Transcript Slide 1

Quiz# 2
deals with answering questions about
(and calculating values for)
charged particle beams being accelerated and deflected by
E and B fields.
Here is a recap of in-class problems/questions dealing with
charged particles accelerating across a potential difference
and/or being steered or deflected by Electric or Magnetic fields.
You might also want to review the textbook’s
Assigned problems:
Ch.19: P 56, 76
Ch.20: P 17,18
Ch.22: P 17
Electron Gun
Heated filament
Positively charged can
If the effective field is E=800,000 N/C over a gap length
of 2.5 cm, what is the electron’s final velocity, v?
1. One method you can use:
F qeE
(1.610-19C)(8105 N/C)

a

-19 kg
9.1110
me me
17
2
 1.405  10 m / sec
2
2
v  v 0  2ax
 0 + 2(1.40510-19m/s2)(0.025 m)
 7.0251015m2/s2
v  8.38  107 m/s
See your class notes for
Friday, September 5
2. Another
method
E = 800,000 N/C
1e = 1.6010-19 C
What voltage difference
do the electrons
experience jumping
the gap?
d = 2.5 cm,
vfinal?
V  Ed  800,000 V/m  0.025m
 20,000 volts
How much energy is given to each electron?
qV  (1e)V  (1.601019 C)(20,000J/C)
 3.2 1015 J
An electron with this much energy, must be moving how fast?
1 2
mv  3.2 10 15 J
2
See your class notes for
Monday, September 7
v 2  2(3.2 1015 J) /(9.111031kg)
 7.0251015m2 /se c2
v  8.3810 m/sec
7
If the beam then enters the center of the 0.3cm gap
between two electrostatic plates, 2cm in length,
what is the
maximum
E field that
will still allow
the electrons to escape the opposite side?
0.3 cm
-e
2 centimeters
See your class notes for
Friday, September 5
vo = 8.4107 m/sec
me = 9.1110-31 kg
1e = 1.6010-19 C
Emax = ?
vo
0.3 cm
-e
2 centimeters
x = voxt +
2
½axt
0
y = voyt +
2
½ayt
0
Time within plates: t= (horizontal distance)/(horizontal speed)=2 cm/vo
Maximum tolerable acceleration up then comes from
d = (1/2)at2 using that same time and noting from the center d=(1/2)0.3cm:
0.15 cm = (1/2)a(2.3810-10 sec)2  a = 5.2961016 m/sec2
Answer: E = 301,541 N/C
Mass spectrometer:
     
R=?
    
     
v= /2qV/m
    
     
  1  
Show that any charged
ions drifting from the
vaporization chamber
(far left) are launched to
this final speed by the
accelerating potential.
Then show that the radius of curvature inside
the B-field should be:
See your class notes for
Wednesday, October 1
R
B
2Vm / q