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Need a great analogy for Maxwells 1st eq

by rockyshephear
Tags: analogy, maxwells
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rockyshephear
#1
Jul23-09, 09:40 PM
P: 232
The amount of electric field coming from a region of space is equal to the total electric charge in that region of space, (divided by a number).

Imagine a point charge all by itself in space. It has a charge of 1C. According to the above statement, the electric field coming from the 1C point charge is equal to the charge density at that point charge/some number. If you have a 1C point charge, the density is not very dense. Just 1 point charge of 1 C. Is that dense or not dense? Numerically with units, what would the charge be at the 1C point charge?

I know I'm dense and that's why I'm asking.
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Born2bwire
#2
Jul23-09, 11:49 PM
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I would state it as the total flux of the electric field over a closed surface is directly proportional to the total electric charge in the enclosed volume. As such, the orientation of the electric field (and its density) with respect to your surface of interest will affect the measured flux. Only the total flux will be invariant as long as you enclose the same amount of charge. I do not think that you can make any precise claims on the field density. I could choose a surface around your charge that has a surface patch very close to the charge and the rest of the surface at a great distance away. The field density on the small patch near the charge will be very high while the field density elsewhere will be very low. Regardless, you will still derive the same relationship for any kind of surface enclosure. So the total flux is not always a good measure of field density, especially since you could place a charge externally to your enclosed surface and it will not affect your total flux despite the fact that the field density near the external charge will be very high.
rockyshephear
#3
Jul24-09, 06:25 AM
P: 232
So a good analogy for an electric field would be a blanket that covers two metal spheres. It is flat in between the two spheres but has curvature as it leaves and approaches the spheres. There is no 'radiation' just a static geometry.

Now if we consider the equation. del dotprod Electric field = rho/sigmasub0

This states that the rate of change of the Electric field wrt the vectors X hat, Y hat, Z hat equals charge density over the numerical value of sigmasub0.

sigmasub0 never changes so it's effect is the lower the value for the Electric field by division.

The rate of change of the Electric field wrt the vectors X hat, Y hat, Z hat is proportional to the charge density. Charge density goe up, so does the Electric field.

So when the charge density goes UP, what's happening in my blanket analogy?

My guesses

Maybe the diameter of a sphere increases.
Maybe we add another sphere right beside one of them.
Maybe we change the original sphere from copper to lead.

I would need an answer that specifically is in terms of my blanket analogy please.
Thx

rockyshephear
#4
Jul24-09, 06:28 AM
P: 232
Need a great analogy for Maxwells 1st eq

Just had a thought. Maybe the larger the charge density the larger the sphere and the more curvature the blanket has to perform to get to flat and that relates the the higher rate of change of the field blanket wrt to the coordinate system xyz. How does this sound?
Born2bwire
#5
Jul24-09, 07:43 AM
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Divergence is not a measure of change, you are thinking of the gradient operator that acts on a scalar function. Divergence is the measure of the amount of source/sink that exists in a vector field at a point in space. Another way of thinking about it is that it is the limit of the integral form of Gauss' Law as the volume enclosed goes to zero. Gauss' Law is not a direct measure of the strength or density of the field. It allows you to relate a known electric field to the charge distribution that creates it. Not so much the other way around, if you know the charge distribution, it is difficult to regain the electric field using Gauss' Law because the law relates the flux of the electric field, not the electric field itself.
rockyshephear
#6
Jul24-09, 07:49 AM
P: 232
The del operator is a measure of rate of change.
rockyshephear
#7
Jul24-09, 07:51 AM
P: 232
Del =d/dx, d/dy. d/dxz
If that's not a measure of change I don't know what is. A derivative is a measure of tendency of change. Slope of a tangent point on a curve.Right?
rockyshephear
#8
Jul24-09, 07:53 AM
P: 232
I did use the wrong term but I meant Del dotprod E as in the equation.
rockyshephear
#9
Jul24-09, 07:55 AM
P: 232
Couldn't you also say that divergence is a measure of change. The change of input vs output? Magnetic flux has zero divergence since the field doesn't discretely start any where or end anywhere and as such has no start/finish change.
Born2bwire
#10
Jul24-09, 07:57 AM
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But the divergence is not a simple derivative, it is an operator. If you had a vector field described as:

[tex]\mathbf{v} = (y^2-27z+3)\hat{x}[/tex]

it would have zero divergence despite having distinct change in the magnitude of the field with respect to the y and z dimensions.

EDIT:

The divergence of the magnetic field is zero because there are no magnetic monopoles assumed. Thus, there are no magnetic sources/sinks, which gives a divergence-less field. But the basic source of a magnetic field is a dipole and it most certainly changes over space.
rockyshephear
#11
Jul24-09, 09:31 AM
P: 232
So a good analogy for an electric field would be a blanket that covers two metal spheres. It is flat in between the two spheres but has curvature as it leaves and approaches the spheres. There is no 'radiation' just a static geometry.

Now if we consider the equation. del dotprod Electric field = rho/sigmasub0

This states that the rate of change of the Electric field wrt the vectors X hat, Y hat, Z hat equals charge density over the numerical value of sigmasub0.

sigmasub0 never changes so it's effect is the lower the value for the Electric field by division.

The rate of change of the Electric field wrt the vectors X hat, Y hat, Z hat is proportional to the charge density. Charge density goe up, so does the Electric field.

So when the charge density goes UP, what's happening in my blanket analogy?

My guesses

Maybe the diameter of a sphere increases.
Maybe we add another sphere right beside one of them.
Maybe we change the original sphere from copper to lead.

I would need an answer that specifically is in terms of my blanket analogy please.
Thx
rockyshephear
#12
Jul24-09, 09:38 AM
P: 232
In the above analogy, what would be representative of rho and sigmasub0. Why is permability of free space important when nothing is being propagated?
Born2bwire
#13
Jul24-09, 10:13 AM
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I have no idea what you are describing here.
rockyshephear
#14
Jul24-09, 10:30 AM
P: 232
I'm describing an electric field in terms of a blanket. One blanket would be a cross section of the 3D field. Layer upon layer would constitute the full field. But for purposes of simplicity, I'm considering one blanket covering up two charges. One is a large sphere the other a small sphere a distance apart.
I'm equating the terms in the eq to real world objects so I can understand the eq.
E is the blanket
Del dotprod E is the way the blanket deforms over the spheres
So what would the perm of free space equate to?
And the charge density

Since charge density increase, increases the electric field, and an increasing electric field means a more severe topology of the blanket, I equate charge density to the diameter of the spheres. Doe this sound about right?
Born2bwire
#15
Jul24-09, 10:58 AM
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The divergence of the electric field is not a measure of the deformity or change in the field over space. I can give you an example of a field distribution that has rapidly changing amplitude over space and yet have zero divergence. The fields resulting from a given charge distribution will be dependent upon both the sign and magnitude of the charges and how they are arranged in space. A lot of this information is lost in Gauss' Law.
rockyshephear
#16
Jul24-09, 11:04 AM
P: 232
But my msg listed as 3:30 pm was not talking about divergence. Can you give me an example of a point charge that does not have a field? Since they all have fields, the fields have some geometrical shape at every instance in time. This shape, I am saying, is based on the charge density which in my analogy would be the size of the sphere. This large size causes a disturbance called a field which is perfectly spherical until it interferes with another point charge. Sound right?
Vanadium 50
#17
Jul24-09, 01:09 PM
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You're writing nonsense, Rocky.

The last time you were confused about vector operators, I suggested the book by Schey, Div, Grad, Curl and All That. I recommend you read it.
rockyshephear
#18
Jul24-09, 03:04 PM
P: 232
I disagree that I'm talking nonsense. There's nothing wrong with understanding something, even if it is not via a method normally used. Some people are visual. You guys only use math. There's nothing wrong with visualization. You guys would be sorely at a loss had you to make a video describing physical phenomenon. 3D artists do it without math. So there's no nonsense in my question. Unless you guys simply cannot do it, just say so. Then I'll seek out a 3D artist who happens to be good at physics to give me the analogy I'm seeking.


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