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Charged Rod

2016-04-18T00:22:10Z

Acowart6: /* References */

'''Edits claimed by Ashton Cowart (Spring 2016)'''

This page provides a brief overview of electric fields created by uniformly charged thin rods. 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 various situations of charged objects.

==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.

[[File:Erod.jpg]]

A VPython image of the approximate electric field for a positively charged uniform rod. The field is directed away from the rod at all locations.

==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>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.

'''First Step'''

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.

'''Second Step'''

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>.

'''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>
\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>

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.

'''Third Step'''

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.

'''Approximation Equation'''

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> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Q/L}{r} </math>.

==Brief History and Applications==

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 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."

[[File:Coulaw.gif]]

The example 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.

==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.

[[Point Charge]]

[[Electric Dipole]]

[[Charged Ring]]

[[Charged Spherical Shell]]

[[Charged Disk]]

[[Charged Capacitor]]

==References==

'''Text Sources'''

Chabay, Ruth and Sherwood, Bruce. ''Matter and Interactions, Volume II.'' 4th ed. New Jersey: Wiley, 2015. Print.

'''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. [http://compphysed.shodor.org/documents/Info%20materials/WrkGrp-1/VPython_EM/index.html 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. [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html Coulomb's Law]

Work in progress

--[[User:Spennell3|Spennell3]] ([[User talk:Spennell3|talk]]) 14:47, 2 December 2015 (EST)

--[[User:Acowart6|Acowart6]] ([[User talk:Acowart6|talk]]) 20:22, 17 April 2016 (EDT)

2016-04-16T20:43:48Z

Acowart6: Acowart6 uploaded a new version of "File:Erod.jpg"

http://compphysed.shodor.org/documents/Info%20materials/WrkGrp-1/VPython_EM/index.html

File:Erod.jpg

2016-04-16T20:43:30Z

Acowart6: Acowart6 uploaded a new version of "File:Erod.jpg"

http://compphysed.shodor.org/documents/Info%20materials/WrkGrp-1/VPython_EM/index.html

2016-04-16T20:06:58Z

Acowart6:

'''Claimed by Ashton Cowart (Spring 2016)'''

This page provides a brief overview of electric fields created by uniformly charged thin rods. 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 various situations of charged objects.

==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, numerical integration, of dividing an object's surface into many pieces and summing the individual pieces' electric field contributions. But what do these electric field patterns look like?

==The Algorithm==

Consider a uniformly charged thin rod of Length <math>L</math> and positive Charge <math>Q</math>.

'''First Step'''

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 Electric Field of the rod. You may recall from your calculus training that this 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.

'''Second Step'''

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>.

'''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 to you 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 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.

'''Expression for <math> \Delta \vec{E}</math>

Substituting what we just found into our previous expression and dividing into 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 dy </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 dy </math>. Note that we have replaced <math> \Delta y </math> with <math> dy</math> in preparation for integration.

'''Third Step'''

The third step is to sum all of our slices. One way is with Numerical summation. 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 dy. </math> Solving this gives the final expression <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^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.

Work in progress.
([[User talk:Spennell3|talk]])([[User talk:Spennell3|talk]]) 14:47, 2 December 2015 (EST)
([[User talk:Acowart6|talk]]) 16:05, 16 April 2016 (EST)