Question 24.70: The chocolate crumb mystery. This story begins with Problem ......
The chocolate crumb mystery. This story begins with Problem 60 in Chapter 23. (a) From the answer to part (a) of that problem, find an expression for the electric potential as a function of the radial distance r from the center of the pipe. (The electric potential is zero on the grounded pipe wall.) (b) For the typical volume charge density p = -1.1 × 10^{-3} C/m³, what is the difference in the electric potential between the pipe’s center and its inside wall? (The story continues with Problem 60 in Chapter 25.)
(a) We use Eq. 24-18 to find the potential: V_{\mathrm{{wall}}}-V=-{\int_{r}}^{R}E d r\,,or
V_{f}-\,V_{i}=\,-\,{\int_{t}^{f}{\vec{E}}\cdot d\,\vec{s},} (24-18)
0-V=-\int_{r}^{R}\left({\frac{\rho r}{2\varepsilon_{0}}}\right)\ \ \Longrightarrow\ \ -V=-{\frac{\rho}{4\varepsilon_{0}}}\left(R^{2}-r^{2}\right).
Consequently, V=\rho(R^{2}-r^{2})/4\varepsilon_{0}.
(b) The value at r = 0 is
V_{\mathrm{center}}={\frac{-1.1\times10^{-3}\,\mathrm{C/m}^{3}}{4(8.85\times10^{-12}\,\mathrm{C/V}\cdot\mathrm{m})}}\Big((0.05\,\mathrm{m})^{2}-0\Big)=-7.8\times10^{4}\mathrm{V}.
Thus, the difference is |V_{\mathrm{center}}|=7.8\times10^{4}\mathrm{V}.