Charged Rod

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Changes: Reformatting to match the template, adding a computational model, adding examples, and some grammatical/spelling fixes. Figures used in the examples were drawn by me.

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]\displaystyle{ \Delta \vec{E} }[/math] contributed by one of the pieces.

2. Choose an origin and the 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. 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]\displaystyle{ L }[/math] and positive charge [math]\displaystyle{ Q }[/math] centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis.

Step 1: Divide the Distribution into Pieces; Draw [math]\displaystyle{ \Delta \vec{E} }[/math]

Imagine dividing the rod into a series of very thin slices, each with the same charge [math]\displaystyle{ \Delta Q }[/math]. This charge [math]\displaystyle{ \Delta Q }[/math] is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, [math]\displaystyle{ \Delta E }[/math]. Summing all these individual slices of [math]\displaystyle{ E }[/math] gives you the total electric field of the rod. This process 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. We use the formula of the electric field for a point charge because we are imagining each slice as a point charge. First, determine [math]\displaystyle{ r }[/math], the vector pointing from the source to the observation location. For our example, this is [math]\displaystyle{ r = obs - source = \lt 0,y,0\gt - \lt x,0,0\gt = \lt -x,y,0\gt }[/math]. Now use this to calculate the magnitude and direction of [math]\displaystyle{ r }[/math]. So [math]\displaystyle{ |\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2} }[/math] and [math]\displaystyle{ \hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{\lt -x,y,0\gt }{\sqrt{x^2 + y^2}} }[/math]. [math]\displaystyle{ \hat{r} }[/math] is the vector portion of the expression for the field. The scalar portion is [math]\displaystyle{ \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2} }[/math]. Thus the expression for one slice of the rod is: [math]\displaystyle{ \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\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 means the integration variable is [math]\displaystyle{ dx }[/math]. We need to put this integration variable into our expression for the electric field. More specifically, we need to express [math]\displaystyle{ \Delta Q }[/math] in terms of the integration variable. Recall that the rod is uniformly charged, so the charge on any single slice of it is: [math]\displaystyle{ \Delta Q = (\frac{\Delta x}{L})\cdot Q }[/math]. This quantity can also be expressed in terms of the charge density.

Expression for [math]\displaystyle{ \Delta \vec{E} }[/math]

Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get [math]\displaystyle{ \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{(\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. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from [math]\displaystyle{ -L/2 }[/math] to [math]\displaystyle{ +L/2 }[/math]. So the expression is [math]\displaystyle{ \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\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

Finally, the fourth step is to check the result. The units should be the same as the units of the expression for the electric field for a single point particle [math]\displaystyle{ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} }[/math]. Our answer has the right units, since [math]\displaystyle{ \frac{1}{(\sqrt{x^2+ (L/2)^2})} }[/math]. has the same units of [math]\displaystyle{ \frac{Q}{r^2} }[/math] Is the direction qualitatively correct? We have the electric field pointing straight away from the midpoint of the rod, which is correct, given the symmetry of the situation. The vertical component of the electric field should indeed be zero.

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

A Computational Model

This simulation depicts the electric field around a charged rod. You can see how when mathematically finding the electric field around a rod, you treat the rod as a series of point charges. This is how it was done in the explanations above, and you can see the difficult example below to put this idea into practice.

Examples

Simple

This problem is meant to give you practice using the pre-derived formulas. While this is good for understanding the differences between the use of the actual formula and the approximation, you will almost always have to derive it yourself on the exam for a different scenario.

Problem Statement

A thin plastic rod of length 2.2 m is rubbed all over with wool, and acquires a charge of 89 nC, distributed uniformly over its surface. Calculate the magnitude of the electric field due to the rod at a location 15 cm from the midpoint of the rod. Do the calculation two ways, first using the exact formula for a rod of any length, and second using the approximate formula for a long rod.

Solution

(a) Exact formula

From the exam formula sheet: [math]\displaystyle{ |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{r \sqrt{r^2 + (L/2)^2}} }[/math] ([math]\displaystyle{ r }[/math] perpendicular from the center)

[math]\displaystyle{ |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{89e^{-9}}{.15 \sqrt{.15^2 + (2.2/2)^2}} }[/math]

[math]\displaystyle{ |\vec{E_{rod}}| = 4810.03 N/C }[/math]

(b) Approximate Formula

From the exam formula sheet: [math]\displaystyle{ |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{2Q/L}{r} }[/math] (if [math]\displaystyle{ r \lt \lt L) }[/math]

[math]\displaystyle{ |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{2(89e^{-9}/2.2)}{.15} }[/math]

[math]\displaystyle{ |\vec{E_{rod}}| = 4854.55 N/C }[/math]

This shows that while the approximation formula can be accurate for rod lengths much greater than the distance from the rod, in this case, the distances are too close for it to be a good approximation.

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]\displaystyle{ -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]\displaystyle{ ‹ 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.

Use the following as necessary: [math]\displaystyle{ x, y, dx, A, Q }[/math]. Remember that the rod has charge [math]\displaystyle{ -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]\displaystyle{ dQ }[/math] on the small piece of length [math]\displaystyle{ 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]\displaystyle{ \lambda = - \frac{Q}{2A} }[/math]

Part B

[math]\displaystyle{ dQ = - \frac{Q}{2A} dx }[/math]

Part C

[math]\displaystyle{ \vec{r} = \lt -x, y, 0\gt }[/math]

Part D

[math]\displaystyle{ d = \sqrt{(-x)^2 + y^2} }[/math]

Part E

[math]\displaystyle{ x }[/math]

Difficult

This problem is meant to combine the first two examples to give you practice doing an actual exam problem, where you're presented with an unfamiliar setup, and have to derive the formula for electric field yourself.

Problem Statement:

A very thin plastic rod of length [math]\displaystyle{ L }[/math] is rubbed with cloth and becomes uniformly charged with a total charge [math]\displaystyle{ +Q }[/math].

(a) Consider an arbitrary piece of rod of length [math]\displaystyle{ dx }[/math] located at a position x on the rod. Determine the electric field [math]\displaystyle{ \vec{E} }[/math] from this piece at observation location "*", a distance [math]\displaystyle{ 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]\displaystyle{ \vec{E} }[/math] of the rod at location "*".

Solution

Part A:

[math]\displaystyle{ \vec{r} = \vec{r_*}-\vec{r_{dx}} = \lt L+w, 0, 0 \gt - \lt x, 0, 0 \gt = \lt L+w-x, 0, 0 \gt = (L+w-x)\hat{x} }[/math]

[math]\displaystyle{ |\vec{r}| = L+w-x }[/math]  ; [math]\displaystyle{ \hat{r} = \lt 1, 0, 0 \gt = \hat{x} }[/math]

[math]\displaystyle{ \lambda = \frac{charge}{length} = \frac{Q}{L} = \frac{dQ}{dx} }[/math]

[math]\displaystyle{ dQ = \frac{Q}{L} dx }[/math]

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

Part B:

[math]\displaystyle{ \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]\displaystyle{ \int \frac{dx}{(x+a)^2} = \frac{-1}{a+x} }[/math]

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

[math]\displaystyle{ \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.

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?

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. 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 which objects are contributing how much charge in certain areas. Charged objects may attract or repel (depending on the signs of their charge), so we often need to know how objects will interact with each other based on their charges. The phenomenon of this interaction, or electric force between charged particles, was finally confirmed and stated as a law in 1785 by French physicist Charles-Augustin de Coulomb, hence "Coulomb's Law." A portrait of Coulomb is shown below.

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, such as 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. 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

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