What would E field addition be if Gauss's law was different

In summary: I'm trying to understand your question but I don't quite follow? The right side seems to have units of charge squared divided by farads per meter, but the lefts units are completely different. What is this supposed to mean?math doesn't care about units
  • #1
Curl
758
0
What if Gauss' Law was
div(E)=q2/eps0

Then if we had two point charges, how could we calculate the resultant E field at an arbitrarily point? Obviously superposition would not work anymore, so how could it be done mathematically?

This is essentially a math question
 
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  • #2
Did you stop to check if it was first dimensionally correct before proceeding? The units for the real gauss's law pertain to a physically interpretable / measurable quantity.
 
  • #3
math doesn't care about units
 
  • #4
I'm trying to understand your question but I don't quite follow? The right side seems to have units of charge squared divided by farads per meter, but the lefts units are completely different. What is this supposed to mean?
 
  • #5
Curl said:
math doesn't care about units

Physics does.

You have received your answer: what you wrote does not have the correct units, so there is no answer to your question. It would be like asking "how long would it take a ball to fall 7 gallons"
 
  • #6
I guess another way to ask this is to say if a particle of special type of matter had the property that it created a vector field given by
F=(c*m^2/r°r)*r/|r| where m is the mass of the particle,

then if I have two particles, each of mass m, one at the origin and one at x=1, then what would be the resultant field F at an arbitrary point r0 ?
 
  • #7
I think you already answered your question regarding the force, since you explicitly stated it.

However, if I'm understanding you correctly, I think you might be getting at the concept of linearity and the superposition of fields. Is your question what the resulting field would be if it varied as the square of the sum of all the masses/charges in some way? If so, you likely could not use Gauss's law at all, since it's a sum which assumes that the total field can be produced by summing each up piece of field, which in turn assumes a sum over each charge/mass. If the field produced by each infinitesimal source cannot be added linearly to produce the net overall field (that is, it is not independent of the other charges/masses present), then this sum does not give you the total flux, and therefore cannot give you the total mass/charge. In other words, by summing the field, you are no longer summing the total charge/mass, since they are not linearly related.

To answer your original question, if the divergence of the electric field were proportional to the square of charge, then nothing would really change other than we'd probably treat the fundamental charge as its squared counterpart (and some units would need to be added or modified). It would still be linear in superposition, though.

(Note: This is intuitive reasoning. Correctness not guaranteed.)
 
Last edited:
  • #8
cwilkins said:
I think you already answered your question regarding the force, since you explicitly stated it.

I don't quite follow this part, how did I explicitly state it? I stated that the masses are at different locations.
 
  • #9
Curl said:
I don't quite follow this part, how did I explicitly state it? I stated that the masses are at different locations.

You gave the expression for the force in your post. The values you gave could have been plugged into it.
 
< H2>What is Gauss's law and how does it relate to E field addition?

Gauss's law is one of the four Maxwell's equations that describes the relationship between electric fields and electric charges. It states that the electric flux through a closed surface is directly proportional to the enclosed electric charge. This law is important because it allows us to calculate the strength and direction of the electric field at a point due to a given charge distribution.

< H2>How is E field addition affected if Gauss's law is changed?

If Gauss's law is changed, it can affect the way we calculate the electric field at a point. Depending on the nature of the change, the strength and direction of the electric field may be altered, leading to different results in our calculations of E field addition.

< H2>Can Gauss's law be violated and still produce accurate results for E field addition?

No, Gauss's law is a fundamental law of electromagnetics and cannot be violated. Any change in the law would result in inaccurate calculations for E field addition. However, there are situations where we may use approximations or simplifications to make calculations easier, but these still follow the principles of Gauss's law.

< H2>What are some examples of situations where Gauss's law may be altered?

One example is when dealing with non-uniform or non-symmetric charge distributions. In these cases, we may have to use different mathematical techniques to calculate the electric field at a point, and the results may differ from those obtained using Gauss's law. Another example is when dealing with electric fields in the presence of magnetic fields, where Gauss's law may need to be modified to account for this interaction.

< H2>How does Gauss's law contribute to our understanding of E field addition in real-world applications?

Gauss's law is essential for understanding and predicting the behavior of electric fields in real-world applications. It allows us to calculate the electric field at a point due to a given charge distribution, which is crucial in designing and analyzing various devices such as capacitors, electric motors, and generators. It also helps us understand the behavior of electric fields in different materials and environments, which is important for practical applications in electronics, telecommunications, and power systems.

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