Physics 1602 5 The Geometry of Electrostatic Fields

• I understand E and φ.
  • Based on that, let’s consider Gauss’s law.
  • The current density $\vec{j}\equiv p_m\vec{v}$
    • (The “current” here refers to “electric current,” not “now.“)
    • $\frac{mass}{lengt{h}^2\cdot time}$
      • If we integrate this over an area, we get flux (mass/time).
  • 
    • Taking the dot product with the normal and integrating over the area is expressed as $d\vec{A}$.
  • The flux of any surface that contains all charges inside it is $\frac{q}{{ϵ}_0}$, not just for a sphere.
    • Why does any shape have the same calculation as a sphere? (blu3mo)
      • When the angle of the surface changes, the normal vector and the area cancel each other out.
      • When R changes, the 1/R^2 of the charge and the R^2 of the area cancel each other out.
      • It’s written on p70.
    • Therefore, we can write the flux as $Q/{\epsilon }_0$.
      • Q is the sum of all charges.
      • If the charge is inside the surface, the position doesn’t matter.
    • Is this Gauss’s Law? (blu3mo)
    • I didn’t understand p(r, r’).
  • 
    • If we consider a continuous charge distribution, we can obtain flux with this.
      • Taking the volume integral of the charge distribution gives the sum of charges, which is obvious.
  • If the charge is not inside the surface, the flux is zero.
    • This is not obvious in the equation for E, but it is clear in the equation for Gauss’s Law.
    • For a donut-shaped surface without charge inside:
      • $E_1:E_2=R_2^2:R_1^2$ can be said.
        • This means that if the inflow and outflow of flux cancel each other out and become zero, does it mean that a larger radius or a larger surface area results in a smaller E?
      • If we generalize this, we get the following:
        • 
  • The problem with the equation in 6407682d79e11300005c2efd:
    • The point charge is ill-defined.
      • That’s true.
    • We can define the point charge as part of the charge distribution using the form $p(r,r_i)$.
      • If we do that, it works well when integrated.
      • It seems like an unnecessary step, I don’t really understand it. (blu3mo)
  • When there is symmetry, the flux can be simplified.
    • Spherical symmetry:
      • The equation for the flux of a sphere can be simplified.
      • Since E(r) is the same everywhere,
      • Furthermore, if there is a uniform density within the sphere’s range, E(r) increases linearly up to r=R and then decreases from there.
        • 
      • We can consider a surface density σ and express E using it.
        • Thanks to symmetry, E has the same value at any point on the surface.
        • So if we distribute the flux value over the entire surface, we can determine E.
    • Infinite cylindrical symmetry:
      • It becomes a function of E(r perp), with z being irrelevant.
        • The direction of E also becomes r perp hat.
      • In this case, we consider a cylindrical surface of length L.
        • There is no z contribution to E, so the flux of the left and right surfaces is zero.
        • At this time, we consider charge per unit length λ and express E using it.
      • Furthermore, $flux=2\pi rLE(r)=\frac{q_{<r}}{{\epsilon }_0}$
        • We can also have the third equation here.
        • Why don’t we need to consider the charge outside?
          • This is Gauss’s law.
    • Planar symmetry:
      • In this case, there are two inversion symmetries, so we can reduce it to a 1D problem.
      • The field remains the same regardless of the distance.
        • Intuitively,
          • When close to the plane, the influence of the nearest point charge is strong.
          • When far from the plane, the influence of the nearest point charge weakens, but the influence of more distant points increases.
        • Another explanation is,
          • Regardless of whether the surface area of the barrel is larger or smaller, if the total charge is the same, E is the same.
  • 
    • This can be considered in terms of surface/plane charge.
    • E*q/area = Force/area = pressure

      • An interesting aspect of this result is that it doesn’t depend on the variation of ρ within the charge layer.

• The concept of potential energy per volume:
  • The local value exists in each location.
    • That is εE^2/2.
  • Is it true that force/area = pressure and energy/volume have the same dimension?

Flux:

• As a premise, electrostatics = ∂E/∂t = 0.
• It shows how the vector changes in the perpendicular direction for each direction.
  • div: If there is a change in that direction, curl is generated.
    • xi, yj, etc.- curl: If there are coefficients like yi or xj that cause changes in directions other than E, curl is generated.
• div(curl) or curl(div) equals 0.
• Stokes’ theorem:
  • Convert curve to area using the right-hand rule.
  • Direction is important.
  • Why are curved faces and flat faces the same?
    • It can be derived from the fact that the div of curl is 0, which is surprisingly simple according to page 111.
• The electrostatic field is actually all - curl = 0.
  • Understanding the background here is important (todo: blu3mo).
  • Therefore, when considering a surface, there is a constraint that the non-normal component does not change before and after the surface.