- #1
MSZShadow
- 9
- 0
Homework Statement
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?
Homework Equations
V_a - V_b = \int^{r_b}_{r_a}Edl
F_{elec}=qE
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
V_{ab}=\frac{50*10^3}{2}=25*10^3
Then I took the integral
V_a-V_b=\int^{r_b}_{r_a}Edl=E(l)^{r_b}_{r_a}=E(r_b-r_a)
Then I solved for the magnitude of E at the specified point.
|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}
Of course, this is wrong. Where did I make a mistake, and how should I have done the problem?
Last edited: