Net Electric Field Defined at a Point with Charges Present

In summary, the electric field is the force/charge ratio on a small "test charge", which is defined by classical electrodynamics. The field at the point we have the charge is based on the fields produced by the surrounding charges. If we calculate the electric field of a charge distribution in matter, the field at the exact position where we have the charge (just based on classical electrodynamics) can only be a macroscopic average.
  • #1
pardesi
339
0
consider a region where charge is distributed then we have by gauss' law
[tex] \nabla \cdot \vec E =\frac{\rho}{\epsilon_{o}}[/tex]
what is [tex]\vec E[/tex] here.if it is the net electric field then how is that the field is defined at a point where charges are itself present
 
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  • #2
pardesi+edits said:
consider a region where charge is distributed then we have by gauss' law
[tex] \nabla \cdot \vec E =\frac{\rho}{\epsilon_{o}}[/tex]
what is [tex]\vec E[/tex] here.if it is the net electric field then how is that the field is defined at a point where charges are itself present


The electric field is defined as the force/charge ratio on a small "test charge", see for instance

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefie.html
 
  • #3
so how come u introduce a test charge at a place where ther i s already presence of charge
 
  • #4
Electric and Magnetic fields in matter are the macroscopic fields ... which means the average over regions large enough to contain many atoms ... The actual microscopic fields will fluctuate strongly inside matter ...

This is what I understand from Griffiths.
 
  • #5
pardesi said:
so how come u introduce a test charge at a place where ther i s already presence of charge

because by definition a test charge's charge is vanishingly small. you need this to argue that the field of the charge itself doesn't interfere with the ambient field at that point.
 
  • #6
ice109 said:
because by definition a test charge's charge is vanishingly small. you need this to argue that the field of the charge itself doesn't interfere with the ambient field at that point.

I think what the issue is trying to get at is that if we calculate the electric field of a charge distribution in matter then what happens to E at the exact position where we have the charge (just based on classical electrodynamics).
 
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  • #7
A charge does not exert a net electric force on itself. Otherwise a charge could set itself in motion by way of its own electric field. Therefore, when calculating the electric force on a particular charge, we include only the fields produced by the other surrounding charges.

ansrivas said:
if we calculate the electric field of a charge distribution in matter then what happens to E at the exact position where we have the charge

An infinitesmal piece of the charge distribution does not exert an electric force on itself. However, the remainder of the charge distribution does. A distribution of (say) positive charge flies apart by mutual repulsion unless there are other forces holding it together.
 
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  • #8
ansrivas said:
I think what the issue is trying to get at is that if we calculate the electric field of a charge distribution in matter then what happens to E at the exact position where we have the charge (just based on classical electrodynamics).
yes this is my question and not whether the charge exerts force on itself or not.
if we have a continious charge distribution then hoe is that w e find the net field at a point(what is the charge we have to leave out)
 
  • #9
jtbell said:
An infinitesmal piece of the charge distribution does not exert an electric force on itself. However, the remainder of the charge distribution does. A distribution of (say) positive charge flies apart by mutual repulsion unless there are other forces holding it together.


I tried to be clear that we are not talking about test charges here at all. The issue is we all know that charge is quantized. So what does one exactly mean by volume density of charge. This is made clear by the question here where we are discussing the field at the very point we have an electron sitting. So as I said without getting into Quantum effects what does classical electrodynamics have to say about this.

So a charge distribution having a volume density [tex]\rho[/tex] is really a macroscopic average density and the electric field that we calculate can only be a macroscopic average. The field calculated using a volume distribution cannot be expected to be uniform in the tiniest scale. This is an idealization to enable ease of calculation.

Griffiths talks about this exact problem in his book "Intro. to electrodynamics". Just look for "macroscopic" in the index if you have the book.
 
  • #10
yes exactly that's what my point is also another question .any book has a derivation of a field inside a solid sphere by using gauss's law on a concentric sphere .But what's the meaning of field inside the sphere .yes one could argue that if we remove some charge and then calculate the field the'r but will that give u the actual net field it does but how do we know that .
for ex had that distribution be surface charge clearly we would have got only half of our desired value
 

What is a net electric field?

A net electric field is the overall electric field experienced at a point due to the presence of multiple charges. It is the vector sum of all the individual electric fields at that point.

How is the net electric field calculated?

The net electric field at a point is calculated by adding up the individual electric fields at that point. This can be done using vector addition, taking into account the magnitude and direction of each electric field.

What factors affect the net electric field at a point?

The net electric field at a point is affected by the magnitude and direction of the individual electric fields, as well as the distance between the point and the charges. It is also affected by the type of charge (positive or negative) and the medium in which the charges are located.

How does the net electric field impact charged particles at a point?

The net electric field at a point determines the force experienced by a charged particle at that point. A positive charge will experience a force in the direction of the net electric field, while a negative charge will experience a force in the opposite direction.

Can the net electric field at a point be zero?

Yes, the net electric field at a point can be zero if the individual electric fields cancel each other out. This can happen if there are equal and opposite charges located at equal distances from the point, or if the individual electric fields are oriented in such a way that their vector sum is zero.

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