Electric Fields due to a charged disk.

In summary, the conversation discusses the question of which has a greater magnitude between Ea and Eb, as shown in a given picture. The website and solution manual both state that Ea > Eb, but the asker is confused as a bigger radius should give a bigger number for Eb, suggesting that Eb > Ea. The suggestion is made to consult with the teacher for clarification. Additionally, there is a discussion about where Ec would lie in magnitude, with the suggestion that it may be the difference between Ea and Eb.
  • #1
help1
4
0

Homework Statement



Why is this answer saying that Ea > Eb?


Homework Equations



Everything pertinent to this question is located in this picture:

http://img94.imageshack.us/img94/1187/ch22q9part1.png
[PLAIN]http://img94.imageshack.us/img94/1187/ch22q9part1.png

The question is asking which has the greater magnitude Ea or Eb, and this website says Ea > Eb, just like the solution manual, but that doesn't make sense!

The Attempt at a Solution



I plugged in numbers, and a bigger radius gives a bigger number for Eb, so shouldn't Eb > Ea?
 
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  • #2
You're right, those don't make sense. Talk to your teacher.
 
  • #3
chronos98 said:
You're right, those don't make sense. Talk to your teacher.

Any ideas as to where Ec would lie in magnitude?
 
  • #4
I'm pretty sure it would be the difference of the two, if c is a donut ring like I think it is. I'll let you figure out how that would compare.
 
  • #5
Seems to me as though the order should be Eb > Ea > Ec instead of Ea > Eb > Ec, as suggested by the text.
 

1. What is an electric field due to a charged disk?

An electric field due to a charged disk is a region in space where electrically charged particles experience a force due to the presence of a charged disk. It is a vector field that is described by the magnitude and direction of the force on a charged particle at any given point.

2. How is the electric field calculated for a charged disk?

The electric field for a charged disk is calculated using the formula E = kσ/2ε0, where k is the Coulomb's constant, σ is the surface charge density of the disk, and ε0 is the permittivity of free space.

3. Is the electric field due to a charged disk uniform?

No, the electric field due to a charged disk is not uniform. It varies with distance from the disk and is strongest near the edges of the disk. This is because the surface charge density is not constant across the disk.

4. How does the electric field change as the distance from the disk increases?

The electric field decreases as the distance from the disk increases. This is because the electric field is inversely proportional to the square of the distance from the source of the field, in this case the charged disk.

5. Can the direction of the electric field due to a charged disk change?

Yes, the direction of the electric field due to a charged disk can change depending on the location of the charged particle. The direction is always perpendicular to the surface of the disk at any given point, but the exact direction may vary depending on the position of the particle.

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