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Homework Help: Electric field in an electrostatic precipitator.

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Electrostatic precipitators use electric forces to remove pollutant particles from smoke, in particular in the smokestacks of coal-burning power plants. One form of precipitator consists of a vertical, hollow, metal cylinder with a thin wire, insulated from the cylinder, running along its axis . A large potential difference is established between the wire and the outer cylinder, with the wire at lower potential. This sets up a strong radial electric field directed inward. The field produces a region of ionized air near the wire. Smoke enters the precipitator at the bottom, ash and dust in it pick up electrons, and the charged pollutants are accelerated toward the outer cylinder wall by the electric field. Suppose the radius of the central wire is 83*10^(-6) m, the radius of the cylinder is 14.0 cm, and a potential difference of 50.0 kV is established between the wire and the cylinder. Also assume that the wire and cylinder are both very long in comparison to the cylinder radius.

    A) What is the magnitude of the electric field midway between the wire and the cylinder wall?

    B) What magnitude of charge must a 34.5*10^(-6) g ash particle have if the electric field computed in part (A) is to exert a force ten times the weight of the particle?

    2. Relevant equations
    [tex]V_a - V_b = \int^{r_b}_{r_a}Edl[/tex]

    3. The attempt at a solution
    The first thing I recognized is that since V is directly proportional to l, if the distance l is halved, so will the potential V. So the V from a to b should be

    Then I took the integral

    Then I solved for the magnitude of E at the specified point.
    [tex]|E|=\frac{V_{ab}}{r_b-r_a}=\frac{25*10^3\ V}{(\frac{14*10^{-2}\ m-83*10^{-6}\ m}{2})-0\ m}=357354.7174\ \frac{V}{m}[/tex]

    Of course, this is wrong. Where did I make a mistake, and how should I have done the problem?
    Last edited: Sep 17, 2011
  2. jcsd
  3. Sep 17, 2011 #2


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    Homework Helper

    It looks like you have assumed E is constant along the radius of the cylinder. I see others agree

    but I don't understand that. The electric field lines spread out radially and are further apart at the outside than at the inside. It seems to me that means the electric field decreases with radius and therefore your integration is incorrect. Gauss' Law indicates the E field is proportional to 1/r.
    You could integrate with E = C/r to get the C related to the potential difference and then you would know E as a function of r.
  4. Sep 17, 2011 #3
    Ah. You're right. E wouldn't be constant...
    That would mean that the graph of V against l wouldn't be linear as well...

    Alright. I'll see where I can get with that. Starting with solving for the linear density...
    Alright. Got it. The answer to part A is 96070.57423 V/m. Thanks. :)
    Last edited: Sep 17, 2011
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