Physics Problems & Solutions: 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, Electric Field

No comments :