Charged Rod: Difference between revisions
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1. Divide the charged object into small pieces. Make a diagram and draw the electric field <math>\Delta \vec{E}</math> contributed by one of the pieces. | 1. Divide the charged object into small pieces. Make a diagram and draw the electric field <math>\Delta \vec{E}</math> contributed by one of the pieces. | ||
2. Choose an origin and axes. Write an algebraic expression for the electric field <math>\Delta \vec{E}</math> due to one piece. | 2. Choose an origin and axes. Write an algebraic expression for the electric field <math>\Delta \vec{E}</math> due to one piece. | ||
3. Add up the contributions of all pieces, either numerically or symbolically. | 3. Add up the contributions of all pieces, either numerically or symbolically. | ||
4. Check that the result is physically correct. | 4. Check that the result is physically correct. | ||
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'''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. Imagine 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 is the same as 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. Imagine 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 is the same as 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''' | '''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>\Delta E</math> contributed by a single slice of the rod. Because we are imagining each slice as a point charge, we use the formula for a point charge. First, determine <math>r</math>, the vector pointing from the source to the observation location. For our example, this is <math> r = obs - source = <0,y,0> - < x,0,0> = <-x,y,0></math>. Now use this to calculate the magnitude and direction of <math>r</math>. So <math>|\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2}</math> and <math>\hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{< -x,y,0>}{\sqrt{x^2 + y^2}} </math>. <math> \hat{r}</math> is the vector portion of the expression for the field. The scalar portion is <math> \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2}</math>. Thus the expression for one slice of the rod is: | The second step is to write a mathematical expression for the field <math>\Delta E</math> contributed by a single slice of the rod. Because we are imagining each slice as a point charge, we use the formula for a point charge. First, determine <math>r</math>, the vector pointing from the source to the observation location. For our example, this is <math> r = obs - source = <0,y,0> - < x,0,0> = <-x,y,0></math>. Now use this to calculate the magnitude and direction of <math>r</math>. So <math>|\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2}</math> and <math>\hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{< -x,y,0>}{\sqrt{x^2 + y^2}} </math>. <math> \hat{r}</math> is the vector portion of the expression for the field. The scalar portion is <math> \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> \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\sqrt{x^2+y^2})^{3/2}} \cdot < -x,y,0> </math>. | <math> \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\sqrt{x^2+y^2})^{3/2}} \cdot < -x,y,0> </math>. | ||
'''Determining <math>\Delta Q</math> and the integration variable''' | '''Determining <math>\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 should signify that the integration variable is <math> dx</math>. We need to put this integration variable into our expression for the electric field. More specifically, we need to express <math>\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> | In the first step, we determined that the changing variable for this rod was its x-coordinate. This should signify that the integration variable is <math> dx</math>. We need to put this integration variable into our expression for the electric field. More specifically, we need to express <math>\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> | ||
\Delta Q = (\frac{\Delta x}{L})\cdot Q</math>. This quantity can also be expressed in terms of the charge density. | \Delta Q = (\frac{\Delta x}{L})\cdot Q</math>. This quantity can also be expressed in terms of the charge density. | ||
'''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}{(\sqrt{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}{(\sqrt{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. | 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}{(\sqrt{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}{(\sqrt{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. 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}{(\sqrt{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. | The third step is to sum all of our slices. 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}{(\sqrt{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. | ||
'''Step 4:Checking the Result''' | '''Step 4:Checking the Result''' | ||
The units should be the same as the units of the expression for the electric field for a single point particle <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} </math>. | The units should be the same as the units of the expression for the electric field for a single point particle <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} </math>. | ||
Our answer has the right units, since <math> \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. has the same units of <math>\frac{Q}{r^2}</math> | Our answer has the right units, since <math> \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. has the same units of <math>\frac{Q}{r^2}</math> | ||
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===Special Case: Uniform Thin Rod At An Arbitrary Location=== | ===Special Case: Uniform Thin Rod At An Arbitrary Location=== | ||
When finding the electric field at an arbitrary location (as in not on the midline as shown above) due to a uniformly charged thin rod, we follow the same procedure that was used in the earlier example. The only difference is found at the observation location, which will have an x- and y-component. | When finding the electric field at an arbitrary location (as in not on the midline as shown above) due to a uniformly charged thin rod, we follow the same procedure that was used in the earlier example. The only difference is found at the observation location, which will have an x- and y-component. | ||
==Practical Experiments== | ==Practical Experiments== | ||
Revision as of 13:47, 25 November 2016
Claimed by Edoardo Moauro (Fall 2016)
Some objects, such as rods, can be modeled as a uniformly charged object in order to calculate the electric field at some observation location. The following wiki page provides an overview of electric fields created by uniformly charged thin rods, briefly presenting its inception, some specific cases, and proof of concept experiments. The case of a uniformly charged thin rod is a fundamental example of electric field patterns and calculations within physics. Its implications can be applied to other charged objects, such as rings, disks, and spheres.
History and Applications
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."Physicists and scientists make use of electric fields and charged objects all the time. Many times we may need to know what 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 model of a uniformly charged rod can be applicable to situations where the rod is interacting with other charged objects, whether they are large or microscopic, in the case of charged particles like protons or electrons travelling in the rod's field.
Electric Field of Distributed Charges
A Uniformly Charged Thin Rod
In many situations, objects, especially conductors, may have charges spread all over their surface. Sometimes it is necessary to calculate the electric fields of these objects, and this requires a process, known as numerical summation or integration, of dividing an object into many pieces and summing the individual pieces' electric field contributions. But what do these electric field patterns look like? As with point charges, the direction of the field is determined by the sign of the object's charge (positive-points away, negative-points toward) and the size of the field is determined by the observation distance and the magnitude of the object's charge.
The process of finding the electric field due to charge distributed over an object has four steps:
1. Divide the charged object into small pieces. Make a diagram and draw the electric field [math]\displaystyle{ \Delta \vec{E} }[/math] contributed by one of the pieces.
2. Choose an origin and axes. Write an algebraic expression for the electric field [math]\displaystyle{ \Delta \vec{E} }[/math] due to one piece.
3. Add up the contributions of all pieces, either numerically or symbolically.
4. Check that the result is physically correct.
The two figures above depict both a two-dimensional and three-dimensional view of the electric field of a uniformly positively charged rod. Below, the process to find the electric field of a uniformly charged thin rod is carried out.
The Algorithm
The process of calculating a uniformly charged rod's electric field may seem quite tedious, but breaking the process down into several steps makes the task much 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. Imagine 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 is the same as 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. Because we are imagining each slice as a point charge, we use the formula for 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}{(\sqrt{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 should signify that 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}{(\sqrt{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}{(\sqrt{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. 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}{(\sqrt{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
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.
Special Case: A Very Long Rod
The above method of calculating the electric field of a uniformly charged rod is most accurate and necessary for problems with rods that aren't very long relative to the observation distance. For problems where the rod is long enough, the electric field for the rod can be approximated as [math]\displaystyle{ E=\frac{1}{4\pi\epsilon_0} \cdot \frac{2Q/L}{r} }[/math].
Special Case: Uniform Thin Rod At An Arbitrary Location
When finding the electric field at an arbitrary location (as in not on the midline as shown above) due to a uniformly charged thin rod, we follow the same procedure that was used in the earlier example. The only difference is found at the observation location, which will have an x- and y-component.
Practical Experiments
The following two DIY experiments are shown to better understand the physical consequences of electric fields.
Charged Rod and Aluminum Can
Consider the situation in the figure below, in which two charged rods are placed on the sides of an aluminum can, separated by a distance d. The can will stay still because the net force action on the can is zero. This is because the positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can, and the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod.
Similarly, lets consider a similar situation with charged rods on sides of a can, but this time both rods are positively charged, as shown in the image below.
In this situation, the can will also stay still because polarization force is always attractive.
Going back to the first setup. Imagine briefly touching the negatively charged rod and the can. Holding the cans at equal distance on either side of the can, the can will now roll toward the positively charged rod. This is because the can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod.
Charged Rod and Pith Ball
The video embedded below depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod. <https://www.youtube.com/watch?v=aeiqw81kGio>
See Also
Along with calculating electric fields of uniformly charged thin rods, many real-world examples require the electric fields of other shapes and objects to be known. Information on this wiki about some of these topics can be found below.
References
Text Sources
Chabay, Ruth and Sherwood, Bruce. Matter and Interactions, Volume II. 4th ed. New Jersey: Wiley, 2015. Print.
Plaza.obu.edu. N. p., 2016. Web. 4 Nov. 2016. <http://plaza.obu.edu/corneliusk/up2/effcr.pdf>
Phys.uri.edu. N. p., 2016. Web. 4 Nov. 2016. <http://www.phys.uri.edu/gerhard/PHY204/tsl31.pdf>
Stmary.ws. N. p., 2016. Web. 4 Nov. 2016. <http://stmary.ws/HighSchool/Physics/home/notes/electricity/staticElectricity/induction.htm>
Image Sources
N.a. "Electric field of a charged rod at many locations." n.d. Shodor Education Foundation, n.a. Shodor Visioning WORKshop, Computational Labs for Electric and Magnetic Interactions (M&I Vol. 2). Web. 16 Apr. 2016. Electric field of a charged rod at many locations
N.a. "Coulomb's Law." n.d. Georgia State University, Atlanta. HyperPhysics. Web. 17 Apr. 2016. Coulomb's Law
25: Video Tutor: Charged Rod and Aluminum Can "25: Video Tutor: Charged Rod And Aluminum Can". Masteringmasteringphysics.blogspot.com. N. p., 2014. Web. 4 Nov. 2016.
Work in progress
--Spennell3 (talk) 14:47, 2 December 2015 (EST)