Electricfield of a Disk using Gauss's law

[perplexabot]

Electricfield of a Disk using Gauss's law!

Hi all. Trying to find electric field of a disk with a charge density of D and a radius R (let us assume it has a thickness h). I know you can do so by breaking up the disk into concentric rings and integrating coulombs law. However, i would like to go about using gauss's law if that is possible. So:

Let k = permittivity of free space (usual symbol: epsilon)

(integral of) (E*dA) = Q_Enc/k

gonna use a cylinder as a gaussian surface (pillbox).

E * (integral of) (dA) = Q_Enc/k

Q_Enc = D*V​

(integral of) (dA) = (2*pi*R^2) + (2*pi*R*h) = 2*pi*R * (R + h)​

=> E * 2*pi*R*(R+h) = (D*V)/k
=> E = (D*V) / (k*2*pi*R*(R+h))

V = h*pi*R^2​

=> E = (D*h*pi*R^2) / (k*2*pi*R*(R+h))
=> E = (D*R*h) / (2*k*(R+h))

this is my final answer, i know it's probably wrong but i have no idea why. Also using coulombs law as mentioned above does not leave an h variable (thickness of the cylinder) in the final equation. Can someone please shed some light as to what my problem is?

P.S: My final goal is to find the electric potential of a point on the disk's axis a distance x from the middle of the disk. My book uses V = (integral of) dQ/r but I'm trying to use V_a - V_b = (integral of E*dL), So that is why I'm trying to find the E-field. Help provided for calculating electric Potential of disk will be greatly appreciated.

 

[Doc Al]

this is my final answer, i know it's probably wrong but i have no idea why.

You assume that the field is uniform across your Gaussian surface, but it's not. (It would be uniform--or zero--if the disk were infinite.)

Also using coulombs law as mentioned above does not leave an h variable (thickness of the cylinder) in the final equation.

Treating the disk as having a non-zero thickness just makes the integral harder to do. If you really want to do it, break the disk into infinitesimal layers.

 

[perplexabot]

You assume that the field is uniform across your Gaussian surface, but it's not. (It would be uniform--or zero--if the disk were infinite.) <----------[1st quote]


Treating the disk as having a non-zero thickness just makes the integral harder to do. If you really want to do it, break the disk into infinitesimal layers. <----------[2nd quote]

Reply to 1st quote:
wow that makes great sense. Thank you. But let's say i still want to go on and use Gauss's law with non-uniform e-field, what would i have to do? or does Gauss's law only apply to symmetrical objects.

Reply to 2nd quote:
I got this question from a problem in a book. And it says that the disk has a uniform DENSITY charge, which i assume means that the cylinder has 3-dimensions, which led me to use h. Why is it that we can disregard an objects thickness?

Once again thank you for the reply.

 

[Meir Achuz]

Gauss's law can't be used for a disc, because it has the wrong symmetry.

 

[Doc Al]

Reply to 1st quote:
wow that makes great sense. Thank you. But let's say i still want to go on and use Gauss's law with non-uniform e-field, what would i have to do? or does Gauss's law only apply to symmetrical objects.

Gauss's law applies in all cases, but is only useful in cases with sufficient symmetry. As clem said, a finite disk has the wrong kind of symmetry.

Reply to 2nd quote:
I got this question from a problem in a book. And it says that the disk has a uniform DENSITY charge, which i assume means that the cylinder has 3-dimensions, which led me to use h. Why is it that we can disregard an objects thickness?

They probably meant a uniform surface charge density, usually symbolized as σ. (Unless of course they specifically said it has thickness h.) If the disk is thin enough, treating it as just a charged surface is good enough.

 

[perplexabot]

The math for electric field of disk is shown here (scroll up to see math for ring and line):

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c3

Potentials here:

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/potlin.html#c3

It's also possible to do this for an infinite plane by segmenting the plane into thin infinitely long rectangles that can be treated as infinitely long lines for an integral. The result is the same as an infinitely wide disc.

thanks for the links. i got one question though.

For calculating Electric potential you can use either of the following equations (assuming all required variables are given)?:

V_a-V_b = (integral of) E*dL​

(equation 1)

V = (integral of) dQ/r​

(equation 2)

in the link you gave for calculating electric potential they used equation 2, i just used equation 1 with the E-field (e-field from link) and achieved the same answer. Is this always the case?

Gauss's law can't be used for a disc, because it has the wrong symmetry.

ok, so gauss's law only works for specific symmetries? And when gauss's law fails i should just resort to coulomb's law.

Thank you all

 

[perplexabot]

Gauss's law applies in all cases, but is only useful in cases with sufficient symmetry. As clem said, a finite disk has the wrong kind of symmetry.


They probably meant a uniform surface charge density, usually symbolized as σ. (Unless of course they specifically said it has thickness h.) If the disk is thin enough, treating it as just a charged surface is good enough.

ok so gauss's concept always applies but is hard to apply mathematically with bad symmetries.
And no the problem says nothing about thickness. Thank you for clarifying that.

 

[Gordianus]

Gauss's law is ALWAYS valid. However, finding the electric field using that law is only possible in three cases:
a) spherically symetric charge distribution. The charge density may vary with r but not with theta and phi.
b) Infinitely long cylindrical distribution. Again, the charge density may depend on r but not on z or phi.
c) Uniformly charged plane.
d) A clever combination of the above.

 

[perplexabot]

Gauss's law is ALWAYS valid. However, finding the electric field using that law is only possible in three cases:
a) spherically symetric charge distribution. The charge density may vary with r but not with theta and phi.
b) Infinitely long cylindrical distribution. Again, the charge density may depend on r but not on z or phi.
c) Uniformly charged plane.
d) A clever combination of the above.

thank you for clearing that out!