Chapter 17
Electric Potential
Electric Potential Energy
Electric Potential
Equipotential surfaces
Conservation of Energy
Capacitors
Dielectrics
Electric Potential
Energy
 Electric Potential Energy
 Gravitational Potential Energy
 Sign
 Work
 Example
 Electric Potential
 Definition
 Point charges
 Electric field and electric potential
 Potential difference
Gravitational
Potential Energy
F =G
m1m2
r122
U = "G
m1m2
r12
But U = m1gh?
!
g=G
mE 1
m
mm
"
"#U = m1G 2E h "
"2 # $G 1 E
!
r122
r12
r12
m1
In step
! 2 the negative sign comes from where
we define zero Potential Energy … r-->∞
!
Recall: Only changes in U are important, not
The path taken.
r12
mE
1
Potential Energy sign
Electric
Gravitational
q q
F =k 122
r12
mm
mm
F = G 1 2 2 U = "G 1 2
r12
r12
U=k
q1q2
r12
•the negative sign on U
• If U is negative it results
indicates that it results from from an attractive force
an attractive force
(opposite !
charges)
!
!
!
we define zero Potential
Energy at r-->∞
•If U is positive, it results
from a repulsive force
q1
r12
q2
U = "G
Potential Energy
m1m2
r12
Electric
Gravitational
!
Increasing U
m
-
+
F
!
Increasing U
+Q
+Q
Earth
q1q2
r12
Increasing U
F
F
U
U=k
U
U
r
r
r
Think of increasing U as increasing stored energy
F always points in the direction of lower U
Work= -ΔU
What happens as the particles are moved closer?
Electric
Gravitational
a
Increasing U
m
Earth
-
+Q
U
ΔU=Ub-Ua
<0
Work is positive
•
•
r
Increasing U
Δr
Increasing U
Δr
Δr
b
+
U
•
ΔU=Ub-Ua
>0
Work is negative
•
+Q
U
r ΔU=U -U
b
a
•
•
r
<0
Work is positive
Positive Work -->Energy is taken from the
Stored U and given to the object (KE)
Negative Work --> Energy is stored in U
2
Electric Potential Energy
Multiple Charges
1
qq
qq
qq
U = k 1 2 + k 2 3 +k 3 1
r31
r12
r23
r12
2
r31
r23
3
1
r12
r
3
q1q2
qq
qq
+k 1 3 +k 1 4 +
r12
r13
r14
q2q3
q2q4
qq
k
+k
+k 3 4
r23
r24
r34
U=k
! 14 r23 ! 2
r31
!
r24
r34
4
!
Electric Potential, V
Electric Potential ≠Electric Potential Energy
F=k
q1 q2
F
= qE " E =
r122
q
U=k
q1q2
U
= qV " V =
r12
q
!
Unit: 1 Volt = 1 J/C
!
Electric Potential
Consider a moveable point charge q
and a collection of other charges
r12
1
r14
r31
3
2
r23
r24
q
r34
q1q2
qq
qq
+k 1 3 +k 1 4 +
r12
r13
r14
q2q3
q2q4
qq
k
+k
+k 3 4
r23
r24
r34
U=k
Since U = qV, and U = 0 when r = ",
we want V = 0 when
qq
U=k 1 4
the moveable charge is at "
r
!
14
+k
q2q4
qq
+k 3 4
r24
r34
Why is this ok?
! Remember we can always add or subtract a constant
From the potential energy since we only care about changes
!
3
Electric Potential due to
point charges
If our movable point charge,q, is a distance r from
another charge, Q, then
r
V=
kQ
r
q
Q
If our movable point charge,q, is a distance ri from lots
of other charges, Qi, then
kQ
V = " i for i = 1,2,3,...,N
ri
kQ1 kQ 2 kQ 3
kQ N
=
+
+
+ ....+
r1
r2
r3
rN
!
!
Example 17.4
There are three Charges
 Q1=+4.0µC at (0.0, 3.0) cm;
1
Q2=+2.0µC at (1.0, 0.0)cm;
Q3=-3.0µC at (2.0, 2.0)cm.
 (a) Find the electric potential
at point A (0.0, 1.0) due to
the three point charges.
A
 (b)A point charge q=-5.0nC
moves from a great distance
to point A. What is the
change in electric potential
energy?
V=
!
kQ
r
3
2
Electric Potential and
the Electric Field
Moveable positive charge Moveable negative charge
F
+
F
E
+Q
E
+Q
Force points in the direction of lower U
Electric field points in the direction of lower V
Note: for a negative charge lower
potential energy means higher potential
4
Electric Potential
Difference
"V = V f # Vi
"U = q"V = #Work
Electric Potential Difference is what we measure
when we measure a voltage.
!
!
Neurons
Impulse on
Dendrite opens
Excess Na+ Na+ protein channel
Protein channel
Outside cell
K+
Cell
membrane
Inside cell
ΔV=-70mV
Excess K+
Na+
ΔV=+50mV
ΔV=-50mV
Sodium channels
Continue to open
Along the membrane
Now the positive
Potential opens
potasium channel
A nerve impulse is a change in potential across the
cell membrane that moves along the axon
Pulse propagation
http://www.bris.ac.uk/synaptic/public/basics_ch1_2.html
5
Axon Pulse propagation
Myelin Sheath
Nodes of Ranvier
6