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Today’s agenda:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ law, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
Equipotentials
Equipotentials are contour maps of the electric potential.
http://www.omnimap.com/catalog/digital/topo.htm
Equipotential lines are another visualization tool. They
illustrate where the potential is constant. Equipotential lines
are actually projections on a 2-dimensional page of a 3dimensional equipotential surface. (“Just like” the contour
map.)
The electric field must be perpendicular to equipotential lines.
Why?
Otherwise work would be required to move a charge along an
equipotential surface, and it would not be equipotential.
In the static case (charges not moving) the surface of a
conductor is an equipotential surface. Why?
Otherwise charge would flow and it wouldn’t be a static case.
Here are electric field and equipotential lines for a dipole.
I’ll discuss in lecture some
implications this figure has
for charged particle motion.
Equipotential lines are shown in red.
“Toy”
http://www.falstad.com/vector2de/
Today’s agenda:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ law, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
Potential Gradient
(Determining Electric Field from Potential)
The electric field vector points from higher to lower potentials.
More specifically, E points along shortest distance from a higher
equipotential surface to a lower equipotential surface.
You can use E to calculate V:
b
Vb  Va   E  d .
a
E
You can use the differential version of this equation to calculate
E from a known V:
dV  E  d   E d
dV
 E 
d
For spherically symmetric charge distribution:
dV
Er  
dr
In one dimension:
dV
Ex  
dx
In three dimensions:
V
V
V
Ex  
, Ey  
, Ez  
.
x
y
z
V ˆ V ˆ V ˆ
or E  
i 
j 
k   V
x
y
z
dV
E 
d
dV
Er  
dr
V
V
V
Ex  
, Ey  
, Ez  
.
x
y
z
Calculate -dV/d(whatever) including all signs. If the result is
+, E points along the +(whatever) direction. If the result is -,
EE points along the –(whatever) direction.
Example (from a Fall 2006 exam problem): In a region of
space, the electric potential is V(x,y,z) = Axy2 + Bx2 + Cx,
where A = 50 V/m3, B = 100 V/m2, and C = -400 V/m are
constants. Find the electric field at the origin
V
E x (0, 0, 0)  
   Ay 2  2Bx  C 
 C
(0,0,0)
x (0,0,0)
V
E y (0, 0, 0)  
 (2Axy) (0,0,0)  0
y (0,0,0)
V
E z (0, 0, 0)  
0
z (0,0,0)
V

E(0,0,0)   400  ˆi
m
