Chapter 27 - Gauss's Law - Exercises and Problems - Page 807: 40

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According to Gauss's law $$\oint EdA=\dfrac{Q_{in}}{\epsilon_0}$$ and since $E$ is constant, $$EA=\dfrac{Q_{in}}{\epsilon_0}$$ Hence, $$Q_{in}=\epsilon_0EA= \epsilon_0E(4\pi r^2)$$ $$Q_{in}=4\pi\epsilon_0Er^2 \tag 1$$ where $r$ is the radius of the Gaussian surface. We can choose $r_1=8$ cm, $r_2=10$ cm, and $r_3=17$ cm. $$\color{blue}{\bf [a]}$$ So to find the charge on the exterior of the inner sphere, $$Q_{1,in}=4\pi\epsilon_0Er_1^2$$ where $E$ here points toward the Gaussian sphere which means that the inner charge had to be negative. Thus, $$Q_{1,in}=4\pi(8.85\times 10^{-12})(-15000)(0.08)^2$$ $$Q_{1,in}=\color{red}{\bf -11.0 }\;\rm nC$$ $$\color{blue}{\bf [b]}$$ To find the charge on the inside surface of the hollow sphere, Now we know that the inner sphere has a -11 nC charge which will push the free electrons away from the outer hollow sphere toward the outer surface of the big sphere. This means that the electrons will move toward the outer surface of the hollow sphere until the net charge of these electrons is equal to the charge of the inner sphere. Thus, the charge of the inner walls of the hollow sphere is then equal in magnitude to the charge of the small sphere. Thus, $$Q_{inner}=\color{red}{\bf +11}\;\rm nC$$ $$\color{blue}{\bf [c]}$$ To find the charge on the exterior surface of the hollow sphere, $$Q_{3,in}=4\pi\epsilon_0Er_3^2$$ where $E$ here points away from the Gaussian sphere which means that the net inner charge had to be positive. Thus, $$Q_{3,in}=4\pi(8.85\times 10^{-12})(15000)(0.17)^2=\bf 48.2 \;\rm nC$$ Note that this net charge includes the charge of the inner sphere plus the charge of the inner walls of the hollow sphere plus the charge of the outer wall of the hollow sphere. Thus, $$Q_{3,in}= Q_{sphere}+Q_{inner}+Q_{outer}$$ As we explained in part (b) above, the net charge from the free electrons that escaped from the charge of the inner sphere to the outer surface of the hollow sphere must be equal to the charge of the inner sphere. Hence, $Q_{outer}=-11\;\rm nC$, $Q_{inner}=+11\;\rm nC$. Therefore, $$Q_{3,in}= -11+11+Q_{outer}=Q_{outer}$$ Therefore, $$Q_{outer}=\color{red}{\bf +48.2}\;\rm nC$$ The whole idea of this problem is the polarization, you need to figure it out with no math or equation manipulation needed.