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anonymous
 one year ago
Question regarding cylinders and radial electric fields
anonymous
 one year ago
Question regarding cylinders and radial electric fields

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You understand that the recent increase in U.S. natural gas prices is expected to shunt demand for gas as utilities use more coal to generate power. Such utilities are interested in the use of electrostatic precipitators, which use electric forces to remove pollutant particles from smoke, in particular in the smokestacks of coalburning power plants. One form of precipitator you are studying 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 9.0 x 10–5 m, the radius of the cylinder is 0.14 m, and the potential difference of 5 x 104 V 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. You are asked: (a) What is the magnitude of the electric field midway between the wire and the cylinder wall? (b) What magnitude of charge must a 3.00 x 10–8 kg ashparticle have if the electric field computed in part (a) is to exert a force ten times the weight of the particle? @Mashy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Mashy Now, there are two approaches that I have found in my research. One claims that the field is uniform and radial, therefore the problem can be solved using \[V_2V_1=\int\limits E dr\] However, another claim is that the field is not uniform. How should I approach this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your questions are very nicely worded :P .. What grade questions are these? Clearly radial cylindrical fields are non uniform.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0They are universityengineering level :) And that would make sense because at the longer ends, the field should be weaker.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay.. you should know the expression for the cylindrical field \[E = \frac{\lambda}{2 \pi \epsilon_0 R}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay and so \[\lambda=\frac{Q}{L}\] right? But is it the length of the whole cylinder?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since it is infinitely long (very long) its better to keep the answer in lamda itself. Now our job is to calculate lambda.. We have the knowledge of potential difference. so you can do the integral.. plug in the values of V2, V1, and calculate Lambda

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would I go about setting up this integral? \[\lambda=\int\limits \frac{ E }{ 2\pi \epsilon_0R }dR\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you already did.. now integrate it from R1 (the radius of the wire) to R2 (radius of the pipe)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lambda=\frac{ E }{ 2\pi \epsilon_0 }\ln(\frac{ R_2 }{ R_1 }) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait.. I overlooked.. why is there an E? :P It should be lamda

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I solved for lambda in the first equation you gave me and then integrated with respect to R >_< lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, I'm not actually sure what I did.. lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I accidentally swapped the variables. It should be E = lambda and the rest

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes.. just swap E and Lamda and you should be fine!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sweet. And for the second part it's jsut F=qE? F=mg, so mg=qE and just solve for q?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well 10mg since it says the F is 10 times the weight

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you are all set on reducing pollution.. (Well theoretically :P)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you! Do you have time for one more problem? I think it should be simple but I'm stuck (probably because it's in the wee hours of the evening :P)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok shoot.. lets do one more before I go for lunch

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Mashy Actually, I just realized that my problem contains actual constants, so I think I'm suppose to have a numerical answer rather than an expression. I don't know lambda. They've also given me the potential difference..
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