Simulating the E-field distribution using the charge distribution

In summary, charge distribution is useful when simulating electric field intensity because it allows for more accurate results than electrostatics.
  • #1
TL;DR Summary
How do I simulate e-field using charge distribution?
Hello everyone,

I am new to this site so I hope this is the right place to ask this. I understand simulating electric field intensity using electrostatics because E=V/d makes sense to me. I do not understand how to consider e-field intensity using charge distribution. When is charge distribution useful compared to electrostatics? I apologize in advance if I'm not even using the correct terminology.

Thank you in advance,
Will
 
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  • #2
What you are asking is the heart of electrostatics. The controlling equation relates the divergence of the Electric field to a given charge distribution. The relationship runs both directions.
For some geometries ("infinite" flat sheets and spheres for instance) simple relations can be derived.
If charges are allowed to move and circulate, the entire power of Maxwell's equations must be used, as well as the notion of Magnetic fields...
 
  • #3
If charges are allowed to move and circulate like the charge in two wires with current flowing through both?

I've seen some papers stating that they used a voltage to define a boundary condition for simulation. I think others used charge? If I have two conductors separated by some distance, why would I apply a charge to them and not a voltage?
 
  • #4
willDavidson said:
If I have two conductors separated by some distance, why would I apply a charge to them and not a voltage?
The relationship between charge and voltage is capacitance. C = Q / V.
 
  • #5
willDavidson said:
Summary:: How do I simulate e-field using charge distribution?

Hello everyone,

I am new to this site so I hope this is the right place to ask this. I understand simulating electric field intensity using electrostatics because E=V/d makes sense to me. I do not understand how to consider e-field intensity using charge distribution. When is charge distribution useful compared to electrostatics? I apologize in advance if I'm not even using the correct terminology.

Thank you in advance,
Will
If you are given an arbitrary charge distribution ρ(x,y,z), you can solve for the potential φ by solving Poisson's equation [itex] \nabla^2 \varphi = -\rho/\epsilon_0 [/itex] . Once you have φ, you can solve for E given E = -∇φ. There are many ways to solve Poisson's equation, some analytic, and some numeric. But the point is, if you know ρ everywhere, you can solve for E everywhere. This assumes things are static. If the charges are moving, then it gets more complicated. But the bottom line is the same. If you know the charges and currents everywhere, you can solve for E and B everywhere.
 
  • #6
phyzguy said:
If you know the charges and currents everywhere, you can solve for E and B everywhere.
I'm still trying to understand why the charge wouldn't be uniform in a conductor. Is this like if a conductor has a higher current density near a cutout or something and there's a much higher current density? If so I guess that would mean the E-field intensity would be higher there as well? I'm trying to understand a practical application of this method.
 
  • #7
willDavidson said:
I'm still trying to understand why the charge wouldn't be uniform in a conductor.
Consider two nearby conductors, at different potentials.
The electron density on the surface will change due to electrostatic forces.
https://en.wikipedia.org/wiki/Electrostatic_induction
 
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