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Electric Potential (III)
- Fields
- Potential
- Conductors
Potential and Continuous Charge Distributions
We can use two completely different methods:
dq
V   ke
r
source
1.
dq
dV  ke
,
r
2.

Or, Find E from Gauss’s Law, then…


dV  E  ds ,
B 

VB VA    E  ds
A
Ex 1: Given V=3x2+12x-1, find where E=0.
Ex 2: The Electric Potential of a Dipole
y
a
-q
a
+q
P
Find: a) Potential V at point P.
b) What if x>>a ?
c) Find E.
x
Solution
Ex 3: Find the potential of a finite line charge at P,
AND the y-component of the electric field at P.
P
r
d
dq
x
L
Solution
Ex 3: Find the potential of a uniformly charged sphere of
radius R, inside and out.
R
Uniformly Charged Sphere,radius R
E
R
r
R
r
V
Recall that the electric field inside a solid conducting sphere
with charge Q on its surface is zero. Outside the sphere
the field is the same as the field of a point charge Q
(at the center of the sphere). The point charge is the
same as the total charge on the sphere.
Find the potential
inside and outside the
sphere.
+Q
R
Solution
-Inside (r<R), E=0, integral of zero = constant, so
V=const
-Outside (r>R), E is that of a point charge, integral gives
V=kQ/r
Solid Conducting Sphere,radius R
E
R
r
R
r
V
Quiz
A charge +Q is placed on a
spherical conducting shell.
What is the potential (relative
to infinity) at the centre?
A)
B)
C)
D)
keQ/R1
keQ/R2
keQ/ (R1 - R2)
zero
+Q
R1
R2
Calculating V from Sources:
i) Point source:
Q
V 
or
4  r
Q
V k
r
(note: V0
as r  )
ii) Several point sources:
Qi
1
V 

4  i ri
(Scalar)
iii) Continuous distribution:
dq
V k
r

OR … I. Find E from Gauss’s Law (if possible)

II. Integrate, V    E  ds (a “line integral”)