Field of a Charged Rod: Difference between revisions

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'''Milo Karnes, Spring 2025'''
''(After finishing this, I found an article called [[Charged Rod]] written by the people of yesteryear. I didn't see that and created this new one instead. The 2 should be merged in the future.)''


== The Main Idea ==
== The Main Idea ==


Previously, we've learned about the electric field of a point particle. Often, when analyzing physical systems, it is the case that we're unable to analyze each individual particle that composes an object and need to therefore generalize collections of particles into shapes (in this case, a rod) whereby the mathematics corresponding to electric field calculations can be simplified. This can essentially be done by adding up the contributions to the electric field made by parts of an object, approximating each part of an object as a point charge.
In earlier studies, we learned about the electric field created by a point charge. However, in the real world, charges are often spread out over objects with shape and structure. One common example is a '''uniformly charged rod'''. To determine the electric field from such an object, we divide the rod into many infinitesimally small charge segments, treat each as a point charge, and integrate their contributions.


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 key idea is to approximate the rod as a continuous line of charge using the principle of superposition. We consider the symmetry of the setup to simplify the problem and focus on the components of the electric field that don't cancel out.


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 from a charged rod involves four main steps:


The process of finding the electric field due to charge distributed over an object has four steps:
# '''Model the rod as many small charge elements''' and draw the electric field vector <math>\Delta \vec{E}</math> from a single element.
# '''Use symmetry''' to argue which components cancel and which remain.
# '''Integrate''' the contributions from all elements to find the net electric field.
# '''Verify''' that the result makes physical sense (units, direction, and limiting behavior).


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 the axes. Write an algebraic expression for the electric field <math>\Delta \vec{E}</math> due to one piece.
== A Mathematical Model ==


3. Add up the contributions of all pieces, either numerically or symbolically.
To calculate the electric field of a uniformly charged rod, we treat the rod as a continuous distribution of charge. Let the rod have total length <math>L</math> and total charge <math>Q</math>, centered along the x-axis. The observation point is located on the y-axis a distance <math>y</math> above the center.


4. Check that the result is physically correct.
=== Step 1: Break the Rod into Pieces ===


=== A Mathematical Model ===
We divide the rod into tiny segments of length <math>dx</math>. Each segment behaves like a point charge:
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.
<math> dq = \lambda \, dx </math>, where <math> \lambda = \frac{Q}{L} </math> is the linear charge density.


'''Step 1: Divide the Distribution into Pieces; Draw <math>\Delta \vec{E}</math>'''
Each <math>dq</math> contributes a small electric field <math> d\vec{E} </math> at the observation point.


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.
<div style="text-align: center;">[[File:ChargedRodBreakdown.png]]</div>


'''Step 2: Write an Expression for the Electric Field Due to One Piece'''
=== Step 2: Write the Field Expression for 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:
Using Coulomb’s Law, the electric field contribution from one element 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'''
<math>
d\vec{E} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{dq}{r^2} \cdot \hat{r}
</math>


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>
The vector from the source to the observation point is:
\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>
<math> \vec{r} = \langle 0, y \rangle - \langle x, 0 \rangle = \langle -x, y \rangle </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.
So:


'''Step 3: Add Up the Contributions of All the Pieces'''
<math> r = \sqrt{x^2 + y^2} \quad \text{and} \quad \hat{r} = \frac{\langle -x, y \rangle}{\sqrt{x^2 + y^2}} </math>


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.
Putting it all together:
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'''
<math>
d\vec{E} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{\lambda \, dx}{(x^2 + y^2)} \cdot \frac{\langle -x, y \rangle}{\sqrt{x^2 + y^2}}
= \frac{\lambda}{4\pi\varepsilon_0} \cdot \frac{\langle -x, y \rangle \, dx}{(x^2 + y^2)^{3/2}}
</math>


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>.
Note: The x-components of the field cancel due to symmetry. Only the y-component adds up.
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.


==== The System in Question ====
=== Step 3: Integrate ===


As discussed in the previous section, we're considering a system
We now integrate from <math>-L/2</math> to <math>+L/2</math>:
abstracted from the particle model we're familiar with, therefore we will
make the generalization that our rod of length L has a total charge of
quantity Q. For this generalization, we will need to assume that the rod
is so thin that we can ignore its thickness.


[[Image:LukasYoder01.jpg|200px|center]]
<math>
E_y = \frac{\lambda y}{4\pi\varepsilon_0} \int_{-L/2}^{L/2} \frac{dx}{(x^2 + y^2)^{3/2}}
</math>


Since the electric field produced by a charge at any given location is
This integral has a standard solution:
proportional to the distance from the charge to that location, we will
need to relate the observation location to the source of the charge, which
we will consider the origin of the rod. To do that, we will need to divide
the rod into pieces of length <math>\Delta y</math> each containing a charge <math>\Delta Q</math>.
In the image below, you can see what this looks like and the relation that
can be found between the observation location and the source, forming the
distance vector <math>\vec{r}</math>.


[[Image:LukasYoder02.jpg|400px|center]]
<math>
E_y = \frac{\lambda}{4\pi\varepsilon_0 y} \cdot \left( \frac{L/2}{\sqrt{(L/2)^2 + y^2}} \right)
</math>


By the pythagorean theorem, we can find the vector <math>\vec{r}</math> as follows:
Final result:


[[Image:LukasYoder03.jpg|400px|center]]
<math>
E_y = \frac{Q}{4\pi\varepsilon_0 L y} \cdot \left( \frac{L/2}{\sqrt{(L/2)^2 + y^2}} \right)
</math>


<div style="text-align: center;">[[File:ElectricFieldVectorFromRod.png]]</div>


And to find the unit vector in the direction of <math>\vec{r}</math>, <math>\hat{r}</math>, we do as
follows:


[[Image:LukasYoder04.jpg|400px|center]]
=== Step 4: Check the Result ===


* '''Units''': The result has units of N/C, as expected.
* '''Direction''': The field points away from the rod if <math>Q > 0</math>, and toward the rod if <math>Q < 0</math>.
* '''Limiting Behavior''': As <math>y \gg L</math>, the result simplifies to the electric field of a point charge:
<math>
E \approx \frac{Q}{4 \pi \varepsilon_0 y^2}
</math>




==== Finding the Contribution of Each Piece to the Electric Field ====
== Computational Models ==


Now that we've set up a model for the system, with the rod broken down
While symbolic solutions give us a deep understanding of how to derive the electric field, computational models allow us to visualize it in action. This is especially useful when dealing with continuous charge distributions like a uniformly charged rod.
into pieces, we can find the contribution of each piece to the electric
field of the system. We will start from the electric field equation you
learned for a point particle but plug in the parameters for the rod system
into the equation.


[[Image:LukasYoder05.jpg|400px|center]]
GlowScript (a VPython-based simulation platform) makes it possible to numerically simulate and animate the electric field from a segmented rod. In these simulations, we treat the rod as a collection of small point charges, and compute the electric field vectors they produce at various locations in space.


=== Field Mapping Along the Rod's Length ===


By mathematically simplifying, we then get the following equation:
In this first simulation, the rod is broken into a series of green spheres, each representing a small segment of uniform charge. At various points near the rod, electric field vectors are plotted to show how the strength and direction of the field change along and around the rod.


[[Image:LukasYoder06.jpg|400px|center]]
[http://www.glowscript.org/#/user/yoderlukas/folder/Public/program/ElectricFieldAlongRodLength '''▶ Run Simulation: Electric Field Along Rod''']


'''Key takeaways:'''
* Observe how the direction of the field changes based on position.
* Near the center of the rod, the field is strongest and most symmetric.
* Near the ends, the field vectors curve — an effect known as ''edge effects''.
* This illustrates why we often assume the rod is “infinitely long” in theory — to ignore those ends and simplify the math.


==== Finding the Net Contribution of all Pieces ====
This tool is particularly helpful for developing intuition before solving test problems involving rods, wires, or even capacitors.


In the previous section, we found out the contribution to the electric
=== Radial Field Symmetry: Positive vs. Negative Charge ===
field at a given location of only one of the pieces constituting the rod.
In order to figure out the net field at any particular location, we need
to add up the electric fields produced by individual pieces along the
length of the rod.


We will switch from vector notation for the electric field to the scalar
The next simulations demonstrate how the electric field behaves around a rod when it is positively or negatively charged. This is where direction matters — not just magnitude.
notation for the x- and y-components. (From the vector in the equation
above, we can see that the z-component of the electric field at any point
is always 0.) The x-component of the electric
field is the sum of the x-components of every <math>\Delta{y}</math> along the rod, and
the y-component of the electric field is the sum of the y-components of
every <math>\Delta{y}</math> along the rod. We can show this mathematically:


[[Image:LukasYoder07.jpg|400px|center]]
* [https://www.glowscript.org/#/user/michaelwise/folder/Public/program/LineofCharge-Positive '''▶ Positive Rod Simulation''']
* [https://www.glowscript.org/#/user/michaelwise/folder/Public/program/LineofCharge-Negative '''▶ Negative Rod Simulation''']


These models break the rod into 40 discrete segments and compute the net electric field throughout a 2D grid surrounding the rod.


To make use of this relation, because we don't know <math>\Delta{Q}</math>, we need to
'''Key observations:'''
relate it to parameters that we already know about the rod system we're
* Field lines from a positively charged rod '''point outward'''.
analyzing. We can express <math>\Delta{Q}</math> as the charge density of the rod
* For a negatively charged rod, field lines '''point inward'''.
(which is Q/L) times the <math>\Delta{y}</math> we've chosen for the system. Thus,
* The symmetry is most noticeable when zoomed out — mimicking how an infinite line of charge behaves.
* Near the ends, the distortion shows the importance of boundary conditions in real systems.


[[Image:LukasYoder08.jpg|200px|center]]
These are powerful tools when preparing for conceptual questions or visual reasoning tasks — they help make the math real.


=== Why It Matters ===


By plugging the above equation into our equations for the x- and
Computational models bridge the gap between the idealized math we do on paper and the messy, real-world systems we encounter in labs and engineering. They show us what electric fields actually look like when we account for discrete steps, edge effects, and variable observation points.
y-components of the electric field at a point, we can find the electric
field at any point in the system. This technique is called numerical
integration and is typically done by computers because the computational
complexity is dependant upon the size of <math>\Delta{y}</math> with respect to L.


==== Simplifying ====
They’re also interactive — you can zoom in, rotate, and change parameters. When studying for exams, use these models to test your intuition: If a question asks you about the direction of a field, imagine running one of these and predicting what it would look like.


Using calculus, we can simplify a lot of the math required to compute the
<small>All simulations above are written using GlowScript (VPython), and are free to edit or remix as part of your own projects or demonstrations.</small>
electric field at any given point. Notationally, all we're doing is switching from the
discretely-sized <math>\Delta{y}</math> to <math>dy</math> and from the sigma notation to
an integral starting from -L/2 (the lower end of the rod) and ending at
L/2 (the upper end of the rod) as follows:


[[Image:LukasYoder09.jpg|400px|center]]
[[Image:LukasYoder10.jpg|400px|center]]


== Examples ==


By evaluating the integral, we can determine that the x-component of the
These examples cover different levels of conceptual and quantitative understanding of the electric field due to a uniformly charged rod. Try using the computational models above to visualize your answers!
electric field at any point is:


[[Image:LukasYoder11.jpg|400px|center]]
=== Example 1: Symmetry and Direction (Conceptual) ===


You are observing a positively charged rod lying along the x-axis, centered at the origin. You stand at a point on the y-axis a distance <math>y</math> above the midpoint of the rod.


Without evaluating the integral for the y-component of the electric field,
'''Question:''' 
we can use symmetry to determine that the y-component of the electric
What direction does the electric field point at your location?
field at any given point is 0. Let's consider the contributions to the
electric field from the top and bottom halves of the rod at any
observation point.


[[Image:LukasYoder12.jpg|200px|center]]
'''Answer:''' 
The horizontal components of the field from symmetric charge elements cancel, and only the vertical (y-direction) components add constructively. Therefore, the electric field points directly upward (in the +y direction) if the rod is positively charged.




Since the y-components of <math>E_{top}</math> and <math>E_{bottom}</math> are of equal magnitude and
=== Example 2: Deriving the Field at a Point (Symbolic) ===
opposite direction, they cancel each other out, and therefore the
y-component of teh electric field at any given point due to the rod is 0.


[[Image:LukasYoder13.jpg|200px|center]]
A rod of total length <math>L = 2.0 \, \text{m}</math> carries a total charge <math>Q = 4.0 \times 10^{-6} \, \text{C}</math>. It lies along the x-axis, centered at the origin. Find the magnitude of the electric field at a point <math>y = 0.5 \, \text{m}</math> directly above its center.


'''Solution:'''


Finally, because the rod is round and can be rotated, as a convenience,
Use the derived formula:
we'll use d (distance from the rod) as opposed to x (distance along the
x-direction) to refer to the electric field.


Thus we can simplify electric field calculations for a rod into a form
<math>
that we can readily use:
E_y = \frac{Q}{4\pi\varepsilon_0 L y} \cdot \left( \frac{L/2}{\sqrt{(L/2)^2 + y^2}} \right)
</math>


[[Image:LukasYoder14.jpg|400px|center]]
Substitute known values:


<math>
\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2
</math>


==== Further Simplification ====
<math>
E_y = \frac{4.0 \times 10^{-6}}{4\pi(8.85 \times 10^{-12}) \cdot (2.0)(0.5)} \cdot \left( \frac{1.0}{\sqrt{1.0^2 + 0.5^2}} \right)
</math>


By noting the contributions of each variable to the equation for the
Evaluate:
electric field, we can make approximations to simplify our math by simply
declaring one variable as insignificant.


For example, if we have a system in which the length of a rod is much
<math>
greater than the magnitude of the distance from the rod (denoted L>>d), we
E_y \approx 2.3 \times 10^5 \, \text{N/C}
can neglect some of the instances in which d is taken into account as
</math>
follows:


[[Image:LukasYoder15.jpg|400px|center]]
---


=== A Computational Model ===
=== Example 3: Using the Simulation to Analyze Field Strength (Numerical/Visual) ===


==== (Finding the Electric Field from a Rod with Code) ====
Use the [Electric Field Along Rod Simulation](http://www.glowscript.org/#/user/yoderlukas/folder/Public/program/ElectricFieldAlongRodLength) to compare electric field strengths at two different points:


Here is some code that you can run which shows the electric field vector
* Point A: Directly above the midpoint of the rod
at a given distance from the rod along its length. The rod is shown as a
* Point B: The same distance above one of the rod’s ends
series of green balls to help emphasize that when using the numerical
integrations mentioned on this page, you are measuring the field produced
by discrete parts of the rod being analyzed.


Notice the edge-effects of the electric field of the rod. For reasons
'''Question:''' 
discussed above, if we used the long rod approximation (L>>d), these
Which point has a stronger electric field? How does the field direction differ?
effects would be negligible.


[http://www.glowscript.org/#/user/yoderlukas/folder/Public/program/ElectricFieldAlongRodLength Click Here to Run the Code]
'''Reflection:''' 
The field at Point A is stronger and points vertically due to symmetry. At Point B, the field is weaker and points diagonally inward toward the rod’s center. This shows the importance of symmetry in superposition, and helps illustrate “edge effects” that are harder to account for with algebra alone.


==Examples==


Although this is not a very difficult topic, some reasonably difficult conceptual questions can be asked about it.
== Connectedness ==


===Simple===
At first glance, finding the electric field of a charged rod might just seem like a physics exercise, but it actually connects to a lot of real things — especially in aerospace.


[[Image:LukasYoderNo.jpg|400px|center]]
In spacecraft and satellites, components like power cables, structural booms, or tethers can hold charge and act basically like long rods. In space, charge builds up from sunlight or the plasma environment, and knowing how that charge creates electric fields helps engineers prevent issues like interference, arcing, or electrostatic discharge — which could damage sensitive electronics.


===Middling===
Even in aircraft, stuff like static wicks or long sensor probes can behave like charged rods, and their fields matter for things like EMC (electromagnetic compatibility). We need to make sure systems don’t interfere with each other, especially in environments where you’ve got a bunch of tightly packed electrical components.


[[Image:LukasYoderMaybe.jpg|400px|center]]
As an AE major, I see this concept show up in ways I didn’t expect — whether it’s power systems on satellites, electrostatics in high-altitude flight, or how fields interact with composite materials. So yeah, we start with this simple charged rod, but the same physics applies all the way up to full-on orbital systems.


===Difficult===


[[Image:LukasYoderYes.jpg|400px|center]]
== History ==


== Connectedness ==
The idea that charged objects create electric fields has been around for a long time, but things really clicked in the 1700s when Charles-Augustin de Coulomb figured out how the force between charges actually worked. His experiments led to Coulomb’s Law, which became the foundation for electric field theory.
 
Back then, most experiments focused on point charges because they were easier to understand and measure. But over time, physicists realized that most real objects — wires, rods, surfaces — aren’t point-like. That’s where the idea of continuous charge distributions came in.


Electric fields are very important to electrical engineering because they can be used to convey signals. In fact, there is an entire sub-field of electrical engineering called digital signal processing that focuses on modulating different characteristics of radio frequency (RF) signals that are produced by electric fields.
The charged rod is one of the first examples you see when learning how to deal with this. It helps build the foundation for understanding more complex things like rings, disks, and eventually 3D shapes. Once you learn how to break up a rod into tiny charge pieces and add up their fields, you can start doing the same for almost any shape.


== History ==
It’s kind of cool to think that this whole section — which starts with drawing a little rod and doing some integrals — is part of the same chain of ideas that eventually leads to things like electric field simulations for spacecraft or antennas.


The history of the electric field can be found in a previous section on electric fields. The equation for the electric field of a charged rod was not a discovery, but simply a mathematical simplification that someone made to model a rod. As such, nobody knows who first made the mathematical simplification, where they made it, or when, though it was certainly made after electric fields were discovered.


== See also ==
== See also ==


The equation for the electric field of a charged rod was derived from the equation for an electric field of a charged particle. See the article "[[Electric Field]]" for more information.
The equation for the electric field of a charged rod was derived from the equation for the electric field of a charged particle. See the article "[[Electric Field]]" for more information.


=== Further Reading ===
=== Further Reading ===
Line 240: Line 223:


=== External Links ===
=== External Links ===
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html


http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRods.asp
http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRods.asp
Line 248: Line 232:


== References ==
== References ==
https://www.glowscript.org/#/
https://rhettallain_gmail_com.trinket.io/intro-to-electric-and-magnetic-fields#/electric-fields/multiple-charges


https://www.youtube.com/watch?v=BBWd0zUe0mI
https://www.youtube.com/watch?v=BBWd0zUe0mI


(For the above reference, I chose to follow the textbook's method in not defining the charge distribution and assuming it was constant, though this was helpful in figuring out a better way to introduce it.)
(For the above reference, the textbook's method is followed in that the charge distribution was left undefined, and assumed to be constant)


Chabay and Sherwood: Matter and Interactions, Fourth Edition, Chapter 15
Chabay and Sherwood: Matter and Interactions, Fourth Edition, Chapter 15


All figures created by author
 
[[Category: Electric Field]]
[[Category: Electric Field]]

Latest revision as of 17:05, 13 April 2025

Milo Karnes, Spring 2025

The Main Idea

In earlier studies, we learned about the electric field created by a point charge. However, in the real world, charges are often spread out over objects with shape and structure. One common example is a uniformly charged rod. To determine the electric field from such an object, we divide the rod into many infinitesimally small charge segments, treat each as a point charge, and integrate their contributions.

The key idea is to approximate the rod as a continuous line of charge using the principle of superposition. We consider the symmetry of the setup to simplify the problem and focus on the components of the electric field that don't cancel out.

The process of finding the electric field from a charged rod involves four main steps:

  1. Model the rod as many small charge elements and draw the electric field vector [math]\displaystyle{ \Delta \vec{E} }[/math] from a single element.
  2. Use symmetry to argue which components cancel and which remain.
  3. Integrate the contributions from all elements to find the net electric field.
  4. Verify that the result makes physical sense (units, direction, and limiting behavior).


A Mathematical Model

To calculate the electric field of a uniformly charged rod, we treat the rod as a continuous distribution of charge. Let the rod have total length [math]\displaystyle{ L }[/math] and total charge [math]\displaystyle{ Q }[/math], centered along the x-axis. The observation point is located on the y-axis a distance [math]\displaystyle{ y }[/math] above the center.

Step 1: Break the Rod into Pieces

We divide the rod into tiny segments of length [math]\displaystyle{ dx }[/math]. Each segment behaves like a point charge: [math]\displaystyle{ dq = \lambda \, dx }[/math], where [math]\displaystyle{ \lambda = \frac{Q}{L} }[/math] is the linear charge density.

Each [math]\displaystyle{ dq }[/math] contributes a small electric field [math]\displaystyle{ d\vec{E} }[/math] at the observation point.

Step 2: Write the Field Expression for One Piece

Using Coulomb’s Law, the electric field contribution from one element is:

[math]\displaystyle{ d\vec{E} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{dq}{r^2} \cdot \hat{r} }[/math]

The vector from the source to the observation point is:

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

So:

[math]\displaystyle{ r = \sqrt{x^2 + y^2} \quad \text{and} \quad \hat{r} = \frac{\langle -x, y \rangle}{\sqrt{x^2 + y^2}} }[/math]

Putting it all together:

[math]\displaystyle{ d\vec{E} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{\lambda \, dx}{(x^2 + y^2)} \cdot \frac{\langle -x, y \rangle}{\sqrt{x^2 + y^2}} = \frac{\lambda}{4\pi\varepsilon_0} \cdot \frac{\langle -x, y \rangle \, dx}{(x^2 + y^2)^{3/2}} }[/math]

Note: The x-components of the field cancel due to symmetry. Only the y-component adds up.

Step 3: Integrate

We now integrate from [math]\displaystyle{ -L/2 }[/math] to [math]\displaystyle{ +L/2 }[/math]:

[math]\displaystyle{ E_y = \frac{\lambda y}{4\pi\varepsilon_0} \int_{-L/2}^{L/2} \frac{dx}{(x^2 + y^2)^{3/2}} }[/math]

This integral has a standard solution:

[math]\displaystyle{ E_y = \frac{\lambda}{4\pi\varepsilon_0 y} \cdot \left( \frac{L/2}{\sqrt{(L/2)^2 + y^2}} \right) }[/math]

Final result:

[math]\displaystyle{ E_y = \frac{Q}{4\pi\varepsilon_0 L y} \cdot \left( \frac{L/2}{\sqrt{(L/2)^2 + y^2}} \right) }[/math]


Step 4: Check the Result

  • Units: The result has units of N/C, as expected.
  • Direction: The field points away from the rod if [math]\displaystyle{ Q \gt 0 }[/math], and toward the rod if [math]\displaystyle{ Q \lt 0 }[/math].
  • Limiting Behavior: As [math]\displaystyle{ y \gg L }[/math], the result simplifies to the electric field of a point charge:

[math]\displaystyle{ E \approx \frac{Q}{4 \pi \varepsilon_0 y^2} }[/math]


Computational Models

While symbolic solutions give us a deep understanding of how to derive the electric field, computational models allow us to visualize it in action. This is especially useful when dealing with continuous charge distributions like a uniformly charged rod.

GlowScript (a VPython-based simulation platform) makes it possible to numerically simulate and animate the electric field from a segmented rod. In these simulations, we treat the rod as a collection of small point charges, and compute the electric field vectors they produce at various locations in space.

Field Mapping Along the Rod's Length

In this first simulation, the rod is broken into a series of green spheres, each representing a small segment of uniform charge. At various points near the rod, electric field vectors are plotted to show how the strength and direction of the field change along and around the rod.

▶ Run Simulation: Electric Field Along Rod

Key takeaways:

  • Observe how the direction of the field changes based on position.
  • Near the center of the rod, the field is strongest and most symmetric.
  • Near the ends, the field vectors curve — an effect known as edge effects.
  • This illustrates why we often assume the rod is “infinitely long” in theory — to ignore those ends and simplify the math.

This tool is particularly helpful for developing intuition before solving test problems involving rods, wires, or even capacitors.

Radial Field Symmetry: Positive vs. Negative Charge

The next simulations demonstrate how the electric field behaves around a rod when it is positively or negatively charged. This is where direction matters — not just magnitude.

These models break the rod into 40 discrete segments and compute the net electric field throughout a 2D grid surrounding the rod.

Key observations:

  • Field lines from a positively charged rod point outward.
  • For a negatively charged rod, field lines point inward.
  • The symmetry is most noticeable when zoomed out — mimicking how an infinite line of charge behaves.
  • Near the ends, the distortion shows the importance of boundary conditions in real systems.

These are powerful tools when preparing for conceptual questions or visual reasoning tasks — they help make the math real.

Why It Matters

Computational models bridge the gap between the idealized math we do on paper and the messy, real-world systems we encounter in labs and engineering. They show us what electric fields actually look like when we account for discrete steps, edge effects, and variable observation points.

They’re also interactive — you can zoom in, rotate, and change parameters. When studying for exams, use these models to test your intuition: If a question asks you about the direction of a field, imagine running one of these and predicting what it would look like.

All simulations above are written using GlowScript (VPython), and are free to edit or remix as part of your own projects or demonstrations.


Examples

These examples cover different levels of conceptual and quantitative understanding of the electric field due to a uniformly charged rod. Try using the computational models above to visualize your answers!

Example 1: Symmetry and Direction (Conceptual)

You are observing a positively charged rod lying along the x-axis, centered at the origin. You stand at a point on the y-axis a distance [math]\displaystyle{ y }[/math] above the midpoint of the rod.

Question: What direction does the electric field point at your location?

Answer: The horizontal components of the field from symmetric charge elements cancel, and only the vertical (y-direction) components add constructively. Therefore, the electric field points directly upward (in the +y direction) if the rod is positively charged.


Example 2: Deriving the Field at a Point (Symbolic)

A rod of total length [math]\displaystyle{ L = 2.0 \, \text{m} }[/math] carries a total charge [math]\displaystyle{ Q = 4.0 \times 10^{-6} \, \text{C} }[/math]. It lies along the x-axis, centered at the origin. Find the magnitude of the electric field at a point [math]\displaystyle{ y = 0.5 \, \text{m} }[/math] directly above its center.

Solution:

Use the derived formula:

[math]\displaystyle{ E_y = \frac{Q}{4\pi\varepsilon_0 L y} \cdot \left( \frac{L/2}{\sqrt{(L/2)^2 + y^2}} \right) }[/math]

Substitute known values:

[math]\displaystyle{ \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 }[/math]

[math]\displaystyle{ E_y = \frac{4.0 \times 10^{-6}}{4\pi(8.85 \times 10^{-12}) \cdot (2.0)(0.5)} \cdot \left( \frac{1.0}{\sqrt{1.0^2 + 0.5^2}} \right) }[/math]

Evaluate:

[math]\displaystyle{ E_y \approx 2.3 \times 10^5 \, \text{N/C} }[/math]

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Example 3: Using the Simulation to Analyze Field Strength (Numerical/Visual)

Use the [Electric Field Along Rod Simulation](http://www.glowscript.org/#/user/yoderlukas/folder/Public/program/ElectricFieldAlongRodLength) to compare electric field strengths at two different points:

  • Point A: Directly above the midpoint of the rod
  • Point B: The same distance above one of the rod’s ends

Question: Which point has a stronger electric field? How does the field direction differ?

Reflection: The field at Point A is stronger and points vertically due to symmetry. At Point B, the field is weaker and points diagonally inward toward the rod’s center. This shows the importance of symmetry in superposition, and helps illustrate “edge effects” that are harder to account for with algebra alone.


Connectedness

At first glance, finding the electric field of a charged rod might just seem like a physics exercise, but it actually connects to a lot of real things — especially in aerospace.

In spacecraft and satellites, components like power cables, structural booms, or tethers can hold charge and act basically like long rods. In space, charge builds up from sunlight or the plasma environment, and knowing how that charge creates electric fields helps engineers prevent issues like interference, arcing, or electrostatic discharge — which could damage sensitive electronics.

Even in aircraft, stuff like static wicks or long sensor probes can behave like charged rods, and their fields matter for things like EMC (electromagnetic compatibility). We need to make sure systems don’t interfere with each other, especially in environments where you’ve got a bunch of tightly packed electrical components.

As an AE major, I see this concept show up in ways I didn’t expect — whether it’s power systems on satellites, electrostatics in high-altitude flight, or how fields interact with composite materials. So yeah, we start with this simple charged rod, but the same physics applies all the way up to full-on orbital systems.


History

The idea that charged objects create electric fields has been around for a long time, but things really clicked in the 1700s when Charles-Augustin de Coulomb figured out how the force between charges actually worked. His experiments led to Coulomb’s Law, which became the foundation for electric field theory.

Back then, most experiments focused on point charges because they were easier to understand and measure. But over time, physicists realized that most real objects — wires, rods, surfaces — aren’t point-like. That’s where the idea of continuous charge distributions came in.

The charged rod is one of the first examples you see when learning how to deal with this. It helps build the foundation for understanding more complex things like rings, disks, and eventually 3D shapes. Once you learn how to break up a rod into tiny charge pieces and add up their fields, you can start doing the same for almost any shape.

It’s kind of cool to think that this whole section — which starts with drawing a little rod and doing some integrals — is part of the same chain of ideas that eventually leads to things like electric field simulations for spacecraft or antennas.


See also

The equation for the electric field of a charged rod was derived from the equation for the electric field of a charged particle. See the article "Electric Field" for more information.

Further Reading

The page on electric fields: Electric Field

External Links

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRods.asp

https://pages.uncc.edu/phys2102/online-lectures/chapter-02-electric-field/2-4-electric-field-of-charge-distributions/example-1-electric-field-of-a-charged-rod-along-its-axis/

http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml

References

https://www.glowscript.org/#/

https://rhettallain_gmail_com.trinket.io/intro-to-electric-and-magnetic-fields#/electric-fields/multiple-charges

https://www.youtube.com/watch?v=BBWd0zUe0mI

(For the above reference, the textbook's method is followed in that the charge distribution was left undefined, and assumed to be constant)

Chabay and Sherwood: Matter and Interactions, Fourth Edition, Chapter 15

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