Topics:
•
Electric Potential of continuous charge
distributions
•
Connecting Potential and Field
•
EquiPotential Surfaces
PHYS 1022: Chap. 30, Pg 2
1
New Topic
PHYS 1022: Chap. 30, Pg 3
Potential difference is equal to the
line integration over field over any
path that connects the two points.
(In practice, pick the line that makes
the integral the easiest to do)
Another unit for electric field: volt / meter
1 V/m = 1 N/C
PHYS 1022: Chap. 30, Pg 4
2
We can obtain the electric field E from the potential V by
inverting our previous relation between E and V:

•  Expressed as a vector, E is the negative gradient of V


E = −∇V
z
Cartesian coordinates:
θ

∂V
∂V
∂V
∇V =
xˆ +
yˆ +
zˆ
∂x
∂y
∂z
€
€
r
o
y
Spherical coordinates:

∂V
1 ∂V ˆ
1 ∂V ˆ
∇V =
rˆ +
θ+
φ
∂r
r ∂θ
r sin θ ∂φ
x
ϕ
PHYS 1022: Chap. 30, Pg 5
€

The field is constant between the plates. So the line integral is easy

The electric field is E = σ/ε0 so it provides a practical way to
measure the surface charge density on the plate
PHYS 1022: Chap. 30, Pg 6
3

A ring of radius R carries a total charge Q distributed uniformly around
it. The electric potential at a point on its axis is
We obtained the same result as that
from vector sums of electric fields.
But it’s much easier here.
PHYS 1022: Chap. 30, Pg 7
Group Work
V (r) =
€
⎛ r + L /2 ⎞
1
ln⎜
⎟
4 πε 0 L ⎝ r − L /2 ⎠
L
+++++++++++++++++
r
PHYS 1022: Chap. 30, Pg 8
4
New Topic
PHYS 1022: Chap. 30, Pg 9
PHYS 1022: Chap. 30, Pg 10
5
PHYS 1022: Chap. 30, Pg 11
Rank in order, from largest to smallest,
the electric potentials at points 1 to 5.
1)
2)
3)
4)
V1=V2=V3=V4=V5
V1>V2=V3=V4>V5
V1=V2=V3>V4=V5
V1=V2>V3>V4=V5
PHYS 1022: Chap. 30, Pg 12
6
PHYS 1022: Chap. 30, Pg 13
PHYS 1022: Chap. 30, Pg 14
7
A metal ball carries a net charge +Q
and is enclosed by a neutral metal
shell. Which surface of the shell has
the higher potential?
1)
2)
3)
Inner surface
Outer surface
They have the same potential
The potential difference can be calculated from the field by:
Since the field is zero inside the conductor, the potential is
the same everywhere on the in conductor.
In general: Conductors in electrostatic equilibrium
have constant potential throughout.
PHYS 1022: Chap. 30, Pg 15
Two metal spheres are attached to a wire with
a switch in it. Sphere 1 is larger than sphere
2 and is given a positive charge; sphere 2 is
neutral. After the switch is closed, which
sphere has greater electric potential?
1)
2)
3)
Sphere 1
Sphere 2
They have the same potential
PHYS 1022: Chap. 30, Pg 16
8
Two metal spheres are attached to a wire with
a switch in it. Sphere 1 is larger than sphere
2 and is given a positive charge; sphere 2 is
neutral. After the switch is closed, which
sphere has greater electric charge?
1)
2)
3)
Sphere 1
Sphere 2
They have the same charge
Since they have the same potential given by kQ/r,
the one with a large r has larger Q.
PHYS 1022: Chap. 30, Pg 17
Two metal spheres are attached to a wire with
a switch in it. Sphere 1 is larger than sphere
2 and is given a positive charge; sphere 2 is
neutral. After the switch is closed, which
sphere has greater electric field on its
surface?
1)
2)
3)
Sphere 1
Sphere 2
They have the same charge
The electric field on the surface is E=V/r. Since
V is the same, the one with the smaller r has the
greater E.
One could also argue using E=kQ/r2,
PHYS 1022: Chap. 30, Pg 18
9
Rank in order, from largest to smallest,
the electric potentials at points 1 to 5.
1)
2)
3)
4)
V1=V2=V3=V4=V5
V1>V2=V3=V4>V5
V1=V2=V3>V4=V5
V1=V2>V3>V4=V5
PHYS 1022: Chap. 30, Pg 19
PHYS 1022: Chap. 30, Pg 20
10
PHYS 1022: Chap. 30, Pg 21
PHYS 1022: Chap. 30, Pg 22
11
Group Work
+++++++++++++++++
L
Q
r
PHYS 1022: Chap. 30, Pg 23
Group Work
PHYS 1022: Chap. 30, Pg 24
12
PHYS 1022: Chap. 30, Pg 25
PHYS 1022: Chap. 30, Pg 26
13
Vba = 1 V
+
+
+
+e
+
+ b
+
–
–
–
+e
–
a –
–
PHYS 1022: Chap. 30, Pg 27
0 V
-
500 V
+
+
+
+
+
+
+
PHYS 1022: Chap. 30, Pg 28
14
PHYS 1022: Chap. 30, Pg 29

The two 2.0-mm-diameter spheres
(almost touching) are released from
rest. What are their speeds when
they are far apart?
PHYS 1022: Chap. 30, Pg 30
15

If the inner shell has a net charge +q, and the outer shell has no net charge,
 how is the charge distributed?
What about electric potential?
 find the E field everywhere.
+q
r<a, E=0
+q
-q
a<r<b, E=0
b<r<c, E=kq/r2
c<r<d, E=0
r>d, E=kq/r2
PHYS 1022: Chap. 30, Pg 31
16