Electric Fields ( i attached a file)

In summary: One for the positive charge, one for the negative charge, and one for the field. The electric field equation is Equation 3. The integral is used to calculate the field at a point. I am not sure where Pie is used in the equation. It seems like it might be in the denominator, but I am not sure.I don't think Pie is used in the denominator, it might be in the numerator. Pie is used to determine the direction of the field. In summary, pie is used to determine the direction of the field.
  • #1
trah22
45
0

Homework Statement


Hi, my question isn't actually over a particular homework problem, I've attached a file of my notes on electric field, electric dipoles, i don't quite understand why the equation of the electric field is setup that way for the negative charge.

Homework Equations





The Attempt at a Solution

 

Attachments

  • physics3.PDF
    31.8 KB · Views: 258
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  • #2
can someone help me out:smile:
 
  • #3
I'd love to help, but I'm unable to open the file
 
  • #4
weird what error comes up? maybe it requires the latest adobe, my file is in pdf. Argh I am not sure what could be wrong

Is there a specific way to attach files in order for them to be posted correctly for this forum?
 
Last edited:
  • #5
There's nothing wrong with how you posted your file; attachments just need to be approved before they can be viewed. Done!

Now, what exactly is your question? What about that expression for the field of the dipole is unclear? It's just expressing the field as a vector sum of the components of the field from each charge. Since the field from a negative charge points towards the charge, both components of its field at point P are negative. (Note that [itex]E_-[/itex] stands for the magnitude of the field at point P contributed by the negative charge.)
 
  • #6
You could also have noted that the components in the [tex]x[/tex] direction cancel each other out by simmetry, without having to worry about cosines ;)
 
  • #7
Well to find the Electric Field u have to add the electric field for the postive and negative charge. For the xaxis i undestand that its in the positive direction so its Ecos(theta)i but howcome for the y direction its -esin(theta)j, is it because its goin down? Thats for the postive charge electric field. For the negative charge, i don't understand why its minus Ecos(theta)i and then minus Esin(theta)j. Basically could someone explain to me why for the negative charge according to the way its graphed, why its -Ecosthetai-Esinthetaj.:smile:
 
  • #8
Ok, i actually just figured it out, thnx for the help anyhow,
 
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  • #9
Im just having some trouble on figuring out how some of the components are and how there being used for this part of the notes ( i attached another file),
 

Attachments

  • physics4.PDF
    33.3 KB · Views: 236
Last edited:
  • #10
Since [itex]E_+ = E_-[/itex], [itex]E_+ + E_- = E_+ + E_+ = 2 E_+ = 2 kq/D^2[/itex].
 
  • #11
ah i c thanks doc
 
  • #12
There was one more thing on that page that i don't quite get, when integrating equation 3 into equation 2, why is the a?d there? because E=F/q=kq/r^2x1/q, I am not quite sure on where and why the a/D is needed.

And then for this one particular problem, its very similar to the above problem, but i ran into a couple problems, 1 during the setup of the electric field equation, another when choosing the correct integral to use and lastly how R-L/Pie is used in the final Electric field equation.

The actual problem is:a uniformly charged insulating rod of length 14 cm is bent into the shape of a semcircle. The rod has a total charge of -7.5uC. Find the magnitude and direction of the electric field at 0, the center of the semicircle.

I wrote down on the scan on the things I am talking. There just 3 parts.
 

Attachments

  • p7.PDF
    46.4 KB · Views: 259
  • p6.PDF
    34 KB · Views: 268

1. What is an electric field?

An electric field is a physical field that is created by electrically charged objects. It is a vector field, meaning it has both magnitude and direction, and it exerts a force on any other charged objects within its influence.

2. How is an electric field calculated?

The strength of an electric field at any given point is calculated by dividing the force exerted on a test charge placed at that point by the magnitude of the test charge itself. This value is also affected by the distance between the charged objects and the type of material between them.

3. What is the difference between an electric field and an electric potential?

An electric field is a vector field that describes the force a charge would experience at a given point, while electric potential is a scalar quantity that describes the potential energy of a charge at a given point. In other words, the electric field determines the direction of the force, while the electric potential determines the magnitude of the force.

4. How does the presence of a charged object affect the electric field around it?

A charged object will create an electric field around it that extends in all directions. The strength and direction of this field will depend on the charge of the object and the distance from it. Other charged objects placed within this electric field will experience a force due to the interaction between their own charges and the electric field.

5. What are some common applications of electric fields?

Electric fields have a wide range of practical applications, including powering electronic devices, charging batteries, and inducing electric currents for various purposes. They are also used in medical technologies such as MRI machines and electrocardiograms. Additionally, electric fields are used in industrial processes such as electroplating and welding.

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