How Do Electrostatic Precipitators Clean Smokestack Emissions?

[bunkergirl198]

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 (Intro 1 figure) . 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 82.0 micrograms, 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.

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


What magnitude of charge must a 28.0 microgram ash particle have if the electric field computed in part (a) is to exert a force ten times the weight of the particle?


http://session.masteringphysics.com/prob ...

I know that the first answer is 9.60×104 V/M
but I'm not sure how to get it.

Any help would be great.

Thanks cuddlemuffins.

 

[jbsteele]

To get the electric field midway between the wire and the cylinder wall, you need to use the equation E = V/d, where V is the potential difference (50,000 V) and d is the distance between the wire and the cylinder wall, which is equal to half the radius of the cylinder (7 cm). So, the electric field is E = 50,000 V/7 cm = 9.60×10^4 V/m. To get the magnitude of charge on the particle, you need to use the equation F = qE, where F is the force on the particle, q is the charge on the particle, and E is the electric field (9.60×10^4 V/m). Since we know that the force is 10 times the weight of the particle, we can rearrange the equation to solve for q: q = F/E = (10*weight of particle)/E. You can calculate the weight of the particle by using the equation W = mg, where m is the mass of the particle (28.0 micrograms) and g is the acceleration due to gravity (9.8 m/s^2). So, the charge on the particle is q = (10*(28.0 micrograms)*(9.8 m/s^2))/(9.60×10^4 V/m) = 2.89×10^-11 C.

 

[baba89]

Hello! I am happy to help you with this physics problem.

To find the electric field midway between the wire and the cylinder wall, we can use the equation E = V/d, where E is the electric field, V is the potential difference, and d is the distance between the two objects. In this case, the distance is the radius of the cylinder, which is 14.0 cm or 0.14 m. The potential difference is given as 50.0 kV, but we need to convert it to volts by multiplying by 1000, giving us 50,000 V. Plugging these values into the equation, we get E = 50,000 V / 0.14 m = 3.60 x 10^5 V/m. However, this is only the electric field at the surface of the cylinder. To find the electric field midway between the wire and the cylinder wall, we need to divide this value by 2, giving us 1.80 x 10^5 V/m.

For the second part, we can use the equation F = qE, where F is the force, q is the charge, and E is the electric field. We know that the force needs to be ten times the weight of the particle, so we can set up the equation 10mg = qE, where m is the mass of the particle and g is the acceleration due to gravity (9.8 m/s^2). We are given the mass of the particle as 28.0 micrograms, or 2.8 x 10^-8 kg. Plugging in the values, we get 10(2.8 x 10^-8 kg)(9.8 m/s^2) = q(1.80 x 10^5 V/m). Solving for q, we get a charge of 1.53 x 10^-14 C.

I hope this helps you with your problem. Remember to always use the correct units and pay attention to the given information in the problem. Good luck!