Derive E field for a disk

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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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Answers and Replies

  • #2
Doc Al
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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
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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
Doc Al
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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
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Learn how to use Coulombs Law. That will give the exact result for the E field of a disk of charge.
 
  • #6
Doc Al
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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
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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
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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
Doc Al
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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.
 

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