prelecture_03_electric_flux_and_field_lines

Electric Flux and Field Lines

Overview

• Gauss' Law
  • Inside a spherical surface
    • [ Flux = \frac{ Q}{ \varepsilon_0} ]
  • Outside a spherical surface
    • [ Flux = 0]

Electric Field Lines

• Electric Field
  • [ E = \kappa \frac{ q}{ r^2}]
    • [ E = \frac{ 1}{ 4 \pi \varepsilon_0} \frac{ q}{ r^2}]
• Density of Field Lines
  • [ D = \frac{ N}{4 \pi r^2}]
    • [ N \equiv \frac{ q}{ \varepsilon_0}]
    • [ E = \frac{ N}{ 4 \pi r^2}]
• Surface Area of a Sphere
  • [ A = 4 \pi r^2]

Electric Flux

• Electric Flux
  • [ \phi = \vec E \cdot \vec A]
  • [ \phi \equiv \int\limits_{\text{surface}}{ \vec E \cdot d \vec A}]
  • The quantity the counts the number of field lines that pass through a surface.

Question 1

Consider the same point charge at the center of two different spherical surfaces, one having a radius three times as big as the other as shown below.

Compare the electric flux through each of the spheres.

• The flux through the two spheres is the same
  • We just learned that for a simple spherical configuration the flux is just the product of the electric field [E] with the surface area A of the sphere. Since [E] decreases as [ \frac{ 1}{ R^2}] and the surface area increases as [R^2], their product remains constant. This answer can also be understood in terms of our graphical representation in which the flux is a measure of the number of lines leaving the surface. Clearly, the same number of lines leave the surface of any sphere centered on the charge.

Flux Example

• Flux through Hemispers
  • [ \phi_{2R} = \phi_R = \frac{ 1}{ 2} \frac{ q}{ \varepsilon_0}]
• Flux through Plane
  • [ \phi_{Plane} = 0]
• Surface Area of a Hemisphere
  • [ A_{hemisphere} = 2 \pi r^2]

Gauss' Law

• Gauss' Law
  • [\oint\limits_{surface}{ \vec E \cdot d \vec A} = \frac{ q_{enclosed}}{ \varepsilon_0}]
    • The toatal flux is the total charge enclosed over varepsilon not.
  • Inside a spherical surface
    • [ Flux = \frac{ Q}{ \varepsilon_0} ]
    • True for all surfaces
  • Outside a spherical surface
    • [ Flux = 0]
    • True for all surfaces

Question 2

Three point charges are fixed at the vertices of an equilateral triangle of side [d] as shown. Two of the charges are positive ([+1 \mu C] and [+2 \mu C]) and one is negative ([-1 \mu C]). Consider a sphere of radius [R = \frac{ d}{ 2}] centered on the equilateral triangle as shown. What is the sign of the total electric flux through this sphere produced by the fields of the three charges?

• The flux is zero.
  • The flux through the sphere is zero since the enclosed charge is zero; all charges are outside of the sphere. All field lines that leave one of the positive charges and enter the sphere at one point must leave the sphere at another point. The contribution to the flux is negative at the entry point and positive at the exit point. The situation is just the reverse for the negative charge. The flux through the sphere is totally determined by the charges inside the sphere. Here there are none; consequently, the flux through the sphere is zero.

Infinite Line of Charge Example

• Infinite Line of Charge
  • [ E = 2\kappa \frac{\lambda }{ r} ]
  • [ E = \frac{ 1}{ 2 \pi \varepsilon_0} \frac{ \lambda}{ r}]

Electric Dipole Field Lines

Question 3

Consider the electric field lines shown below, produced by two point charges, one positive and one negative. Let [E_A], [E_B], and [E_C] denote the magnitudes of the electric fields at points [A], [B], and [C], respectively. Which of the following statements correctly ranks the strengths of the fields at these points?

• [E_C > E_B > E_A ]
  • In the fieldline picture, the magnitude of the field is represented by the density (i.e. closeness) of the field lines. In the example shown, the field lines are most dense at point [C] and least dense at point [A]. Consequently, [E_C > E_B > E_A ].

Gauss' Law and Dipoles

Main Points

• Electric Field
  • [ E = \kappa \frac{ q}{ r^2}]
    • [ E = \frac{ 1}{ 4 \pi \varepsilon_0} \frac{ q}{ r^2}]
• Density of Field Lines
  • [ D = \frac{ N}{4 \pi r^2}]
    • [ N \equiv \frac{ q}{ \varepsilon_0}]
    • [ E = \frac{ N}{ 4 \pi r^2}]
• Surface Area of a Sphere
  • [ A = 4 \pi r^2] *Gauss' Law
  • [\oint\limits_{surface}{ \vec E \cdot d \vec A} = \frac{ q_{enclosed}}{ \varepsilon_0}]
    • The toatal flux is the total charge enclosed over epsilon not.