Next Week
Tutorial and Test 1 on chapter 18
Tutorial problem numbers updated on the web
Week of January 26 - 28
Experiment 2: Wheatstone Bridge
Chemistry 1310 midterm
Is DEFINITELY on March 4 !
Friday, January 15, 2010
20
Electric Field, Potential Energy
Electric field: Es = -!V/!s, V/m or N/C
Electron-volt (eV)
Potential energy of 1 electron charge at a potential of 1 V.
PE = qV = eV = (1.6!10-19 C)!(1 V) = 1.6!10-19 J
or, PE = 1 eV (electron-volt)
1 eV = 1.6!10-19 J
As the charge moves,
!KE = –!PE
(i.e., KE + PE = constant)
An electron charge gains a kinetic energy of 1 eV in falling through
a potential drop of 1 V.
Friday, January 15, 2010
21
Clicker Question: A positive charge is moving from point A to point B
in a uniform electric field, as shown in the drawing.
The electric force does positive/negative
__________ work on the charge and, as a
consequence, its electric potential energy increases/decreases
__________ .
A) negative, does not change
B) positive, decreases
C) positive, increases
D) negative, increases
E) negative, decreases
F = qE
Answer: D
Force in direction opposite to displacement, does negative work
KE + PE = constant: KE decreases, PE increases
Friday, January 15, 2010
22
y
Prob. 19.41/56:
C
�E
10 cm
A
6 cm
x
8 cm
�E
B
E = 3600 V/m.
Find VB – VA,
VC – VB,
VA – VC
Friday, January 15, 2010
0V
288 V
-288 V
23
Prob. 19.2/3: Point A is at a potential of 250 V, and point B is
at a potential of –150 V.
An "-particle is a helium nucleus that contains 2 protons and 2
neutrons. The neutrons are electrically neutral.
An "-particle starts from rest at A and accelerates toward B.
When the "-particle arrives at B, what kinetic energy (in
electron volts) does it have?
800 eV, or 1.28x10-16 J
Friday, January 15, 2010
24
The anode (+ terminal) of an x-ray tube is at a potential
of +125,000 V with respect to the cathode (– terminal).
a) How much work is done by the electric force when an
electron is accelerated from the cathode to the
anode?
b) If an electron starts from rest, what KE does it have
just before it hits the anode?
a) 125,000 eV, or 2x10-14 J
b) 125,000 eV
Friday, January 15, 2010
25
Prob. 19.-/31: The spark to the door knob from the end of
the finger problem. Due to charge build-up when humidity is
low when the air is a poor conductor of electricity.
A spark occurs when the electric field is so great that the
air breaks down (“dielectric breakdown”), and the air then
becomes conducting.
This happens when E " 3 ! 106 V/m for dry air (3 MV/m).
Suppose the spark is 3 mm long, electric field is uniform.
Calculate potential difference between finger and door knob.
Assume electrons travel from hand to door knob.
Vdoor - Vfinger = 9000 V
Friday, January 15, 2010
26
Prob. 19.40/35: Equipotential surface A has a potential of 5650 V,
while equipotential surface B has a potential of 7850 V.
A particle has a mass of 0.05 kg and a charge of +4 ! 10-5 C.
The particle has a speed of 2 m/s on surface A. An outside force is
applied to the particle, and it moves to surface B, arriving there with a
speed of 3 m/s.
How much work is done by the outside force in moving the particle
from A to B?
What is the PE of the particle at A and at B?
What is its kinetic energy?
0.213 J
Friday, January 15, 2010
27
Flow of Current
I
I
Current is the rate of flow of charge.
I=
!q
amps (A), 1 A = 1 C/s
!t
“Conventional” flow of current – as if positive charges were
flowing around the circuit from + terminal to – terminal.
Current is really a flow of negatively-charged electrons
around the circuit in the opposite direction, from – to +
terminal.
The electrons change PE by:
!PE = (−e) × (12 V) = −12 eV
Friday, January 15, 2010
28
Example:
Determine the number of electrons that pass between the
terminals of a 12 V battery when a 60 W headlight burns for 1
hour.
• How much energy is used in 1 hour?
• How much PE does 1 electron lose in travelling between – and +
terminals of the 12 V battery?
• How many electrons are needed to account for the energy
provided by the battery?
1.15x1023
Friday, January 15, 2010
29
Potential and Potential Energy
As F =
kqq0
r2
the work done by the Coulomb force
in pushing the charges apart is:
kqq0 kqq0
WAB =
−
rA
rB
rA
(integral
calculus)
q0
rB
= q0(VA −VB)
( = PEA - PEB)
Potential energy of a pair of charges: P E =
Potential due to q alone: V =
kqq0
r
(PE = 0 at infinity)
PE
kq
=
q0
r
Friday, January 15, 2010
30
Electric Potential
Potential at r = 1 km from a charge q = 1 C
kq 9 × 109 × 1
V=
=
= 9 × 106 V
r
1000
= 9,000 kV
= 9 MV !!!
Charge built up in clouds during thunderstorm " 10 to 30 C
huge electric field, eg 10 C at 1 km:
kq 9 × 109 × 10
E= 2=
= 90, 000 V/m
r
10002
V = 90 MV
Friday, January 15, 2010
31
Prob. 19.31/29: Two equipotential surfaces surround a
1.5!10-8 C point charge.
How far is the 190 V surface from the 75 V surface?
1.8 - 0.7 = 1.1 m
Friday, January 15, 2010
32
Prob. 19.30/60: A positive charge +q1 is located 3 m to the left of a
negative charge –q2. The charges have different magnitudes.
On the line between the charges, the net electric field is zero at a
spot 1 m to the right of the negative charge.
On this line there are also two spots where the potential is zero.
Locate these two spots relative to the negative charge.
+q1
16q2
3m
–q2
–q2
1m
E=0
• Find the relation between the charges from the info on the electric
field.
0.2 m to right and 0.176 m to left of -q2
Friday, January 15, 2010
33