Electromagnetic Fields and WavesYear 1

Electromagnetic Fields and Waves: Electrostatic field and potential difference

Electromagnetic Fields and Waves: Electrostatic field and potential difference

We know how to find the potential difference between two points in the electrostatic field:

φ1φ2=12Eldl

Let’s find the reverse dependence. Charge q movement to the distance dl:

dA=q*El*dl

and

dA=q*dφ

Then

Eldl=dφ

and

El=dφdl

This means that field projection along the dl direction is equal to the potential rate from point 1 to 2 in the electrostatic field. “-” means that electric field E is directed in the same direction with potential φ decreasing. For Cartesian coordinate system

Ex=dφdx,Ey=dφdy,Ez=dφdz

And the vector of electric field

E¯=dφdxex¯dφdyey¯dφdzez¯

or

E ¯=gradφ

Task 5: Determine the electric field

E¯ (r¯) 

on the axis of uniform charged circle with radius R. Its potential is known

φ(r)

is a charge of the circle.

Conductors in the electric field

For simplicity, let’s consider the conductor as a metal. This conductor contains an enormous number of free charges, that are normally in the equilibrium state.

  1. The electric field in the conductor in the equilibrium is 0. Electric field in the conductor causes electric current, and free charges are ‘directly flowing’ only until the moment when the electric field will become 0.
  2. Field potential for all the points in the electric field is the same
    φ=constThis is a conclusion from the formula

    E¯=gradφ
  3. Excessive charge in the conductor is distributed along the conductor surface. Total charge of the conductor is 0,
    ρ=0 (except the thin surface layer). Figure 14. This statement can be proved from the Gauss theorem. At any point of the conductor

    E¯=0 so

    EldS~q=0 then q = 0.
  4. Electric field lines close to the surface of the conductor are perpendicular to the surface. Conductor surface is an equipotential surface.
  5. Electric field of a charged conductor close to its surface is proportional to the surface charge density. To show this statement, let’s take a cylindrical piece of a conductor with charge density σ Figure 14
    ΦE=EdS Total charge inside the surface

    Σ:E*dS=1ε0σdS Then

    E=σε0
Figure 14. Charged conductor with a small surface element of charge
Figure 14. Charged conductor with a small surface element of charge

6.  Surface density of a conductor charge depends on the surface curvative. Let’s replace the real conductive body with its model, where minimal curvative is characterised by radius R1, maximum – R2. Then, having the following formulas for the charged system:

φ1=kσ14πR12R1,φ2=kσ24πR22R2φ1=φ2

Then

σ1σ2=R2R1

and

σ~1R

And according to Gauss theorem

E~1R

#4 Conductors in the external electric field. Faraday theorem

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