Physics 1602 3 Electrostatic Fields

• However, there is an issue of causality in Physics 1602 2 Electric charge and electrostatic interactions.

  • It contradicts special relativity that all influences propagate instantaneously.
  • We need to confirm the emergence of q from E (blu3mo).

• So, let’s consider the Electric Vector Field, E(r).

  • $\vec{F}=q\vec{E}(r)$
    • Since this only determines E at a single location, there is no issue of causality (?).
  • Fields can be added.
  • It was mentioned that the field can be separated into xyz components.

• For now, let’s consider the situation of Electrostatics (constant charge function, no charge movement).

  • We want to define the field.
    • 
    • By rearranging this, we can determine E(r).
      • That’s obvious (blu3mo).

• Understanding the concept of the E field:

  • Two charges, or from a distance, they can be seen as a single consolidated charge.
    • We should understand the method of deriving it using binomial approximation.
  • The imagined situation:
    • There is a charge somewhere on the z-axis, like (0, 0, ±d).

    • 1. x = 0 - i.e. we evaluate the field on the z-axis.

      • We consider the field on the z-axis.

    • 2. z = 0 - i.e. we evaluate the field along the x-axis.

      • We consider the field along the x-axis.
      • This behaves quite non-trivially, with the field being maximum at x=±d/√2.
  • The quadrupole field is even more complex.

• Continuous Distribution:

  • The infinitesimal limit of the discrete case.
  • λ
    • Assuming a uniform charge density, charge / unit length = λ.
  • The line of charge.
    • By separating it into x and y components, we can perform complex integrals (using substitutions, etc.) to calculate it.
    • Important points:
      • If x >> L, it can be considered as a single charge.
      • If x << L, the change is not 1/x^2 but 1/x (?)(blu3mo).
      • This result, using cylindrical symmetry, can be applied to any axis (r-perpendicular).
  • The ring charge.
    • The r-perpendicular term cancels out in the opposite direction, so it can be ignored.
    • In the z-direction:
      • The circumference of the circle, 2πR, appears in the integral.
    • As usual, if z >> R, it changes with 1/z^2.

• Electric potential:

  • Conservative $F=-\nabla U$.
    • It is conservative when the same force is always applied at a certain r.
    • That’s obvious (blu3mo).
  • Conservative E field $E=-\nabla \phi$.
    • Since it is the field instead of the force, the unit differs by C.
    • Therefore, φ is not potential energy!
    • $\Delta \phi =\int \vec{E}d\vec{r}$
      • The unit of φ is length * force / charge.
        • This becomes volts, I see~ (blu3mo).
    • 
      • As we did in APMAE2000 Multivariable,
        • ∇Φ+c and ∇Φ are the same.
        • E points in the direction of the greatest change in Φ.
        • Even if the field is 0, the potential may not be 0.
          • That’s obvious, but don’t mistake it for potential being 0 when the forces cancel out, for example.
    • 
      • Simply put, potential is the value of charge multiplied by 1/distance.
        • This can be understood as the derivative being the force.
      • It only depends on distance = There is rotational symmetry.
        • So, calculating the potential of a ring charge, for example, is very simple.
  • I want to read about the ring charge in 4.3.1.
  • Symmetry:
    • Inversion symmetry: When p(x)=p(-x), for example.
      • $\phi (x)=\phi (-x)$;
        • From this, we can see that at Φ(0), $-E=\nabla \Phi =0$.
        • In other words, the force is 0.
      • $E_x(-x)=E_x(x)$
        • If the potential is reversed, it makes sense that the slope is also reversed.
        • From this, we can see that $E_x(0)=0$.
          • However, the y-component is not 0 (https://gyazo.com/dd873a6e13e5916a636f74efab6e391d).
          • This can be seen from the figure.
    • Reverse inversion symmetry: When p(x)=-p(-x), for example.
      • At x=0, the y and z components of E become 0.
        • I see, it makes sense that the field lines only point left or right in that area.
    • Translational symmetries: When p(x) = p(x+Δx), for example.
      • In this case, p is invariant with respect to x.
        • As a result, Φ and E are also invariant with respect to x.
          • E becomes the gradient of ${\phi }_x$ = 0.
      • This is also inversion symmetry.
    • Rotational symmetry:
      • For a ring, it becomes a function of the radial component and the z component.- If the radial component is zero, it means that the radial direction of the electric field is also zero, which can be understood from the concept of inversion symmetry.

• Ah, I see. This is due to the “infinite set of inversion symmetry.”

• As a result, all components of the electric field in every direction become zero, so $\vec{E}=\vec{0}$.

• When we combine this with infinite cylindrical symmetry, we get the following:

  • Rotational symmetry in the direction perpendicular to r.
  • Translational symmetry in the z direction.

• In the case of a disk charge, the derivative of Φ becomes proportional to |x|.

• Since the term includes σ = q/πR^2, there are no issues with dimensions.

• At this point, the gradient (E-Force) becomes constant.

  • Is this what we call a uniform electric field? [blu3mo] [blu3mo]

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