In an electrostatic air cleaner (“precipitator”), the strong n...

Question

In an electrostatic air cleaner (“precipitator”), the strong nonuniform electric field in the central region of a cylindrical capacitor (with outer and inner cylindrical radii Rₐ and R₆ ) is used to create ionized air molecules for use in charging dust and soot particles (Fig. 24–22). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength Eₛ ≈ 3.0 x 10⁶ N/C, air molecules will dissociate into positively charged ions and free electrons. In a precipitator, the region within which air is ionized (the corona discharge region) occupies a cylindrical volume of radius R that is typically five times that of the inner cylinder. Assume a particular precipitator is constructed with R₆ = 0.10 mm and Rₐ = 10.0 cm. In order to create a corona discharge region with radius R = 5.0 R₆, what potential difference V should be applied between the precipitator’s inner and outer conducting cylinders? [Besides dissociating air, the charged inner cylinder repels the resulting positive ions from the corona discharge region, where they are put to use in charging dust particles, which are then “collected” on the negatively charged outer cylinder.]

Diagram of a cylindrical capacitor illustrating the corona discharge region with ionized air and labeled dimensions.

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with capacitors. This problem says consider a cylindrical capacitor composed of two concentric conducting cylinders with inner and outer radii are in and are out. A strong non uniform electric field is created in the space between these cylinders designed to ionize the air in this region. When the electric field strength exceeds the dielectric breakdown threshold of air, which is e breakdown is equal to 3.0 multiplied by 10 to the six noons per column. Air molecules in the region become ionized. This ionization occurs in a specific region around the inter conductor extending to a radius of R which is typically five times the radius of the inner cylinder. In this set up, the inner radius of the cylinder is RN is equal to 0.15 millimeters and the outer radius is R out is equal to 12.0 centimeters to generate the required electric field strength for ionization at a radius of R is equal to 5.0 multiplied by RN. What potential difference V should be applied between the inner and outer conducting cylinders? Although the question were given a diagram of what was described in the problem. We're also given four possible choices as our answers. For choice A, we have 11 kilovolts for choice B, we have 15 kilovolts for choice C, we have 18 kilovolts and for choice D, we have 27 kilovolts. Now, we need to calculate the potential difference between these two cylinders. And so we call your formula for the potential difference between two conducting cylinders. And that is V is equal to Lambda divided by the quantity of two pi absolutely not. And that fraction is multiplied by the natural log of the quantity of R out divided by RN. Where V here is our potential difference. Lambda is the charge per unit length of the cylinder R out is the outer radius and RN is the inner radius. Now, we're given values for both of the radii, but we weren't given any information about the charge per unit length. Lambda. The only other piece of information that we're given is the electric field strength at which the dielectric breakdown of air occurs. But we can actually use that to find our charge per unit length. And so we call for your electric fields, you'll have E breakdown, which will denote as EB and that's gonna be equal to. And here we're going to use our electric field or an infinitely long line of charge. And so recall for that formula, we have LAMBDA divided by the quantity of two pi epsilon, not R where EB here is the electric field strength at which the dielectric breakdown occurs. Lambda is my charge per unit length and R is the distance from the center of the cylinder at which this dielectric breakdown occurs. So we can actually solve this equation for Lambda by multiplying both sides by the two pi epsilon knot R and then I'll get LAMBDA is equal to two pi multiplied by epsilon not R multiplied by E breakdown. And so we can actually replace the lambda in our potential difference formula with this expression. And so when we do that, we will get V is equal to two pi multiplied by epsilon, not multiplied by R multiplied by eb divided by the quantity of two pi absolutely not. And this fraction is multiplied by the natural log of the quantity of or out divided by RN. Several things cancel the two pi epsilon knots will actually cancel out. And that will leave me with V is equal to R, but we can write R in terms of RN. So that means that R is gonna be equal to five multiplied by RN. So we'll go ahead and plug that in. So we'll have five multiplied by RN multiplied by E breakdown or EB and that's gonna be multiplied by the natural log of the quantity of our out divided by RN. So we now have values for all the quantities on the right hand side of our equation So we'll go ahead and plug them in. So we'll have V is equal to five, multiplied by four RN. That was 15 millimeters or 0.15 millimeters. And we'll need to convert that into um meters. So to convert millimeters to meters, we need to multiply by 10 to the negative three. So we'll have 0.15 multiplied by 10 to the negative 3 m. And that's gonna be multiplied by e breakdown, which we were told in the problem was three multiplied by 10 to the six newtons per column. And that's gonna be multiplied by the natural log of the quantity of R out, which was 12 centimeters. Again, we'll need to convert this into meters by, divided by 100. So that'll be 0.12 m. That can be divided by RN, which is the 0.15 multiplied by 10 to the negative 3 m. So at this point, I can plug everything into my calculator and I'll get the is equal to 1.5 multiplied by 10 to the four vaults. And here have just kept two significant figures. However, if I look at my choices, they're all in K volt, I need to multiply this by the conversion factor of one kalt divided by 1000 bolts. And when I do that, I'll get V is equal to 15 kilovolts. So that's the answer that I'm looking for. And if I look at my answer choices, this corresponds to answer choice B. So just to recap what we've done here, we started off by looking at our formula for the potential difference between two conducting cylinders. But in order to use this formula, we needed to have the chart per unit length. The only other piece of information that we were given in the problem was the electric field strength at which the dielectric breakdown of error occurred. But we actually used our formula for a long line of charge in order to write that electric field strength in terms of the charge per unit length, we then substituted back in and calculated the value for a potential difference. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Electric Field and Dielectric Strength

Cylindrical Capacitor

Potential Difference and Ionization

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