Consider a soot particle of m kg that is traveling upwards in a smokestack. Let’s say that a scrubber adds 69 electrons to the particle, and its charge density q is 1.0 * 10^{-12} Coulombs/kg. Assume the electric field in the scrubber is produced by two parallel square plates of width L = 1.0 m separated by a distance of d = 0.010 m, and carrying a charge Q each.

1. What must be the value of the electric field between the plates, so that force on the soot particle is equal to the weight of the particle?
(Hint: A real scrubber would use a collection of many pairs of such plates in parallel. )

2. What charge Q on the scrubber’s plates is required to produce the above electric field?

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2014-09-07T04:20:52+05:30

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See diagram.

C = \frac{\epsilon_{0}\ Area}{d} = \frac{\epsilon_{0}\ L^2}{d} \\ \\ E = \frac{V}{d} = \frac{Q}{C\ d} = \frac{Q}{\epsilon_0\ L^2} \\

Electrical force on soot particle = field * charge.

So,\ \ \ E * (q m) = weight = m g \\ \\ E = \frac{g}{q} = \frac{9.8 }{1*10^{-12}} = 9.8*10^{12} Volt/meter \\ \\ Q = E * \epsilon_0 * L^2 = 9.8 * 10^{12} * 8.85 * 10^{-12} * 1^2 \\ = 86.73 Coulombs. \\

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