1. No category

Chapter 28
Electric Potential
Phys 133 – Chapter 30
1
Looking ahead
New force (Chapter 25)
Kq1q2
Felec
describe as
Felec = 2 r̂ ¾¾¾¾
® Eelec =
(only depends on source charge q1 )
r
q2
New potential energy (Chapter 28)
Kq1q2
U elec
describe as
Uelec =
¾¾¾¾
® V=
(only depends on source charge q1 )
r
q2
Phys 133 -- Chapter 29
2
Electric Potential Energy (Uelec)
• Potential energy
~ Gravitational
~ Uniform electric field
~ Point charge
Phys 133 -- Chapter 29
3
Gravitational potential (energy)
Defined
f
Wconservative = ò F × dr
i
= F ×∆r
(constant F)
Then defined
For the gravitational force
In general
Phys 133 -- Chapter 29
4
Uniform field: Electric potential energy
where
Uelec (uniform field) = qEDs
Phys 133 -- Chapter 29
5
Uniform field: Electric potential energy
Direction of lower potential energy depends on type of charge
Positive charge
Negative charge
Lower potential energy
Lower potential energy
s
s
Phys 133 -- Chapter 29
6
Do Workbook 29.6
Phys 133 -- Chapter 29
7
Problem 29.3
A proton is released from rest at the positive plate of
a parallel plate capacitor. It crosses the capacitor
and reaches the negative plate with a speed of
50,000 m/s. What will be the speed if the
experiment is repeated with double the amount of
charge on the capacitor plates?
(Ans: 7.07x104 m/s)
Phys 133 -- Chapter 29
8
Electrical Potential Energy of Two Point Charges
f
W elec = ò F1on 2 dx
i
f
æ Kq1q2 ö
=ò
dx
2
è x ø
i
Kq1q2 Kq1q2
=+
xf
xi
where
U elec
Phys 133 -- Chapter 29
Kq1q2
=
x
9
Question
a) Two point charges (both +Q) have their separation increased. Does
the electric potential energy of the system (A) increase, (B) decrease
or (C) remain the same?
b) Two point charges (+/- Q) have their separation increased. Does the
electric potential energy of the system (A) increase, (B) decrease or
(C) remain the same?
Phys 133 -- Chapter 29
10
Question ans
U elec
Like charges
U elec
K(+Q)(+Q)
KQ
=
=+
r
r
2
Kq1q2
=
r
Opposite charges
U elec
Phys 133 -- Chapter 29
K(+Q)(-Q)
KQ2
=
=r
r
11
Do Workbook 29.3
Phys 133 -- Chapter 29
12
Problem
An electron (q=-e) is sent from very far away towards a
charged sphere (Q=-0.25 nC). It gets to a distance of 1.10 mm
from the center of the sphere before turning around.
What is the initial speed of the electron?
(Ans: 2.68x107m/s)
What was the speed of the same electron with same initial
conditions when it was 2.20 mm away?
e=1.6x10-19C, me=9.1x10-31 kg, K=9x109 Nm2/C2
Phys 133 -- Chapter 29
13
Problem ans
r1 = 0.0011m
v1 = 0
r0 = ¥
v0 = ?
E 0 = E1
K 0 + U 0 = K1 + U1
Kqblueqgreen 1 2 Kqblue qgreen
2
1
= 2 mv1 +
2 mv 0 +
r0
r1
Kqblueqgreen 1
Kqblue qgreen
2
2
1
= 2 m(0)1 +
2 mv 0 +
¥
r1
Kqblue qgreen
2
1
2 mv 0 =
r1
2Kqblue qgreen
Þv =
mr1
Þ v 0 = 2.68 ´ 107 m s
2
0
Phys 133 -- Chapter 29
14
Electric Potential (Velec)
• Potential
~ Gravitational
~ Uniform electric field
~ Point charge
Phys 133 -- Chapter 29
15
Electric Potential (Velec)
Charge creates potential
define
Utest (x, y, z)
V (x, y, z) º
qtest
Potential only depends on
source charges
Uelec (x, y, z) = qV(x, y, z)
Look at
•Gravitational
•Uniform electric field
•Point charge
Phys 133 -- Chapter 29
16
Gravitational Potential
Vgrav
U grav
=
m
object
m gy
=
m
= gy
environment
Environment
only
Ugrav = mVgrav
= m(gy)
Phys 133 -- Chapter 29
17
Electric potential capacitor
Uelec = qEs (from before)
Velec
U elec
=
q
object
environment
q Es
=
q
Environment
= Es only
s
0
Uelec = qVelec
Phys 133 -- Chapter 29
18
Capacitor: potential representations
Phys 133 -- Chapter 29
19
Draw V(x) for the two equipotential
surfaces
Draw U(x) for a proton in these potentials
Draw U(x) for an electron in these potentials
Phys 133 -- Chapter 29
20
Workbook 29.18b extension
A proton is at x=0, moving to the right with 5x10-18 J of
kinetic energy. What is its kinetic energy at point a?
What if it were an electron?
Phys 133 -- Chapter 29
21
Workbook 29.18b extension ans
proton
electron
E0 = Ea
K 0 + U 0 = K1 + U1
K 0 + (qV0 ) = K1 + (qV1 )
E0 = Ea
K 0 + U 0 = K1 + U1
K 0 + (qV0 ) = K1 + (qV1 )
5 ´ 10 -18 J + 0 = K1 + (1.6 ´ 10 -19 C)(40V )
5 ´ 10
-18
J = K1 + 6.4 ´ 10
-18
J
5 ´ 10 -18 J + 0 = K1 + (-1.6 ´ 10 -19 C)(40V )
5 ´ 10 -18 J = K1 - 6.4 ´ 10 -18 J
Not enough energy to reach a
Phys 133 -- Chapter 29
K1 = 11.4 ´ 10 -18 J
22
Problem 29.21 Modified
a) In the figure which capacitor plate, left
or right, is the positive plate?
b) What is the electric potential energy of
a proton at the midpoint of the
capacitor?
c) What is the electric field strength
inside the capacitor?
In general
d) What is the electric field strength inside the capacitor in
terms of  and o?
e) What is the electric field strength inside the capacitor in
terms of ∆V and d (the distance between the plates?
Phys 133 -- Chapter 29
23
Only potential difference matters
Phys 133 -- Chapter 29
24
Electric potential two point charges
U elec
Velec
Kqqtest
=
(from before)
r
U elec
=
q
Kqqtest
r
=
qtest
Kq
=
r
Uelec = qVelec
Phys 133 -- Chapter 29
25
Point particle: potential representations
Phys 133 -- Chapter 29
26
Do Workbook 29.21
Phys 133 -- Chapter 29
27
Electric Potential: multiple charges
Vnet = å Vi
i
Vnet = å Vi
i
= Vblue + Vred
No vectors
Phys 133 -- Chapter 29
28
Do Workbook 29.25
Phys 133 -- Chapter 29
29
Problem
a) Find an expression for the electric potential
V(x,y) for a dipole with point charges +/- Q at y=
+/- a.
b) Use the result for the dipole to find the electric
potential at x=2 m, y=7 m when a=2 m, and Q=2 nC.
c) What is the potential energy of a proton placed there?
Phys 133 -- Chapter 29
30
Problem ans
V po int
r+
Kq
=
r
r+ = x 2 + (y - a) 2
y-a
x
r-
y+a
Kq+
V+ =
r+
=
KQ
x 2 + (y - a) 2
r- = x 2 + (y + a) 2
KqV- =
r-
x
V = å Vi = V+ +Vi
=
KQ
x 2 + (y - a) 2
+
K(-Q)
x 2 + (y + a) 2
Phys 133 -- Chapter 29
=
K(-Q)
x 2 + (y + a) 2
31
Problem plot
Phys 133 -- Chapter 29
32
Problem ans
b) Use the result for the dipole to find the electric potential at x=7 m,
y=2 m when a=2 m, and Q=2 nC.
é
ù
1
1
V (x, y) = KQê 2
ú
2
2
2
x + (y + a) úû
êë x + (y - a)
ù
Jé
1
1
= 18 ê 2
ú
2
2
C êë x + (y - 2) 2
x + (y + 2) úû
ù
Jé
1
1
V (2,7) = 18 ê
ú
2
2
C êë (2) 2 + (7 - 2) 2
(2) + (7 + 2) úû
J
= 18 ´ 10
C
-3
c) What is the potential energy of a proton placed there?
U(2,7) = qV(2,7) = (1.6 ´10-19 C)()
Phys 133 -- Chapter 29
33
Potential of continuous charge distribution
Vnet = å Vi
i
where
Then let
solve
Phys 133 -- Chapter 29
34
Problem 29.69
The figure shows a thin rod of length L and charge Q. Find
an expression for the electric potential a distance x away
from the center of the rod on the axis of the rod.
Phys 133 -- Chapter 29
35
Problem 29.69
Sum up contribution from each charge at P
V (d,0) = å Vi
y
i
d
ith
P (d,0)
x
xi
ri
ri = d - x i
Find potential at P due to ith bit of charge
L
V (d,0) =
2
ò
-L 2
KQdx
L( d - x )
L
é KQln(d - x) ù 2
= -ê
úû L
L
ë
- 2
KQ æ d + L 2 ö
=
lnç
÷
L èd-L ø
2
Phys 133 -- Chapter 29
36