Download Lecture #36 12/02/05

Alpha-Decay
•Very heavy nuclei are “crowded” – nucleons want to leave
•Although it is possible for them to emit single nucleons, this is
very rare
•Although it is possible for them to emit large particles, it is
easier for them to emit small well-bound particles
•Such a particle is a 4He nucleus
238
92
U
234
90
Th  He
4
2
•Because the 4He nucleus has four nucleons, two of which are
protons, Z decreases by 2 and A decreases by 4
•The 4He nucleus is also called an  - particle
•This process is called alpha decay
Beta Decay
•A neutron inside a nucleus is spontaneously decays into a
proton, an electron, and an antineutrino.
1
0
n  p  e 
1
1
-
C 147N  e   v
14
6
•The number of protons changes, so the element changes.
•Though energy, momentum, angular momentum, nucleon
number, and charge is conserved.
•A dominate mechanism for light nuclei.
Beta: Electron Capture
•Another mechanism for light nuclei decay.
•A proton inside a nucleus absorbs an electron, and becomes a
neutron
7
4

Be  e  Li  v
7
3
•The number of protons changes, so the element changes.
•Though energy, momentum, angular momentum, nucleon
number, and charge is conserved.
Gamma Decay
•After a nucleus undergoes a radioactive decay, the nucleus is
often in an excited state.
•The nucleus can lose the energy by emitting a gamma ray
(high energy photon)
12
5
*
C * 126C  
12
6

B C  e  v
12
6
Quiz
Two test charges are brought separately into the vicinity of a charge +Q.
First,• test
T charge +q is brought to point A a distance r from charge +Q.
Next, the +q charge is removed and a test charge +4q is brought to point B
a distance 2r from charge +Q. Compared with the electric field of the
charge at A, the electric field of the charge at B is:
+Q
+q
+Q
A
+2q
B
A) Greater
B) Smaller
C) The same.
Coulomb’s Law
•Like charges repel, unlike charges attract
•Force is directly along a line joining the two charges
q1
ke q1q2
Fe 
rˆ
2
r
q2
r
q1q2
Fe 
rˆ
2
4 0 r
0 = 8.85410-12 C2/ (N●m2)
•Permittivity of free space
•An inverse square law, just like gravity
•Can be attractive or repulsive – unlike gravity
•Constant is enormous compared to gravity
Electric Fields
•Electric Field is the ability to extert a force at a distance on a
charge
•It is defined as force on a test charge divided by the charge
•Denoted by the letter E
•Units N/C
+ +
– ––
+ +
+
F
Small test
charge q
E  F /q
Electric Field Lines
Consider the four field patterns below:
Assuming that there are no charges in the region of space depicted,
which field pattern(s) could represent electrostatic field(s)?
Electric Field Lines
•Graphical Illustration of Electrical Fields
•Lines start on positive charges and end on negative
•Number of lines from/to a charge is proportional to that charge
•Density of lines tells strength of field.
+
-
+
-
Electric Fields and Forces
E  F /q
F  qE
F  ma
a  qE / m
A region of space has an electric field of 104 N/C, pointing in the plus
x direction. At t = 0, an object of mass 1 g carrying a charge of
1C is placed at rest at x = 0. Where is the object at t = 4 sec?
A) x = 0.2 m
C) x = 20 m
B) x = 0.8 m
D) x = 80 m
Quizzes 2
A cube with 1.40 m edges is oriented as shown in the figure
Suppose there is a charge situated in the middle of
the cube.
• What is the magnitude of the flux through the
whole cube?
• What is the magnitude of the flux through any
one side?
A) q/o
B) q/4o
C) 0
D) q/6o
Electric Flux
•Electric Flux is the amount of electric field flowing through a
surface
•When electric field is at an angle, only the part perpendicular
E
to the surface counts
En

•Multiply by cos 
E = EnA= EA cos 
•For a non-constant electric field,
or a curvy surface, you have to
integrate over the surface
 E   E  dA   E cos dA
•Usually you can pick your surface so that the integration doesn’t
need to be done given a constant field.
Electric Flux
•What is electric flux
E
through surface
surrounding a charge q?
R
charge q
ke q
 4 R E  4 R 2  4 ke q
R
 E  4 ke q
2
2
 E  2 ke q  2 ke q
 4 ke q
Answer is
always 4keq
Quiz
– An electron is accelerated from rest through a
potential difference V. Its final speed is
proportional to:
– A) V2
– B) V
– C) V1/2
– D) 1/V
Quiz
• Points R and T are each a distance d from each of
two equal and opposite charges as shown.
required to move a negative charge q from R to T
is:
•
• A) kQq / (2d)
• B) kqQ / d
• C) kqQ / d2
• D) zero
Potential Energy of charges
charge q
•Suppose we have an electric field
•If we move a charge within this
field, work is being done
•Electric Fields are doing work on Electric Field E
the charge
W  F  s  qE  s

U  W  qE  s  q E  s

•If path is not a straight line, or electric field varies you can
rewrite this as an integral
U   q  E  ds
Electric Potential
Point A
•Path you choose does not matter.
(conservative)
B
U  q  E  ds
A
•Factor out the charge – then you
have electric potential V
Electric Field E
Point B
U
V 
   E  ds
q
A
B
1 1
1
3 1
 


C 12 24 24 8
12
C  12  8  20
12
8
C  12 12  24
2412
12
Each of the Capacitors above has a capacitance of 12 pF.
What is the combined capacitance of the whole system?
A) 12 pF
C) 8 pF
B) 4 pF
D) 20 pF
Combining Capacitors: Series
wire (conductor)
capacitor
switch
–
+
battery
Charges are the same on each capacitor
Voltages add
In Series:
C1
C2
C3
1 1
1
1
 

C C1 C2 C3
Combining Capacitors: Parallel
wire (conductor)
capacitor
switch
–
+
battery
Voltages are the same across each capacitor
Charges add Same Voltages
In Parallel:
C1
C2
C3
C  C1  C2  C3
Capacitors and Dielectrics
Dielectric
constant 
C
 0 A
d
area A
•A dielectric changes the capacitance
•Cause a breakdown potential, Vmax to exist.
d
•Beyond the breakdown potential the dielectric
starts to conduct!
Ohm’s Law
V  IR
•The resistance R is a constant irregardless of the applied potential
•This is equivalent to saying that the resistivity of the material is
independent of the applied field
J  E
Area A
Kirchoff’s Rules
•The total current flowing into a point must equal the total
current flowing out of a point [conservation of charge]
•The total voltage change around a loop must total zero
I2
I3
V1
V2
–
+
I1
I3=I2+I1
V3
V1 + V2 + V3 = 0
Odd Circuit
What is the current through the
resistor?
A) 3.6 A
B) 1.8A
C) 90 A
D) 0 A
–
+
9V
5
9V
– +
Four circuits have the form shown in the diagram. The capacitor is
initially uncharged and the switch S is open.
The values of the emf , resistance R, and the capacitance C for each of
the circuits are
circuit 1:
18 V, R = 3 , C = 1 µF
circuit 2:
circuit 3:
18 V, R = 6 , C = 9 µF
12 V, R = 1 , C = 7 µF
circuit 4:
10 V, R = 5 , C = 7 µF
Which circuit has the largest current right after the switch is closed?
Which circuit takes the longest time to charge the capacitor to
½ its final charge?
Which circuit takes the least amount of time to charge the capacitor to
½ its final charge?
RC circuits
•Capacitor/resistor systems charge or discharge over time
Charging:



q(t )  Qmax 1  e t / RC  Qmax 1  e t /

is the time constant, and equals RC.
Discharging:
q(t )  Qe
t /
Qualitatively: RC controls how long it takes to charge/discharge
completely. This depends on how much current can flow (R)
and how much charge needs to be stored (C)
[As an exercise, show that RC has units of secs]
RC circuits: Prior to Steady-State
•Recall: the voltage across a capacitor is: V=q/C
E
–
+
S1
•When the capacitor is fully charged the voltage
is ( e.g. it acts like a broken wire)
•Prior, the voltage is V, i.e. there is a voltage
drop.
Close S1
Apply the loop rule:
C
R
q
  iR   0
c
dq
q

R 0
dt
c
The result is a differential
equation.