Derive E field for a disk

In summary: The electric field is a vector. By symmetry, you know that only the z-component of the field will remain; other components will cancel out.To find the field due to a ring of charge, use Coulomb's law. You'll need to find the area of the ring, and then use that to find Q.
  • #1
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Homework Statement


I came across an expression in the following pdf at the bottom of page two:
http://iweb.tntech.edu/murdock/books/v4chap2.pdf

Homework Equations



The electric field for a disk:
[itex]\vec{E} = \frac{\sigma}{2{\epsilon}_{0}} ( 1 - \frac{z}{\sqrt{{z}^{2} + {r}^{2}}})[/itex]

Now logically It will be via gauss's law:
[itex] {\phi}_{E} = \iint \vec{E} \cdot d\vec{S} = \frac{{Q}_{tot}}{{\epsilon}_{0}} [/itex]

The Attempt at a Solution


I'm not best sure how to approach this, it is in the z direction which suggests the Ex and Ey fields are zero so this disk would lie in the z plane - which is fine. But i can't quite put my finger on the trick here.

Thank you,
 
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  • #2
imagemania said:
The electric field for a disk:
[itex]\vec{E} = \frac{\sigma}{2{\epsilon}_{0}} ( 1 - \frac{z}{\sqrt{{z}^{2} + {r}^{2}}})[/itex]
Note that that's the field along the axis of the disk (z axis).

Now logically It will be via gauss's law:
[itex] {\phi}_{E} = \iint \vec{E} \cdot d\vec{S} = \frac{{Q}_{tot}}{{\epsilon}_{0}} [/itex]
There's not enough symmetry to make use of Gauss's law. (If it were an infinite sheet of charge, then you could.)

The Attempt at a Solution


I'm not best sure how to approach this, it is in the z direction which suggests the Ex and Ey fields are zero so this disk would lie in the z plane - which is fine. But i can't quite put my finger on the trick here.
No trick. You have to integrate.
 
  • #3
Hi Doc Al,
If i cannot use gauss's law what method should i use in order to get an appropriate form of an integral that integrates to the above equation?

Thanks
 
  • #4
One way is to divide the disk into rings of charge. Find the field due to each ring, then integrate to find the total field.
 
  • #5
Learn how to use Coulombs Law. That will give the exact result for the E field of a disk of charge.
 
  • #6
Doc Al said:
One way is to divide the disk into rings of charge. Find the field due to each ring, then integrate to find the total field.
In case it wasn't clear, when I speak of finding the field, I mean using Coulomb's law.
 
  • #7
Ok, I have looked into this before i saw these replies and found that the standard integral that i should "meet" during the process (though this is missing a factor in front of it):
[itex] \int^{r}_{0} \frac{ r dr}{{({z}^{2} + {r}^{2})}^{3/2}} = \frac{1}{|z|} - \frac{1}{\sqrt{{z}^{2} + {r}^{2}}} [/itex]

Given what you said below, i think i need to use:

[itex] \vec{E} = \int \frac{Q}{4\pi {\epsilon}_{0}{r}^{3}} dr[/itex]

though not so sure where the [itex]\frac{ r dr}{{({z}^{2} + {r}^{2})}^{3/2}}[/itex]
would be constructed from. Granted r does not have to be defined at the origin, but then it would be:
[itex] \vec{r} = \vec{r}_{2} - \vec{r}_{1} = z - r [/itex] ( final r being the comparison in integral i should get).
Im not too sure where the "pluss" rather than "minus" generates from.

Thanks
 
  • #8
After thinking about this more,
I know [itex] \sigma = \frac{Q}{A} [/itex]
Where the [itex] \pi [/itex] must be absorbed into this area as it does not appear in the final answer. i.e. [itex] A = \pi {r}^{2} [/itex]

Though i am still struggling on the plus and power in [itex] ({z}^{2} + {r}^{2})^{3/2} [/itex]

Any help is appreciated :)
 
  • #9
A few things to keep in mind:

The electric field is a vector. By symmetry, you know that only the z-component of the field will remain; other components will cancel out. Express that mathematically.

What's the area of your thin ring of charge? That should allow you to express Q in terms of σ.
 

1. What is the formula for calculating the electric field for a disk?

The electric field for a disk can be calculated using the formula E = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space.

2. How is the electric field distributed around the disk?

The electric field is symmetrically distributed around the disk, with the highest magnitude at the center and decreasing as you move towards the edges.

3. Can the electric field for a disk be negative?

Yes, the electric field for a disk can be either positive or negative, depending on the direction of the surface charge density. A positive E field points away from the disk, while a negative E field points towards the disk.

4. What factors affect the strength of the electric field for a disk?

The strength of the electric field for a disk is affected by the magnitude of the surface charge density, the distance from the disk, and the permittivity of the surrounding medium.

5. How does the electric field for a disk change as the distance from the disk increases?

The electric field for a disk decreases as the distance from the disk increases, following an inverse square relationship. This means that the electric field decreases rapidly at first, but then levels off as the distance becomes significantly larger than the radius of the disk.

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