Field of a Charged Rod: Difference between revisions
(I revised what I believe was an error. The formula previously had \sqrt{term}^(3/2) when it should have been (term)^3/2 or \sqrt{term}^3. It was "double counting" the square root.) |
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''' | '''Sajan Thomas, Spring 2024''' | ||
== The Main Idea == | == The Main Idea == | ||
Previously, we've learned about the electric field of a point particle. | Previously, we've learned about the electric field of a point particle. While this skill does build the fundamentals of E&M physics, the real world is not simply made up of isolated particles. In fact, the world is consisted of many geometric shapes our one equation for the electric field of a particle cannot account for. One such example of these geometric shapes is a '''rod'''. | ||
For the purposes of this course's scope, we assume that the charge is uniformly distributed along the rod. This is an important assumption to make since if it weren't the case, a lot of the calculations would change. This in mind, let's start with what we know: the electric field of a point particle. By itself, we can't represent a rod. But, if we stack a lot of charged particles together closely in a line, what do we get? a rod. That is the basis of how we calculate the electric field of a charged rod. | |||
That's the basis of it but for now, we'll keep that for later when we're actually deriving how to find it. First, let's discuss the factors that can contribute to the electric field of the charged rod. Firstly, as with any electric field, we have to know the charge of the source-- in this case, the rod. Secondly, we have to know the length of the rod. Thirdly, we also should know the observation location to find the radius from the rod. | |||
The process of finding the electric field due to charge distributed over an object has four steps: | The process of finding the electric field due to charge distributed over an object has four steps: | ||
1. | 1. Consider the rod, as previously discussed, as a collection of many, many, infinitesimally small charged particles. Make a diagram and draw the electric field <math>\Delta \vec{E}</math> contributed by one of the pieces. | ||
2. | 2. By drawing the vectors for the <math>\Delta \vec{E}</math> contributed to one piece, and repeating it along the rod, you should realize by symmetry that several of the vector components will cancel out. | ||
3. Add up the contributions of all pieces, either numerically or symbolically. | 3. Add up the contributions of all pieces, either numerically or symbolically. While there is a formula we can just apply simply later, this step requires integration as you are adding all of the <math>\Delta \vec{E}</math> contributions of many, many, charged particles. | ||
4. Check that the result is physically correct. | 4. Check that the result is physically correct. | ||
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=== A Mathematical Model === | === A Mathematical Model === | ||
The process of calculating a uniformly charged rod's electric field is tedious, but breaking the process down into several steps makes the task at hand easier. Consider a uniformly charged thin rod of length <math>L</math> and positive charge <math>Q</math> centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis. | The process of calculating a uniformly charged rod's electric field is tedious, but breaking the process down into several steps makes the task at hand easier. Consider a uniformly charged thin rod of length <math>L</math> and positive charge <math>Q</math> centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis. | ||
[[Image:SajanThomas1.png|800px|center|thumb|Figure 2: Problem 2]] | |||
'''Step 1: Divide the Distribution into Pieces; Draw <math>\Delta \vec{E}</math>''' | '''Step 1: Divide the Distribution into Pieces; Draw <math>\Delta \vec{E}</math>''' | ||
Imagine dividing the rod into a series of very thin slices, each with the same charge <math>\Delta Q</math>. This charge <math>\Delta Q</math> is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, <math>\Delta E</math>. Summing all these individual slices of <math>E</math> gives you the total electric field of the rod. This process approaches taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate. | Imagine dividing the rod into a series of very thin slices, each with the same charge <math>\Delta Q</math>. This charge <math>\Delta Q</math> is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, <math>\Delta E</math>. Summing all these individual slices of <math>E</math> gives you the total electric field of the rod. This process approaches taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate. | ||
[[Image:SajanThomas2.png|800px|center|thumb|Figure 2: Problem 2]] | |||
'''Step 2: Write an Expression for the Electric Field Due to One Piece''' | '''Step 2: Write an Expression for the Electric Field Due to One Piece''' | ||
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'''Expression for <math> \Delta \vec{E}</math> | '''Expression for <math> \Delta \vec{E}</math> | ||
Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get <math> \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{( | Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get <math> \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{(x^2+y^2)^{3/2}} \cdot dx </math> and <math> \Delta \vec{E_y} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(x^2+y^2)^{3/2}} \cdot dx </math>. Note that we have replaced <math> \Delta x </math> with <math> dx</math> in preparation for integration. | ||
'''Step 3: Add Up the Contributions of All the Pieces''' | '''Step 3: Add Up the Contributions of All the Pieces''' | ||
The third step is to sum all of our slices. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from <math>-L/2</math> to <math>+L/2</math>. So the expression is <math> \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{( | The third step is to sum all of our slices. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from <math>-L/2</math> to <math>+L/2</math>. So the expression is <math> \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(x^2+y^2)^{3/2}} \cdot dx. </math> Solving this gives the final expression <math> E_y = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{x} \cdot \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. Note that the field parallel to the x axis is zero. This can be observed due to the symmetry of the problem. | ||
This equation can be written more generally as <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} </math> where r represents the distance from the rod to the observation location. | This equation can be written more generally as <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} </math> where r represents the distance from the rod to the observation location. | ||
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== Connectedness == | == Connectedness == | ||
The | The electrical field of a charged rod has many real world applications. Even within other areas of physics, you can extend what we know about a charged rod to find out the electric field of other objects. For instance, a ring is also just a rod that is bent in a circle. Further, a disk is a collection of many concentric rings. This method of thinking can lead us to some very interesting derivations for physics. | ||
Even if we are to only think about real life, rods exist everywhere. Since everything has an electrical field, any rod or even simply cylindrical shape or anything made of rods is an application. A very useful example that also comes up in physics is the concept of wires. Wires are, in essence, thin rods in which electrons can flow through. This categorizes them as a charged rod. Wires are not only an important subject in physics, but also real life as the device you are reading this on probably needed a wire for charge! What happens inside of the wire is important, but we also what happens outside of it is equally important. If we consider static electricity, a misunderstanding or a failure to account for static could be detrimental at a large scale as static can very easily cause malfunctioning to anything that uses wires to operate. | |||
I, the most recent editor of this page, see the use of the electric field of charged rods as a CS major constantly for a similar reason: wires. Computer Science and any form of software really only exists with accompanying hardware. Almost always, the connection between software and hardware is wires and the electrical pulses that serve as the data that travels along the wire. Especially as someone looking into robotics, it is important to understand the outward effect that wires have because static can be very tricky if you don't know how to handle it! | |||
== History == | == History == | ||
Latest revision as of 02:11, 15 April 2024
Sajan Thomas, Spring 2024
The Main Idea
Previously, we've learned about the electric field of a point particle. While this skill does build the fundamentals of E&M physics, the real world is not simply made up of isolated particles. In fact, the world is consisted of many geometric shapes our one equation for the electric field of a particle cannot account for. One such example of these geometric shapes is a rod.
For the purposes of this course's scope, we assume that the charge is uniformly distributed along the rod. This is an important assumption to make since if it weren't the case, a lot of the calculations would change. This in mind, let's start with what we know: the electric field of a point particle. By itself, we can't represent a rod. But, if we stack a lot of charged particles together closely in a line, what do we get? a rod. That is the basis of how we calculate the electric field of a charged rod.
That's the basis of it but for now, we'll keep that for later when we're actually deriving how to find it. First, let's discuss the factors that can contribute to the electric field of the charged rod. Firstly, as with any electric field, we have to know the charge of the source-- in this case, the rod. Secondly, we have to know the length of the rod. Thirdly, we also should know the observation location to find the radius from the rod.
The process of finding the electric field due to charge distributed over an object has four steps:
1. Consider the rod, as previously discussed, as a collection of many, many, infinitesimally small charged particles. Make a diagram and draw the electric field [math]\displaystyle{ \Delta \vec{E} }[/math] contributed by one of the pieces.
2. By drawing the vectors for the [math]\displaystyle{ \Delta \vec{E} }[/math] contributed to one piece, and repeating it along the rod, you should realize by symmetry that several of the vector components will cancel out.
3. Add up the contributions of all pieces, either numerically or symbolically. While there is a formula we can just apply simply later, this step requires integration as you are adding all of the [math]\displaystyle{ \Delta \vec{E} }[/math] contributions of many, many, charged particles.
4. Check that the result is physically correct.
A Mathematical Model
The process of calculating a uniformly charged rod's electric field is tedious, but breaking the process down into several steps makes the task at hand easier. Consider a uniformly charged thin rod of length [math]\displaystyle{ L }[/math] and positive charge [math]\displaystyle{ Q }[/math] centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis.
Step 1: Divide the Distribution into Pieces; Draw [math]\displaystyle{ \Delta \vec{E} }[/math]
Imagine dividing the rod into a series of very thin slices, each with the same charge [math]\displaystyle{ \Delta Q }[/math]. This charge [math]\displaystyle{ \Delta Q }[/math] is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, [math]\displaystyle{ \Delta E }[/math]. Summing all these individual slices of [math]\displaystyle{ E }[/math] gives you the total electric field of the rod. This process approaches taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate.
Step 2: Write an Expression for the Electric Field Due to One Piece
The second step is to write a mathematical expression for the field [math]\displaystyle{ \Delta E }[/math] contributed by a single slice of the rod. We use the formula of the electric field for a point charge because we are imagining each slice as a point charge. First, determine [math]\displaystyle{ r }[/math], the vector pointing from the source to the observation location. For our example, this is [math]\displaystyle{ r = obs - source = \lt 0,y,0\gt - \lt x,0,0\gt = \lt -x,y,0\gt }[/math]. Now use this to calculate the magnitude and direction of [math]\displaystyle{ r }[/math]. So [math]\displaystyle{ |\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2} }[/math] and [math]\displaystyle{ \hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{\lt -x,y,0\gt }{\sqrt{x^2 + y^2}} }[/math]. [math]\displaystyle{ \hat{r} }[/math] is the vector portion of the expression for the field. The scalar portion is [math]\displaystyle{ \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2} }[/math]. Thus the expression for one slice of the rod is: [math]\displaystyle{ \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(x^2+y^2)^{3/2}} \cdot \lt -x,y,0\gt }[/math].
Determining [math]\displaystyle{ \Delta Q }[/math] and the integration variable
In the first step, we determined that the changing variable for this rod was its x-coordinate. This means the integration variable is [math]\displaystyle{ dx }[/math]. We need to put this integration variable into our expression for the electric field. More specifically, we need to express [math]\displaystyle{ \Delta Q }[/math] in terms of the integration variable. Recall that the rod is uniformly charged, so the charge on any single slice of it is: [math]\displaystyle{ \Delta Q = (\frac{\Delta x}{L})\cdot Q }[/math]. This quantity can also be expressed in terms of the charge density.
Expression for [math]\displaystyle{ \Delta \vec{E} }[/math]
Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get [math]\displaystyle{ \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{(x^2+y^2)^{3/2}} \cdot dx }[/math] and [math]\displaystyle{ \Delta \vec{E_y} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(x^2+y^2)^{3/2}} \cdot dx }[/math]. Note that we have replaced [math]\displaystyle{ \Delta x }[/math] with [math]\displaystyle{ dx }[/math] in preparation for integration.
Step 3: Add Up the Contributions of All the Pieces
The third step is to sum all of our slices. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from [math]\displaystyle{ -L/2 }[/math] to [math]\displaystyle{ +L/2 }[/math]. So the expression is [math]\displaystyle{ \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(x^2+y^2)^{3/2}} \cdot dx. }[/math] Solving this gives the final expression [math]\displaystyle{ E_y = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{x} \cdot \frac{1}{(\sqrt{x^2+ (L/2)^2})} }[/math]. Note that the field parallel to the x axis is zero. This can be observed due to the symmetry of the problem. This equation can be written more generally as [math]\displaystyle{ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} }[/math] where r represents the distance from the rod to the observation location.
Step 4:Checking the Result
Finally, the fourth step is to check the result. The units should be the same as the units of the expression for the electric field for a single point particle [math]\displaystyle{ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} }[/math]. Our answer has the right units, since [math]\displaystyle{ \frac{1}{(\sqrt{x^2+ (L/2)^2})} }[/math]. has the same units of [math]\displaystyle{ \frac{Q}{r^2} }[/math]
Is the direction qualitatively correct? We have the electric field pointing straight away from the midpoint of the rod, which is correct, given the symmetry of the situation. The vertical component of the electric field should indeed be zero.
Computational Models
Finding the Electric Field from a Rod with Code While computation done by hand does have its merits, and certainly was the methodology used when these ideas were conceived, there are much more efficient, powerful, and most importantly, pretty ways to go about finding and showing the electric field. This is of course referring to computers, specifically computers running glowscript code in the case of Physics 2212. With that idea in mind, here are some demonstrations of said methodology:
This is some code that you can run which shows the electric field vector
at a given distance from the rod along its length. The rod is shown as a
series of green balls to help emphasize that when using the numerical
integrations mentioned on this page, you are measuring the field produced
by discrete parts of the rod being analyzed. At each point of analysis,
eight field arrows are shown so as to visualize the electric field.
Notice the edge-effects of the electric field of the rod. For reasons discussed above, if we used the long rod approximation (L>>d), these effects would be negligible.
For a more precise model, the two links below lead to code that generates a forty element line of charge with given magnitude, and length, and then iterates vectors representing the electric field in the space all around said line of charge. Note that the vectors are small, but for the positively charged rod, they lead radially outward, and for the negative, radially inward. This is due to the fact that the rod(s) are treated as lines of positive or negative charge, and the electric field behaves as such.
Try zooming in and out! You can really see the symmetry of the field far out, and the edge effects when zoomed in.
Examples
Although this is not a very difficult topic, some reasonably difficult conceptual questions can be asked about it.
Simple
Middling
Difficult
Connectedness
The electrical field of a charged rod has many real world applications. Even within other areas of physics, you can extend what we know about a charged rod to find out the electric field of other objects. For instance, a ring is also just a rod that is bent in a circle. Further, a disk is a collection of many concentric rings. This method of thinking can lead us to some very interesting derivations for physics.
Even if we are to only think about real life, rods exist everywhere. Since everything has an electrical field, any rod or even simply cylindrical shape or anything made of rods is an application. A very useful example that also comes up in physics is the concept of wires. Wires are, in essence, thin rods in which electrons can flow through. This categorizes them as a charged rod. Wires are not only an important subject in physics, but also real life as the device you are reading this on probably needed a wire for charge! What happens inside of the wire is important, but we also what happens outside of it is equally important. If we consider static electricity, a misunderstanding or a failure to account for static could be detrimental at a large scale as static can very easily cause malfunctioning to anything that uses wires to operate.
I, the most recent editor of this page, see the use of the electric field of charged rods as a CS major constantly for a similar reason: wires. Computer Science and any form of software really only exists with accompanying hardware. Almost always, the connection between software and hardware is wires and the electrical pulses that serve as the data that travels along the wire. Especially as someone looking into robotics, it is important to understand the outward effect that wires have because static can be very tricky if you don't know how to handle it!
History
Physicists and scientists make use of electric fields and charged objects all the time. Many times, we may need to know which objects are contributing how much charge in certain areas. Charged objects may attract or repel (depending on the signs of their charge), so we often need to know how objects will interact with each other based on their charges. The phenomenon of this interaction, or electric force between charged particles, was finally confirmed and stated as a law in 1785 by French physicist Charles-Augustin de Coulomb, hence "Coulomb's Law."
See also
The equation for the electric field of a charged rod was derived from the equation for the electric field of a charged particle. See the article "Electric Field" for more information.
Further Reading
The page on electric fields: Electric Field
External Links
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html
http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRods.asp
http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml
References
https://www.youtube.com/watch?v=BBWd0zUe0mI
(For the above reference, the textbook's method is followed in that the charge distribution was left undefined, and assumed to be constant)
Chabay and Sherwood: Matter and Interactions, Fourth Edition, Chapter 15