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Electric Potential (III)
Text sections 25.5, 25.6
Fields, potential, and conductors
Practice: Chapter 25, problems 29, 33, 35, 39, 57, 61
Read page 710, Van de Graaff generator
We can use two completely different methods:
1. 
2. 
Find
from Gauss’s Law, then…
1
O
+++++++++++++++++
L
b
Total charge Q, uniform linear charge density
(So
)
Find: V at point O
y
dq
b
r
x
x
dx
Charge/unit length:
2
dq
+
+
dq
+
+
+
+
+
+
+
+ +
Total charge Q, uniform
Find: V at centre C
R
+
+
+
C
(Homework exercise: review the
calculation for the electric field E,
which is harder.)
+
+ +
Total charge Q, uniform linear density
Find: V at centre C
R
+
C
(Homework exercise: review the
calculation for the electric field E)
+
+
At the center of the semicircle, the potential is:
A)  less than kQ/R
B)  equal to kQ/R
C)  greater than kQ/R
3
dq
+
+
+
+
+
+ +
Total charge Q, uniform
r
R
+
+
+
C
Find: V at centre C
Solution:
dV (at C) due to dq:
r=R is a constant (same for each dq)
Example: Spherical Charges
1) Find the electric field as a function of r using
Gauss’s Law.
2) Imagine pushing a “test charge” in from infinity
along a radial line: the potential change with each small
change dr in distance is
dV = - E(r) dr.
3) Integrate from R to infinity to find V(R) (relative to
infinity) at any position R.
4
Solid Conducting Sphere,radius R
E
R
r
R
r
V
Solid Conducting Sphere,radius R
E
R
r
R
r
V
5
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
Fields > 3 x 106 V/m will cause a spark in dry air.
What is the maximum potential to which an isolated
metal sphere of radius 1 cm can be charged, without
causing a spark?
A) 300V
B) 3kV
C) 30kV
D) 300 kV
E) 3MV
6
Fields
3 x 106 V/m will cause a spark in dry air.
Find the maximum potential on a metal sphere of
radius… a) 1 mm
b) 1 m
1) 
inside
conductor is an equipotential.
3) 
surface (just outside.)
4)  Excess charge is on the surface; and
_______________________________________
4) Empty cavity inside a conductor,
as well.
5)
7
+
Large
-
-
Weak
-
-
+
+
+
+
-
+
Metal foil
wrapper
$ 500
Computer chip
8
A conducting sphere of radius R1,
carrying charge Q, is surrounded
by a thick conducting shell with
no net charge. What is the
potential of the inner sphere,
relative to infinity?
A) 
B) 
C) 
D) 
+Q
V = zero
0 < V < keQ/R1
V = keQ/R1
V > keQ/R1
The dashed green line
represents a spherical
gaussian surface inside the
conducting material. The
total electric flux through
this surface (in units of Q/
ε0) is
-3Q
R3
R1
+Q
R2
A) 0
B) - Q/ε0
C) +Q/ε0
D) -2Q/ε0
E) 3Q/ε0
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So, the charge on the inner
surface of the outer shell is
A)  0
B)  - Q
C)  +Q
D)  -2Q
E)  +3Q
R3
R1
+Q
R2
-3Q
?
+
+
+
r
+
a
Radius a, total charge Q
r
Uniform volume charge
density
Find: V(r) for r<a
+
+
+
+
answer:
10
-Q
+Q
R1
R2
R3
Find: Potential difference
E
R1
V
V1
R2
R3
r
Find:
V2
R1
R2
R3
r
11
Solution (cont’d):
+
+
+
+
+
E=0
+
R
V = constant
= Vo
+
+
+
Charge Q
+
+
+
Outside (r > R):
(Just outside)
Vo (on sphere) = V (just outside) =
So…
Quiz:
A large conducting sphere has a net charge Q.
A second, smaller conducting sphere with no
net charge is now connected to it by a
conducting wire. When the system comes to
equilibrium, which of the following are true?
A)  the charges on the spheres will be equal
B)  the surface charge densities (charge per unit
area) on the surfaces are equal
C)  the potentials on the spheres are equal
D)  the electric fields just outside the spheres
are equal.
12
Solution (Cont’d):
Thus…
Therefore…
i)  Point source:
(choose V0
as r )
or
ii) Several point sources:
(Scalar)
iii) Continuous distribution:
OR … I. Find
from Gauss’s Law (if possible)
II. Integrate,
(a “line integral”)
13
+
Total charge Q, uniform
r
r
+
+
+
Find: V at centre C
C
Solution:
dV (at C) due to dq:
r is a constant (same for each dq)
r
+
a
+
+
+
+
+
+
+
+
+
+
+ +
+
dq
Radius a, total charge Q
r
Uniform volume charge
density
Find: V(r)
14
Solution (cont’d):
Inside (r > a):
Now, Potential:
i.e.
So for r > a :
Same as for
point charge!!
What about r < a? Can we write
Solution (cont’d):
For r < a:
Note: This is NOT equal to
15