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A point charge of $-2.00 \mu \mathrm{C}$ is located in the center of a
spherical cavity of radius 6.50 cm inside an insulating charged
solid. The charge density in the solid is $\rho=7.35 \times 10^{-4} \mathrm{C} / \mathrm{m}^{3}$ .
Calculate the electric field inside the solid at a distance of 9.50 cm
from the center of the cavity.

$r=6.5 \mathrm{cm} \quad \mathrm{R}=9.5 \mathrm{cm}$

$q_{\text { solid }}=\rho_{V} =\rho\left(\frac{4}{3} \pi R^{3}-\frac{4}{3} \pi r^{3}\right)= \rho\left(\frac{4 \pi}{3}\right)\left(R^{3}-r^{3}\right)$

$q_{\text { solid }}= 7.35 * 10^{-4}\left(\frac{4 \pi}{3}\right) \left(0.095^{3}-0.065^{3}\right)$

$=1.744 * 10^{-6} \mathrm{C}=1.794 \mathrm{MC}$

$q_{total}= q_{solid} + q_{point}= 1.794Mc+(-2Mc)$

$=-0.2059 \mathrm{Mc}$

$\rightarrow E = \frac{|q|}{\varepsilon_{0} A}=\frac{|q|}{\varepsilon_{0} 4 \pi r^{2}}$

$∴ \quad E=2.05 * 10^{5} \mathrm{N} / \mathrm{C}$

The electric field at a distance of 0.145 m from the sur-
face of a solid insulating sphere with radius 0.355 m is 1750 N / C .
(a) Assuming the sphere's charge is uniformly distributed, what is
the charge density inside it? (b) Calculate the electric field inside
the sphere at a distance of 0.200 m from the center.

$E A = \frac{q}{\varepsilon_0} \longrightarrow q * E A \cdot \varepsilon_{0}$

$\rho=\frac{q}{v}=\frac{q}{4 / 3 \pi R^{3}}$

$q=1750 * 4 \pi * 0.5^{2} * 8.85 * 10^{-12}=4.866*10^{-8} c$

$\rho=\frac{4 \cdot 866 * 10^{-8}}{\frac{4}{3} \pi(0.355)^{3}}=2.6 * 10^{-7} \mathrm{c} / \mathrm{m}^{3}$

$q_{\text { enc }}=\rho v=\rho\left(\frac{4}{3} \pi r^{3}\right) =2 \cdot 6 * 10^{-7}\left[\frac{4}{3} \pi(0.2)^{3}\right]=8.7 * 10^{-9} C$

$E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{\text { enc }}}{1^{2}} =9 * 10^{9} * \frac{8.7 * 10^{-9}}{(0.2)^{2}}=1.96 * 10^{3} \mathrm{N/c}$