{dt}=\frac{dmv}{dt}=m\frac{dv}{dt}+v \frac{dm}{dt}=0 \\\\\rightarrow m\frac{dv}{dt}=-v_\text{exhaust}\frac{dm}{dt}$

The equation of motion:

$m \frac{dv}{dt}+u \frac{dm}{dt}=0 \rightarrow m \frac{dv}{dt}= -u \frac{dm}{dt}$

→ u = v_exhaust relative to the rocket.

Answer: E

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