Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 29 - Potential and Field - Exercises and Problems - Page 866: 79

Answer

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Work Step by Step

$$\color{blue}{\bf [a]}$$ The easiest way to solve this problem is to use the Gaussian surfaces. Assume that there is a Gaussian sphere inside the sphere of radius $r$ where $r\lt R$. Now we need to find $\Delta V$ between the center of the sphere $V_0$ and a point at $r$ distance from this center and inside the sphere $V_r$. $$\Delta V=V_r-V_0=-\int E_{\rm in}dr$$ Plug from the given $E_{\rm in}$ formula, $$ V_r-V_0=-\int_0^r \dfrac{rQ}{4\pi \epsilon_0 R^3}dr$$ $$ V_r =V_0- \dfrac{r^2Q}{8\pi \epsilon_0 R^3} \tag 1$$ Now we need to find $V_0$ where we know that at the surface of the sphere, the potential is considered outside the sphere and is then given by $$V_R=\dfrac{Q}{4\pi \epsilon_0R}\tag 2$$ Using (1), $$V_R-V_0=-\dfrac{R^2Q}{8\pi \epsilon_0 R^3}$$ $$V_0=V_R+\dfrac{ Q}{8\pi \epsilon_0 R }$$ Plug from (2), $$V_0=\dfrac{Q}{4\pi \epsilon_0R}+\dfrac{ Q}{8\pi \epsilon_0 R }$$ $$V_0= \dfrac{ 3Q}{8\pi \epsilon_0 R }\tag 3$$ Plug into (1), $$ V_r =\dfrac{ 3Q}{8\pi \epsilon_0 R }- \dfrac{r^2Q}{8\pi \epsilon_0 R^3} $$ $$ \boxed{V_r =\dfrac{ Q}{8\pi \epsilon_0 R }\left[3- \dfrac{r^2 }{ R^2} \right] }\tag 4$$ $$\color{blue}{\bf [b]}$$ The ratio is given by $$\dfrac{V_{\rm center}}{V_{\rm surface}}=\dfrac{V_0}{V_R}$$ Plug from (2) and (3); $$\dfrac{V_{\rm center}}{V_{\rm surface}}=\dfrac{ \dfrac{ 3Q}{8\pi \epsilon_0 R }}{\dfrac{Q}{4\pi \epsilon_0R}}=\color{red}{\bf \dfrac{3}{2}} $$ $$\color{blue}{\bf [c]}$$ At outside the sphere when $r=3R$, using (4), $$ V_{3R}=\dfrac{ 1}{4\pi \epsilon_0 }\dfrac{Q}{3R}$$ $$ V_{R}=\dfrac{ 1}{4\pi \epsilon_0 }\dfrac{Q}{R}$$ $$ V_{0}=\dfrac{ 1}{4\pi \epsilon_0 }\dfrac{3Q}{2R}$$ See the graph below,
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