Transcript Electric Fields

Gauss’s Law
Electric Flux
• Measure of the number of electric field lines
passing through a surface.
• For a uniform electric field
 
  E  A  E  A cos
• Non-uniform electric field
 
   E  dA
surface
Calculate Flux
• A square with an area of 5 m² has a surface
normal of 1 5 iˆ  2 5 ˆj. What is the flux
through this surface if the square is
embedded in a uniform electric field of
6.0 ˆjN / C ?
Closed Surface
• Consider the flux on a closed surface.
Tp
Lf
Bk
Rt
Fr
Bt
 total   Fr   Bk   Lf   Rt  Tp   Bt  0
Gauss’ Law
• Net electric flux through any closed
gaussian surface is equal to the net charge
inside divided by e0.
  qin
 E   E  dA 
e0
Point Charge
• Choose a sphere as your gaussian surface.
• Electric field is perpendicular to the surface
and of uniform strength at the surface.
 
2
   E  dA  E  4r
q
2
 E  4r 
e0
q
or E  k 2
r
q
Line Charge
• Choose a cylinder as your gaussian surface.
• Electric field is perpendicular to the walls
and parallel to the end caps.
   Fr   Bk  W all
q
 E  2rl 
e0
q

or E  2k  2k
lr
r
Sheet Charge
• Choose a box as your gaussian surface.
• Electric field is perpendicular to the top &
bottom, but parallel to the sides.
   Lf   Rt   Fr   Bk   Tp   Bt
 EA  EA 
q
e0
q

or E 

2 Ae 0 2e 0
Volume Charge
• Choose a sphere as your gaussian surface.
• Gaussian surface may be inside the volume.
  E  4r
3
4
qin  3 r 
2

E 
r
3e 0
q
Conductors
• Electric field is zero inside a conductor
• Any excess charge must be on the surface
• Electric field leaves perpendicular to the
surface.
• Irregular shapes: more charge accumulates
where the radius of curvature is smallest.
• Near the surface E  
e0
Irregular
Conductors
• How would you draw the electric field on
the following conductors?