Electric field of a point charge

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SUMMARY

The discussion centers on the concept of a point charge, defined mathematically as a charge density represented by a Dirac delta function, ##\rho(\vec{r}) = q\delta^{3}(\vec{r} - \vec{r}_0)##. It is established that a point charge is not merely an electron but a theoretical construct used in physics. The relationship between point charges and electric fields is demonstrated through Gauss' law, leading to the derivation of Coulomb's law, expressed as ##E_r = \frac{q}{4\pi \varepsilon_0 R^2}##. The point charge is characterized as infinitely small, serving primarily as a conceptual tool in electrostatics.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with Maxwell's equations
  • Knowledge of Gauss' law and its applications
  • Basic calculus, particularly volume integrals and the divergence theorem
NEXT STEPS
  • Study the derivation of Coulomb's law from Gauss' law
  • Explore the implications of the Dirac delta function in physics
  • Learn about electric field calculations for different charge distributions
  • Investigate advanced topics in electrostatics, such as potential theory
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Students and professionals in physics, particularly those focusing on electrostatics, electrical engineering, and theoretical physics, will benefit from this discussion.

willDavidson
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I am trying to understand what a point charge is.
I am trying to understand what a point charge is. Is it just an electron? Or is it just an idea?
 
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A point charge ##q## at position ##\vec{r}_0## is described by a charge density consisting of a Dirac delta function, ##\rho(\vec{r}) = q\delta^{3}(\vec{r} - \vec{r}_0)##.

Edit: N.B. you can also check that this recovers the Coulomb law, i.e. from Maxwell I,$$\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}$$Perform a volume integral over a region ##\Omega## on both sides, and then use the divergence theorem on the LHS,$$\int_{\partial \Omega} \vec{E} \cdot d\vec{S} = \frac{1}{\varepsilon_0} \int_{\Omega} q\delta^3(\vec{r} - \vec{r}_0) dV = \frac{q}{\varepsilon_0}$$That's Gauss' law in integral form, which you can convert to Coulomb's law by choosing ##\partial \Omega## to be a spherical surface concentric with the point charge, of radius ##R##, $$4\pi R^2 E_r = \frac{q}{\varepsilon_0} \implies E_r = \frac{q}{4\pi \varepsilon_0 R^2}$$
 
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It's just a theoretical concept. It's infinitely small. There's not much more than that to it. And yes, if you want to describe it as a charge distribution you'd use a three dimensional dirac delta function for it.
 

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