Electric field magnitude between two charged disks problem

In summary: So at a/2, ##x=a/2##. The variable ##a## represents the distance between the two disks, and in this case, it is equal to the radius of the disks, ##R##. So in your general formula, you would substitute ##a=R## and ##x=a/2## to get the electric field at the point between the two disks.
  • #1
Brystephor
2
0

Homework Statement


Consider two thin disks, of negligible thickness, of radius R oriented perpendicular to the x axis such that the x axis runs through the center of each disk. (Figure 1) The disk centered at x=0 has positive charge density η, and the disk centered at x=a has negative charge density −η, where the charge density is charge per unit area.
What is the magnitude E of the electric field at the point on the x axis with x coordinate a/2?

dRoCanK.jpg


Homework Equations


We will need the equation for the electric field along the x-axis of a disk. I believe it is this:
[itex] \frac η {2∈_0} * \left( 1 - \frac x {\sqrt {(x^2+R^2)} } \right) [/itex]

I'm not sure that we will need anything else since we will be solving symbolically.

The Attempt at a Solution


We will need to find the electric fields emitted from both disks, and then add them, correct? So it seems like we should be able to just double the equation up top since the electric fields will be equivalent but in different directions, resulting in:
[itex] \frac η {∈_0} * \left( 1 - \frac x {\sqrt {(x^2+R^2)} } \right) [/itex]

However, this is incorrect. I am not sure what else to do. MasteringPhysics hint gave me a 'general form' equation of the electric field between the disks that is:

[itex] \frac η {2∈_0} * \left( 2 - \frac 1 { \sqrt {1+R^2/(x-a)^2} } - \frac 1 {\sqrt {1+R^2/x^2} } \right) [/itex]

I can see that the [itex] R^2 / x^2 = \arctanθ [/itex] but I do not understand where the [itex] \left( x - a^2 \right) [/itex] comes from or how to continue from here. Thank you.
 

Attachments

  • dRoCanK.jpg
    dRoCanK.jpg
    4.1 KB · Views: 2,914
Physics news on Phys.org
  • #2
Hello brystephor, :welcome:

did you try to substitute your a/2 in the general formula ? surprise !
 
  • #3
BvU said:
Hello brystephor, :welcome:

did you try to substitute your a/2 in the general formula ? surprise !

I did not do this. I am not sure where to substitute it in for or why I would do such a thing. Clearly a/2 is the point between two disks, but I do not know which of the variables in my general equation I would replace with a/2 or the formula that Mastering Physics gave me.
 
  • #4
Brystephor said:
I did not do this. I am not sure where to substitute it in for or why I would do such a thing. Clearly a/2 is the point between two disks, but I do not know which of the variables in my general equation I would replace with a/2 or the formula that Mastering Physics gave me.
What does ##x## represent in the formulas you have written?
 

1. What is the formula for calculating the electric field magnitude between two charged disks?

The formula for calculating the electric field magnitude between two charged disks is E = (σ/2ε0)(1-1/√2) where σ is the surface charge density of the disks and ε0 is the permittivity of free space.

2. How do the radius and charge of the disks affect the electric field magnitude?

The electric field magnitude between two charged disks is directly proportional to the surface charge density of the disks and inversely proportional to the distance between them. Therefore, increasing the radius or charge of the disks will result in a stronger electric field magnitude between them.

3. What is the unit of measurement for electric field magnitude?

The unit of measurement for electric field magnitude is volts per meter (V/m).

4. How does the electric field magnitude change as the distance between the disks increases?

The electric field magnitude decreases as the distance between the disks increases. This is because the electric field strength follows an inverse square law, meaning that it decreases by the square of the distance.

5. Can the electric field magnitude between two charged disks be negative?

Yes, the electric field magnitude between two charged disks can be negative. This indicates that the direction of the electric field is in the opposite direction of the positive charge. It is important to consider both the magnitude and direction of the electric field when solving problems involving charged disks.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
64
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
250
  • Introductory Physics Homework Help
Replies
5
Views
614
  • Introductory Physics Homework Help
Replies
5
Views
706
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
26
Views
424
  • Introductory Physics Homework Help
Replies
32
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
472
Back
Top