Question 23.60: The chocolate crumb mystery. Explosions ignited by electrost......
The chocolate crumb mystery. Explosions ignited by electrostatic discharges (sparks) constitute a serious danger in facilities handling grain or powder. Such an explosion occurred in chocolate crumb powder at a biscuit factory in the 1970s. Workers usually emptied newly delivered sacks of the powder into a loading bin, from which it was blown through electrically grounded plastic pipes to a silo for storage. Somewhere along this route, two conditions for an explosion were met: (1) The magnitude of an electric field became 3.0 × 10^{6} N/C or greater, so that electrical breakdown and thus sparking could occur. (2) The energy of a spark was 150 mJ or greater so that it could ignite the powder explosively. Let us check for the first condition in the powder flow through the plastic pipes.
Suppose a stream of negatively charged powder was blown through a cylindrical pipe of radius R = 5.0 cm. Assume that the powder and its charge were spread uniformly through the pipe with a volume charge density ρ. (a) Using Gauss’ law, find an expression for the magnitude of the electric field \vec{E} in the pipe as a function of radial distance r from the pipe center. (b) Does E increase or decrease with increasing r? (c) Is E directed radially inward or outward? (d) For \rho=1.{{{1}}}\times10^{-3} C/m³ (a typical value at the factory), find the maximum E and determine where that maximum field occurs. (e) Could sparking occur, and if so, where? (The story continues with Problem 70 in Chapter 24.)
(a) We consider the radial field produced at points within a uniform cylindrical distribution of charge. The volume enclosed by a Gaussian surface in this case is L\pi r^{2}\,. Thus, Gauss’ law leads to
E={\frac{\left|q_{\mathrm{enc}}\right|}{\varepsilon_{0}A_{\mathrm{cylinder}}}}={\frac{\left|\rho\right|\left(L\pi r^{2}\right)}{\varepsilon_{0}(2\pi r L)}}={\frac{\left|\rho\right|r}{2\varepsilon_{0}}}.
(b) We note from the above expression that the magnitude of the radial field grows with r.
(c) Since the charged powder is negative, the field points radially inward.
(d) The largest value of r that encloses charged material is r_{\mathrm{max}} = R. Therefore, with |\rho|=0.0011\ \mathrm{C/m^{3}} and R = 0.050 m, we obtain
E_{\operatorname*{max}}={\frac{|\rho|R}{2\varepsilon_{0}}}={\frac{(0.0011\ \mathrm{C/m^{3}})(0.050\ \mathrm{m})}{2(8.85\times10^{-12}\ \mathrm{C^{2}/N\!\cdot m^{2}})}}=3.1\times10^{6}\ \mathrm{N/C}.
(e) According to condition 1 mentioned in the problem, the field is high enough to produce an electrical discharge (at r = R).