Version One – Homework 1 – Juyang Huang – 24018 – Jan 16 ...

Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 18
supplied by the ignition circuit
Correct answer: 1800 V.
Explanation:
Let : E = 3 × 10 6 V/m and
d = 0.06 cm = 0.0006 m .
Assuming the electric field between the two
electrodes is constant, then the potential difference
between the electrodes is
keywords:
V = E d
= ( 3 × 10 6 V/m ) (0.0006 m)
= 1800 V .
Accelerating an Electron
25:05, trigonometry, numeric, > 1 min, normal.
040 (part 1 of 1) 10 points
Through what potential difference would an
electron need to be accelerated for it to
achieve a speed of 4 % of the speed of light
(2.99792 × 10 8 m/s), starting from rest
Correct answer: 408.799 V.
Explanation:
Let : s = 4% = 0.04 ,
c = 2.99792 × 10 8 m/s ,
m e = 9.10939 × 10 −31 kg ,
q e = 1.60218 × 10 −19 C .
The speed of the electron is
v = 0.04 c
= 0.04 ( 2.99792 × 10 8 m/s )
= 1.19917 × 10 7 m/s ,
By conservation of energy
1
2 m e v 2 = −(−q e ) ∆V
and
v 2
∆V = m e
2 q e
= ( 9.10939 × 10 −31 kg )
(
1.19917 × 10 7 m/s ) 2
×
2 (1.60218 × 10 −19 C)
= 408.799 V .
keywords:
Point Charge
25:05, trigonometry, numeric, > 1 min, normal.
041 (part 1 of 1) 10 points
At distance r from a point charge q, the electric
potential is 600 V and the magnitude of
the electric field is 200 N/C.
Determine the value of q.
Correct answer: 2.00277 × 10 −7 C.
Explanation:
Let : k e = 8.98755 × 10 9 N · m 2 /C 2 ,
V = 600 V , and
e = 200 N/C .
E = k e q
r 2 and V = k e q
r , so that V E = r.
The potential is
V = k e q
r
= k e q
V
E
= k e q E
V
q = V 2
k e E
(600 V) 2
=
(8.98755 × 10 9 N · m 2 /C 2 ) (200 N/C)
= 2.00277 × 10 −7 C .
keywords:
Conducting Spheres 02
25:09, trigonometry, multiple choice, > 1 min,
wording-variable.
042 (part 1 of 4) 10 points
Consider two “solid” conducting spheres with
radii r 1 = 4 R and r 2 = 3 R ; i.e.,
r 2
r 1
= 3 R
4 R = 3 4 .