Question

(2). a charge q is uniformly distributed throughout a non - conducting sphere of radius r. (a) what is the magnitude of the electric field at a distance r/2 from the center of the sphere? (b) what is the magnitude of the electric field at a distance 2r from the center of the sphere?

Explanation:

Step1: Recall Gauss's law

Gauss's law is $\varPhi_E=\frac{Q_{enclosed}}{\epsilon_0}$, and for a spherically - symmetric charge distribution, $\varPhi_E = E\cdot4\pi r^{2}$.

Step2: Calculate the charge enclosed at $r = \frac{R}{2}$

The volume charge density $
ho=\frac{Q}{\frac{4}{3}\pi R^{3}}$. The volume of the sphere of radius $r=\frac{R}{2}$ is $V=\frac{4}{3}\pi(\frac{R}{2})^{3}$. So the charge enclosed $Q_{enclosed}=
ho V=\frac{Q}{\frac{4}{3}\pi R^{3}}\cdot\frac{4}{3}\pi(\frac{R}{2})^{3}=\frac{Q}{8}$.

Step3: Apply Gauss's law to find the electric field at $r=\frac{R}{2}$

$E\cdot4\pi(\frac{R}{2})^{2}=\frac{Q_{enclosed}}{\epsilon_0}=\frac{Q/8}{\epsilon_0}$. Solving for $E$, we get $E = \frac{Q}{8\pi\epsilon_0R^{2}}$.

Step4: Calculate the electric field at $r = 2R$

For $r = 2R$, the charge enclosed $Q_{enclosed}=Q$. Using Gauss's law $E\cdot4\pi(2R)^{2}=\frac{Q}{\epsilon_0}$. Solving for $E$, we get $E=\frac{Q}{16\pi\epsilon_0R^{2}}$.

Answer:

(a) $\frac{Q}{8\pi\epsilon_0R^{2}}$
(b) $\frac{Q}{16\pi\epsilon_0R^{2}}$