Physics Problems & Solutions: Solid Sphere

Showing posts with label Solid Sphere. Show all posts

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

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

Inside the sphere:

$E(4\pi r^2)=\frac{\rho V}{\varepsilon_0}=\frac{\rho}{\varepsilon_0}\frac{4}{3}\pi r^3$
$E=\frac{\rho }{3\varepsilon_0}r$

E is linearly proportional to r.

Answer: C

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