Class 7: More Gauss’s Law and Electric Potential
Strategy
Step 1. Construct a Gaussian surface passing through the point you want to calculate the E field. Most likely the Gaussian surface is parallel to source charge distribution.
Step 2. For this method to work, the magnitude of the E field has to be constant on the Gaussian surface. Most likely the flux through the Gaussian surface is just E=EA, with E as the unknown. (If this is not the case, Gauss’s Law is still correct, but this method just does not work. )
Step 3. Calculate the charge enclosed by the Gaussian surface, qin.
Step 4. Now apply Gauss’s Law:
 0  E  q in   0 EA  q in
and E can be solved.
Uniform spherical distribution
For r>R
E
r
 0  E  Q   0  E  4 r 2  Q
 E
R
Q
Q
4 0 r 2
Note that point charge belongs to this case.
For r<R
Depends on the actual charge distribution.
Uniform cylindrical (infinite long) distribution
R
Linear charge density = 
For r>R
 0  E  q in   0  E  2 r    

Note that a line point  E
2 0 r charge belongs to this case.
For r<R
Depends on the actual charge distribution.
Uniform distribution in an infinite plane
Surface charge density = 
 0  E  q in   0  2  E  A   A

 E
Note field is constant
20
More on Conductors (as a Source of Electric Field)
The following are true for any shape of a conductor, including the ones with cavities inside it (but assume there is no charge inside the cavities).
1.If the conductor has a net charge, all charges will stay only on the surfaces of the conductor.
2.There is no electric field inside the conductor.
3.The electric field outside the outer surface always perpendicular to the surface in the proximity of the conductor.
All on Conductors (as a Source of Electric Field)
(Con’t)
4. Electric field is stronger at the sharper part (smaller radius of curvature) of the outer surface.
Uniform distribution in an infinite conducting
plane
If this is a conducting plate
Surface charge density = 
 0  E  q in   0  E  A   A

 E
 0 Note field is constant
Dynamics in a Constant Electric Field
This is essentially a projectile problem because the acceleration is constant.
E




F  qE  ma  qE
 q 
 a E
m
Test charge
Charge q mass m
Source charges
y
q
E
m
x
ay  0
ax 


Initial condition : r (0)  (x 0 , y 0 ) and v(0)  v x0 î  v y0 ĵ
Projectile problem:
vy  vy0  a y t
1
y  y 0  v y0 t  a y t 2
2
2
2
v y  v y0  2a y (y - y 0 )
v x  v x0
x  x 0  v x0 t
Circular Motion in an Electric Field II
E
Test charge
Charge ‐q mass m
Note that the electric force has to be attractive for this to work
r


| F |  q | E |  ma c  qE
R
q
v2
m
 E
r
m
Source charges Q
For uniform spherical (source) charge distribution: 1
Q
E
4 o r 2
Circular Motion in an Electric Field II
Note that the electric force has to be attractive for this to work


| F |  q | E |  ma c  qE
r
v2
q
m
 E
r
m
E
For uniform cylindrical (source) charge distribution: Test charge
Charge ‐q mass m
Source charges
Linear charge density 
E
1

2 o r
Potential Energy U
If F(r) is conservative, the potential energy change U is defined as the negative work done by the force F(r), which is path independent. B1
`
f
 U-
  
F( r )  d r
i
Pay attention to the negative sign
A
1
`
Three common conservative forces
Spring
Earth surface
y
Uniform spherical object
M
a
m
r
g
For r >a only:
Hooke’s Law:
F = ‐kx
y = 0
F(r)  G
Mm
r2
F = ‐mg
U = ½ kx2
U =mgy
Mm
U(r)  - G
r
(U at natural length = 0 )
(U =0 at y = 0 )
U() = 0