'''Claimed by Edoardo Moauro (Fall 2016)'''
'''Claimed by Carter Gillon (Spring 2017)'''
Changes: Reformatting to match the template, adding a computational model, adding examples

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.

==The Main Idea: Electric Field of Distributed Charges==
===A Uniformly Charged Thin Rod===

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. 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>\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.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

[[File:2D_Charged_Rod.png]]
[[File:3D_Charged_Rod.png]]

The two figures above depict both a two-dimensional and three-dimensional view of the electric field of a uniformly positively charged rod. The red figure represents a rod and the green arrows represent the electric field direction and magnitude.

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 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.

'''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 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>\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>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 means 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.

'''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}{(\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.

'''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> 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>
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===

For special cases, an approximated formula to find the electric field will be provided. The steps to derive this formula are the same as those outlined above. The 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>.

===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 instead of only an x-component as in the example above.

=== A Computational Model===
[https://trinket.io/embed/glowscript/c59cde2427 This simulation] depicts the electric field around a charged rod.

==Examples==

===Simple===

===Middling===

This example is meant to walk you through the steps of setting up an integration problem.

'''Problem Statement'''

A thin rod lies on the x-axis with one end at -A and the other end at A, as shown in the diagram. A charge of <math> -Q </math> is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location <math> ‹ 0, y, 0 › </math> due to the rod. Following the procedure discussed in the textbook, we have cut up the rod into small segments, each of which can be considered as a point charge. We have selected a typical piece, shown in red on the diagram.

[[File:Middling_.jpg]]

Use the following as necessary: <math> x, y, dx, A, Q </math>. Remember that the rod has charge <math>-Q</math>.

(a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod?

(b) What is the amount of charge <math>dQ</math> on the small piece of length <math> dx </math>?

(c) What is the vector from source to observation location?

(d) What is the distance from the source to the observation location?

(e) When we set up an integral to find the electric field at the observation location due to the entire rod, what will be the integration variable?

'''Solution'''

'''Part A'''

<math> \lambda = - \frac{Q}{2A} </math>

'''Part B'''

<math> dQ = - \frac{Q}{2A} dx </math>

'''Part C'''

<math> \vec{r} = < -x, y, 0> </math>

'''Part D'''

<math> d = \sqrt{(-x)^2 + y^2} </math>

'''Part E'''

<math> x </math>

===Difficult===

'''Problem Statement:'''

A very thin plastic rod of length <math>L</math> is rubbed with cloth and becomes uniformly charged with a total charge <math>+Q</math>.

[[File:Hard_problem_update.jpg]]

(a) Consider an arbitrary piece of rod of length <math>dx</math> located at a position x on the rod. Determine the electric field <math>\vec{E}</math> from this piece at observation location "*", a distance <math>w</math> to the right of the end of the rod and on the x-axis as indicated in the diagram.

(b) Write down and solve the integral to determine the net electric field <math>\vec{E}</math> of the rod at location "*".

'''Solution'''

'''Part A:'''

<math>\vec{r} = \vec{r_*}-\vec{r_{dx}} = < L+w, 0, 0 > - < x, 0, 0 > = < L+w-x, 0, 0 > = (L+w-x)\hat{x}</math>

<math> |\vec{r}| = L+w-x </math> ; <math>\hat{r} = < 1, 0, 0 > = \hat{x} </math>

<math>\lambda = \frac{charge}{length} = \frac{Q}{L} = \frac{dQ}{dx} </math>

<math> dQ = \frac{Q}{L} dx </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{dQ}{|\vec{r}|^2} \hat{r} = \frac{1}{4\pi\epsilon_{0}} \frac{(Q/L)dx}{(L+w-x)^2} \hat{x} </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x} </math>

'''Part B:'''

<math> \int d\vec{E} = \int_{0}^{L}\frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x}
= \frac{Q}{4\pi\epsilon_{0}L} \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x}</math>

To solve the integral, use the following (found on your test formula sheet): <math>\int \frac{dx}{(x+a)^2} = \frac{-1}{a+x} </math>

<math> \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x} = \frac{1}{L+w-x} (x_1 = 0, x_2 = L) =
\frac{1}{L+w-L} - \frac{1}{L+w} = \frac{1}{w} - \frac{1}{L+w} = \frac {L+w-w}{w(L+w)} = \frac{L}{w(L+w)} </math>

<math> \vec{E} = \frac{Q}{4\pi\epsilon_{0}L} \frac{L}{w(L+w)} \hat{x} </math>

<math> \vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{w(L+w)} \hat{x} </math>

==Connectedness: Practical Experiments==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

===Charged Rod and Aluminum Can===

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? 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, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:Can_and_Two_Charged_Rods_(1).png]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:Can_and_Two_Charged_Rods_(2).png]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods 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===

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

==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]]

==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. A portrait of Coloumb is shown below.

[[File:coloumb.jpg]]

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.

==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. [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]

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

--[[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)

--[[User:Emoauro3|Emoauro3]] ([[User talk:Emoauro3|talk]]) 22:22, 2 November 2016 (EDT)

File:Middling .jpg

2017-04-09T17:17:59Z

Cgillon3:

Charged Rod

2017-04-09T17:14:26Z

Cgillon3: /* Middling */

'''Claimed by Edoardo Moauro (Fall 2016)'''
'''Claimed by Carter Gillon (Spring 2017)'''
Changes: Reformatting to match the template, adding a computational model, adding examples

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.

==The Main Idea: Electric Field of Distributed Charges==
===A Uniformly Charged Thin Rod===

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. 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>\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.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

[[File:2D_Charged_Rod.png]]
[[File:3D_Charged_Rod.png]]

The two figures above depict both a two-dimensional and three-dimensional view of the electric field of a uniformly positively charged rod. The red figure represents a rod and the green arrows represent the electric field direction and magnitude.

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 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.

'''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 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>\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>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 means 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.

'''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}{(\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.

'''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> 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>
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===

For special cases, an approximated formula to find the electric field will be provided. The steps to derive this formula are the same as those outlined above. The 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>.

===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 instead of only an x-component as in the example above.

=== A Computational Model===
[https://trinket.io/embed/glowscript/c59cde2427 This simulation] depicts the electric field around a charged rod.

==Examples==

===Simple===

===Middling===

A thin rod lies on the x-axis with one end at -A and the other end at A, as shown in the diagram. A charge of <math> -Q </math> is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location <math> ‹ 0, y, 0 › </math> due to the rod. Following the procedure discussed in the textbook, we have cut up the rod into small segments, each of which can be considered as a point charge. We have selected a typical piece, shown in red on the diagram.

<showhide>
Some text (usually title) which will not be hidden __HIDER__
<hide>Text which will be hidden</hide>
</showhide>

Use the following as necessary: <math> x, y, dx, A, Q </math>. Remember that the rod has charge <math>-Q</math>.

===Difficult===

'''Problem Statement:'''

A very thin plastic rod of length <math>L</math> is rubbed with cloth and becomes uniformly charged with a total charge <math>+Q</math>.

[[File:Hard_problem_update.jpg]]

(a) Consider an arbitrary piece of rod of length <math>dx</math> located at a position x on the rod. Determine the electric field <math>\vec{E}</math> from this piece at observation location "*", a distance <math>w</math> to the right of the end of the rod and on the x-axis as indicated in the diagram.

(b) Write down and solve the integral to determine the net electric field <math>\vec{E}</math> of the rod at location "*".

'''Solution'''

'''Part A:'''

<math>\vec{r} = \vec{r_*}-\vec{r_{dx}} = < L+w, 0, 0 > - < x, 0, 0 > = < L+w-x, 0, 0 > = (L+w-x)\hat{x}</math>

<math> |\vec{r}| = L+w-x </math> ; <math>\hat{r} = < 1, 0, 0 > = \hat{x} </math>

<math>\lambda = \frac{charge}{length} = \frac{Q}{L} = \frac{dQ}{dx} </math>

<math> dQ = \frac{Q}{L} dx </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{dQ}{|\vec{r}|^2} \hat{r} = \frac{1}{4\pi\epsilon_{0}} \frac{(Q/L)dx}{(L+w-x)^2} \hat{x} </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x} </math>

'''Part B:'''

<math> \int d\vec{E} = \int_{0}^{L}\frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x}
= \frac{Q}{4\pi\epsilon_{0}L} \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x}</math>

To solve the integral, use the following (found on your test formula sheet): <math>\int \frac{dx}{(x+a)^2} = \frac{-1}{a+x} </math>

<math> \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x} = \frac{1}{L+w-x} (x_1 = 0, x_2 = L) =
\frac{1}{L+w-L} - \frac{1}{L+w} = \frac{1}{w} - \frac{1}{L+w} = \frac {L+w-w}{w(L+w)} = \frac{L}{w(L+w)} </math>

<math> \vec{E} = \frac{Q}{4\pi\epsilon_{0}L} \frac{L}{w(L+w)} \hat{x} </math>

<math> \vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{w(L+w)} \hat{x} </math>

==Connectedness: Practical Experiments==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

===Charged Rod and Aluminum Can===

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? 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, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:Can_and_Two_Charged_Rods_(1).png]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:Can_and_Two_Charged_Rods_(2).png]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods 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===

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

==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]]

==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. A portrait of Coloumb is shown below.

[[File:coloumb.jpg]]

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.

==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. [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]

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

--[[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)

--[[User:Emoauro3|Emoauro3]] ([[User talk:Emoauro3|talk]]) 22:22, 2 November 2016 (EDT)

Charged Rod

2017-04-09T15:46:32Z

Cgillon3: /* Difficult */

'''Claimed by Edoardo Moauro (Fall 2016)'''
'''Claimed by Carter Gillon (Spring 2017)'''
Changes: Reformatting to match the template, adding a computational model, adding examples

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.

==The Main Idea: Electric Field of Distributed Charges==
===A Uniformly Charged Thin Rod===

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. 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>\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.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

[[File:2D_Charged_Rod.png]]
[[File:3D_Charged_Rod.png]]

The two figures above depict both a two-dimensional and three-dimensional view of the electric field of a uniformly positively charged rod. The red figure represents a rod and the green arrows represent the electric field direction and magnitude.

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 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.

'''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 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>\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>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 means 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.

'''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}{(\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.

'''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> 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>
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===

For special cases, an approximated formula to find the electric field will be provided. The steps to derive this formula are the same as those outlined above. The 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>.

===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 instead of only an x-component as in the example above.

=== A Computational Model===
[https://trinket.io/embed/glowscript/c59cde2427 This simulation] depicts the electric field around a charged rod.

==Examples==

===Simple===

===Middling===

===Difficult===

'''Problem Statement:'''

A very thin plastic rod of length <math>L</math> is rubbed with cloth and becomes uniformly charged with a total charge <math>+Q</math>.

[[File:Hard_problem_update.jpg]]

(a) Consider an arbitrary piece of rod of length <math>dx</math> located at a position x on the rod. Determine the electric field <math>\vec{E}</math> from this piece at observation location "*", a distance <math>w</math> to the right of the end of the rod and on the x-axis as indicated in the diagram.

(b) Write down and solve the integral to determine the net electric field <math>\vec{E}</math> of the rod at location "*".

'''Solution'''

'''Part A:'''

<math>\vec{r} = \vec{r_*}-\vec{r_{dx}} = < L+w, 0, 0 > - < x, 0, 0 > = < L+w-x, 0, 0 > = (L+w-x)\hat{x}</math>

<math> |\vec{r}| = L+w-x </math> ; <math>\hat{r} = < 1, 0, 0 > = \hat{x} </math>

<math>\lambda = \frac{charge}{length} = \frac{Q}{L} = \frac{dQ}{dx} </math>

<math> dQ = \frac{Q}{L} dx </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{dQ}{|\vec{r}|^2} \hat{r} = \frac{1}{4\pi\epsilon_{0}} \frac{(Q/L)dx}{(L+w-x)^2} \hat{x} </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x} </math>

'''Part B:'''

<math> \int d\vec{E} = \int_{0}^{L}\frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x}
= \frac{Q}{4\pi\epsilon_{0}L} \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x}</math>

To solve the integral, use the following (found on your test formula sheet): <math>\int \frac{dx}{(x+a)^2} = \frac{-1}{a+x} </math>

<math> \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x} = \frac{1}{L+w-x} (x_1 = 0, x_2 = L) =
\frac{1}{L+w-L} - \frac{1}{L+w} = \frac{1}{w} - \frac{1}{L+w} = \frac {L+w-w}{w(L+w)} = \frac{L}{w(L+w)} </math>

<math> \vec{E} = \frac{Q}{4\pi\epsilon_{0}L} \frac{L}{w(L+w)} \hat{x} </math>

<math> \vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{w(L+w)} \hat{x} </math>

==Connectedness: Practical Experiments==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

===Charged Rod and Aluminum Can===

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? 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, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:Can_and_Two_Charged_Rods_(1).png]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:Can_and_Two_Charged_Rods_(2).png]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods 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===

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

==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]]

==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. A portrait of Coloumb is shown below.

[[File:coloumb.jpg]]

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.

==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. [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]

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

--[[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)

--[[User:Emoauro3|Emoauro3]] ([[User talk:Emoauro3|talk]]) 22:22, 2 November 2016 (EDT)

File:Hard problem update.jpg

2017-04-09T15:46:26Z

Cgillon3:

Charged Rod

2017-04-09T15:20:22Z

Cgillon3: /* Difficult */

'''Claimed by Edoardo Moauro (Fall 2016)'''
'''Claimed by Carter Gillon (Spring 2017)'''
Changes: Reformatting to match the template, adding a computational model, adding examples

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.

==The Main Idea: Electric Field of Distributed Charges==
===A Uniformly Charged Thin Rod===

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. 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>\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.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

[[File:2D_Charged_Rod.png]]
[[File:3D_Charged_Rod.png]]

The two figures above depict both a two-dimensional and three-dimensional view of the electric field of a uniformly positively charged rod. The red figure represents a rod and the green arrows represent the electric field direction and magnitude.

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 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.

'''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 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>\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>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 means 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.

'''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}{(\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.

'''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> 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>
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===

For special cases, an approximated formula to find the electric field will be provided. The steps to derive this formula are the same as those outlined above. The 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>.

===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 instead of only an x-component as in the example above.

=== A Computational Model===
[https://trinket.io/embed/glowscript/c59cde2427 This simulation] depicts the electric field around a charged rod.

==Examples==

===Simple===

===Middling===

===Difficult===

'''Problem Statement:'''

A very thin plastic rod of length <math>L</math> is rubbed with cloth and becomes uniformly charged with a total charge <math>+Q</math>.

[[File:Hard_problem.jpg]]

(a) Consider an arbitrary piece of rod of length <math>dx</math> located at a position x on the rod. Determine the electric field <math>\vec{E}</math> from this piece at observation location "*", a distance <math>w</math> to the right of the end of the rod and on the x-axis as indicated in the diagram.

(b) Write down and solve the integral to determine the net electric field <math>\vec{E}</math> of the rod at location "*".

'''Solution'''

'''Part A:'''

<math>\vec{r} = \vec{r_*}-\vec{r_{dx}} = < L+w, 0, 0 > - < x, 0, 0 > = < L+w-x, 0, 0 > = (L+w-x)\hat{x}</math>

<math> |\vec{r}| = L+w-x </math> ; <math>\hat{r} = < 1, 0, 0 > = \hat{x} </math>

<math>\lambda = \frac{charge}{length} = \frac{Q}{L} = \frac{dQ}{dx} </math>

<math> dQ = \frac{Q}{L} dx </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{dQ}{|\vec{r}|^2} \hat{r} = \frac{1}{4\pi\epsilon_{0}} \frac{(Q/L)dx}{(L+w-x)^2} \hat{x} </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x} </math>

'''Part B:'''

<math> \int d\vec{E} = \int_{0}^{L}\frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x}
= \frac{Q}{4\pi\epsilon_{0}L} \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x}</math>

To solve the integral, use the following (found on your test formula sheet): <math>\int \frac{dx}{(x+a)^2} = \frac{-1}{a+x} </math>

<math> \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x} = \frac{1}{L+w-x} (x_1 = 0, x_2 = L) =
\frac{1}{L+w-L} - \frac{1}{L+w} = \frac{1}{w} - \frac{1}{L+w} = \frac {L+w-w}{w(L+w)} = \frac{L}{w(L+w)} </math>

<math> \vec{E} = \frac{Q}{4\pi\epsilon_{0}L} \frac{L}{w(L+w)} \hat{x} </math>

<math> \vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{w(L+w)} \hat{x} </math>

==Connectedness: Practical Experiments==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

===Charged Rod and Aluminum Can===

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? 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, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:Can_and_Two_Charged_Rods_(1).png]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:Can_and_Two_Charged_Rods_(2).png]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods 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===

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

==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]]

==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. A portrait of Coloumb is shown below.

[[File:coloumb.jpg]]

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.

==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. [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]

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

--[[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)

--[[User:Emoauro3|Emoauro3]] ([[User talk:Emoauro3|talk]]) 22:22, 2 November 2016 (EDT)

2016-11-28T02:50:37Z

Cgillon3: Cgillon3 uploaded a new version of "File:Exchangerequation.gif"

Energy Transfer due to a Temperature Difference

2016-11-28T02:49:14Z

Cgillon3: /* Connectedness */

This section will be sharing how to calculate energy transfer in systems that are affected by temperature changes. This related to Chapter 7 in the Text Book and Exam 3.

Woong Jun Park 
wpark39

Carter Gillon (cgillon3), Fall 2016

==The Main Idea==

[[File:blockheat.jpg]]

When two objects that have different temperatures come into contact with one another, energy will be transferred from the object with the higher temperature to the object with the lower temperature. Although work is occurring at a microscopic level, since at the point of contact, the individual atoms on the surfaces of the objects are doing work on each other, this energy transfer is not referred to as work, W, because there are no macroscopic forces or displacements. Instead:

:''Energy Transfer Due to a Temperature Difference'' = <math>Q </math>

Therefore, for an open system, the energy principle is:
:<math>∆E_{system} = Q + W + </math> ''other energy transfers''

*Q is used to describe the amount of energy that is exchanged between the system and surroundings when there is a temperature difference between them (microscopic work)
*W is used to describe the energy transfer due to other forces (macroscopic, mechanical work)
*other energy transfers could include matter transfer, mechanical waves, electricity, electromagnetic radiation, etc...

Sign of ''Q''

<math>Q</math> can be positive or negative depending on the energy change.

[[File:Qflow.jpg]]

If energy is transferred from the system to the surroundings (the system is getting colder), then <math>Q</math> is negative.
If energy is transferred from the surroundings to the system (the system is heating up), then <math>Q</math> is positive.

The Steady State
When a system is described as being in "steady state", that means that the overall energy of the system, <math>E_{system}</math>, does not change despite the other energy transfers occurring in the system. Also, for a system to be at steady state, the energy flow flowing into the system must be doing so at the same rate as the energy flowing out of the system.

===A Mathematical Model===

The following are mathematical models that you will use in these calculations:

The Energy Principle
*<math>∆E_{system} = Q + W + </math> ''other energy transfers''

Calculating Heat Transfer
*<math>∆E_{thermal} = mC(∆T)</math>
::<math>∆E_{thermal}</math> = Heat Added
::<math>m</math> = mass
::<math>C</math> = Specific Heat
::<math>∆T</math> = change in Temperature

Calculating the Temperatures of Two Objects in Contact
*<math>∆E_{system} = Q + W + </math> ''other energy transfers''
::<math>W = 0</math>, other = 0, so <math>∆E_{system} = Q = 0</math>
*<math>∆E_{system} = 0 = ∆E_{thermal,1} +∆E_{thermal,2}</math>
*<math>0 = mC_1(T_f - T_1) + mC_2(T_f - T_2)</math>

===A Computational Model===

Check this video out for a brief understanding of how Heat & Temperature relate to Physics. 
[https://www.youtube.com/embed/bODiX2PjCPE Heat Transfer]

Check this video out for a detailed look into the topic. 
[https://www.youtube.com/embed/XmXr58ySUhI Relationship between deltaE and Q+W]

==Examples==

===Simple===
5000 <math>J</math> of electric energy is put into an electric heater. If the heating element is maintained at a constant temperature throughout this time period, what is <math>∆E_{thermal}</math> for the system, and what is <math>Q</math>, the energy transfer between the heater's heating element and the cool air?

''Solution:''
System: heater

Surroundings: air, electric input

<math> ∆E_{system} = W + Q</math> + ''other energy inputs''

There is no internal energy change since the heater is maintained at a constant temperature, so <math> ∆E_{thermal} = ∆E_{system} = 0</math>

<math> ∆E_{system} = W + Q</math> + ''other energy inputs''

<math> 0 = 0 + Q + 5000 J</math>

<math>Q = -5000 J </math>

===Middling===
You put a thin metal pot containing 3787 <math>g</math> of room-temperature (20 °C) water on a hot electric
stove. You also stir the water vigorously with an electric beater, which does 100 <math>J</math> of work on
the water every second. You observe that after 10 minutes the water starts to boil (100 °C).
The specific heat capacity of water is 4.2 <math>J(g·C)^{−1}</math>. What is the thermal transfer of energy, <math>Q</math>, into the water from the surroundings?

''Solution:''

<math>∆E_{system} = W_{surr} + Q</math>

<math>∆E_{thermal} = W_{surr} + Q</math>

<math>Q = ∆E_{thermal} - W_{surr}</math>

:<math>∆E_{thermal} = mC(∆T) = (3787 g)(4.2 J(g·C)^{−1})(100 - 20 °C)</math>

::<math>∆E_{thermal} = 1.272 </math> x <math> 10^6 J</math>

:<math>W_{surr} = W*t = (100 J*s^{-1})(10 * 60s)</math>

::<math>W_{surr} = 6 </math> x <math> 10^4 J</math>

<math>Q = ∆E_{thermal} - W_{surr} = 1.272 </math> x <math> 10^6 J - 6 </math> x <math> 10^4 J </math>

<math>Q = 1.212 </math> x <math> 10^6 J </math>

===Difficult===
You have two cups of coffee. Each cup contains 350 grams of coffee at 93 °C. The specific heat of coffee is 4.2 <math>J(g·°C)^{−1}</math>. Assume that no energy is transferred between the coffee and the cup.

*A. If you want to bring the coffee down to a temperature of 82 °C, how much cream do you have to add if the cream's initial temperature is 5 °C and it has a specific heat of 3.8 <math>J(g·°C)^{−1}</math>?

''Solution:''

1 = the cup of coffee

2 = the cream

need to find the mass of the cream you need to add:

<math>∆E = ∆E_{thermal, 1} + ∆E{thermal, 2}</math>

<math>m_1C_1(T_f - T_1) + m_2C_2(T_f - T_2) = 0</math>

<math>m_2C_2(T_f - T_2) = -m_1C_1(T_f - T_1)</math>

<math>m_2 = -m_1C_1(T_f - T_1)/C_2(T_f - T_2) = (350 g)(4.2 J(g·°C)^{−1})(82°C - 93 °C) / (3.8 J(g·°C)^{−1})(882°C - 5 °C)</math>

<math>m_2 = 55.3 g</math>

*B. You add 16 grams of sugar (heat capacity of 1.2 <math>J(g·°C)^{−1}</math> and initial temperature of 20 °C) to the coffee and stir, doing 15 <math>J</math> of work. When the liquid stops moving, what is the final temperature of the coffee?

''Solution:''

<math>∆E = ∆E_{thermal, c} + ∆E{thermal, s} = W</math>

<math>m_cC_c(T_f - T_c) + m_sC_s(T_f - T_s) = W</math>

<math>m_cC_cT_f - m_cC_cT_c + m_sC_sT_f - m_sC_sT_s = W</math>

<math>T_f(m_cC_c + m_sC_s) = W + m_cC_cT_c + m_sC_sT_s</math>

<math>T_f = W + m_cC_cT_c + m_sC_sT_s / (m_cC_c + m_sC_s) = (15 + (350)(4.2)(93) + (16)(1.2)(20)) / ((350)(4.2) + (16)(1.2))</math>

<math>T_f = 92.07 °C</math>

*C. After letting the coffee sit for a long time, the coffee has reached thermal equilibrium with the air and is at 20 °C. Taking the coffee as the system, what was the thermal transfer of energy, <math>Q</math>, between the system and surroundings?

''Solution:''
<math> ∆E = mC∆T = Q</math>

<math>Q = mC(T_f - T_i) = (350 g)(4.2 J(g·°C)^{−1})(20 °C - 93°C)</math>

<math>Q = -107310 J</math>

==Connectedness==

Interest

I am a chemical and biomolecular engineering major, so energy and heat transfer is one of the most fundamental principles behind reactor process design. Having a solid understanding of this topic will be very beneficial in future ChE courses.

Major Connection

As a chemical engineering major, I will see this topic in nearly every course that I will take. The concept of <math>Q = mC∆T</math> was introduced to us in our basic chemistry classes, we take two courses on thermodymanics, which heavily emphasize calculating the changes in thermal energy, and we will continue to see this topic in later courses such as kinetics, transport, and reactor design.

Industrial Application

Energy transfer due to a temperature difference is the key principle that makes heat exchangers work. Heat exchangers are used in many chemical process industries in order to cool down a flow stream, which is essential for regulating temperatures in chemical processes to ensure that the flow streams do not get too hot, which could cause serious damage. For a co-current heat exchanger, a hot stream enters a tube at a high temperature, and a cold stream enters a jacket surrounding the inner tube at a much lower temperature. As the two fluids travel down the tube, thermal energy is transferred between the two streams, causing the hot stream to decrease in temperature and the cold stream to increase in temperature. The cooled down hot stream can now move to the next step in the chemical process, and the heated up cold stream can either be used somewhere else in the process, or discarded. Below is a diagram of a co-current heat exchanger.

[[File:Heatexchanger.jpg]]

Using the principle of energy transfer due to a temperature difference, you can write a differential equation to describe the change in temperature of stream 1 as a function of length, which comes out to be the following when simplified:

[[File:Exchangerequation.gif]]

Where:

:<math>A_s</math> = surface area

:<math>q</math> = flow rate

:<math>ρ</math> = fluid density

:<math>C_p</math> = heat capacity

==History==

The history of heat and work can be dated back to 1789 when the French scientist Antoine Lavoisier created a new theory - the phlogiston theory on Chemistry that basically negated all previous findings of combustion. He gave heat a meaning and definition that led to the interpretation of heat that is accepted today.

The modern interpretation is what we hold true in the example questions that we used earlier Q = mCdeltatT.

In 1798, Benjamin Thompson - minister for war and police in the German state of Bavaria - wanted to figure out where all the heat from the cannons were coming from. He observed that the surroundings of the cannon got hotter and not colder. He hypothesized that some of the mechanical work done on the cannon was converted to heat.

In 1849, English physicist James Prescott Joule published his work and findings on the conversion of work to heat that Thompson started. He formulated work equivalent of heat. 1 newton meter of work = 0.241 calories of heat.

In 1850, German physicist Clausius published his works on how conserved quantity is neither heat nor work, but a combination of both. He named this Energy and that is where we get the macroscopic equation deltaE = Q - W.

== See also ==

Relevant material and additional reading can be looked into the history of Joule and the laws of thermodynamics.

===Further reading===
[http://www.britannica.com/science/specific-heat Specific Heat: Britannica] 
[http://www.nature.com/nature/journal/v181/n4609/abs/181642a0.html Dependence of Heat Capacity on Thermal History: Nature Publishing Group] 
[http://environmentalresearchweb.org/cws/article/news/51788 Environmental Research Web | Uneven Climate Change Due to Atmospheric Heat Capacity]

===External links===
[http://water.usgs.gov/edu/heat-capacity.html USGS Water Science School] 
[http://www.sciencedirect.com/science/article/pii/S2095263513000630 Science Direct | Determination of specific heat capacity by transient plane source] 
[https://www.youtube.com/embed/bODiX2PjCPE Heat Transfer] 
[https://www.youtube.com/embed/XmXr58ySUhI Relationship between deltaE and Q+W] 

==References==

"A Brief History of Heat and Work." A Brief History of Heat and Work. N.p., n.d. Web. 10 Nov. 2015.\

"Application of Specific Heat Capacity" : Application of Specific Heat Capacity. N.p., n.d. Web. 10 Nov. 2015.

"CHAPTER 4: HEAT." : 4.2 Specific Heat Capacity. N.p., n.d. Web. 10 Nov. 2015.

Division of Building Technology, Chalmers University of Technology, Gothenburg SE-412 96, Sweden
Received 23 August 2013, Revised 3 September 2013, Accepted 10 September 2013, Available online 22 October 2013

"Re: What Are Some Practical Uses of Determining the Specific Heat of a Metal?" Re: What Are Some Practical Uses of Determining the Specific Heat of a Metal? N.p., n.d. Web. 10 Nov. 2015.

[[Category:Energy]]

Cgillon3: /* Difficult */

Energy Transfer due to a Temperature Difference

2016-11-28T01:04:07Z

Cgillon3: /* Difficult */