This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 65. We denote the electron with subscript e
and the proton with p. From the
ﬁgure to the right we see that
Ee = Ep = e
4πε0 d 2 where d = 2.0 × 10−6 m. We note
that the components along the
y axis cancel during the vector
summation. With k = 1/4πε0
and θ = 60◦ , the magnitude of
the net electric ﬁeld is obtained
as follows: Enet ...
....
..
..
..
..
..
.. Ep
..
.
..
..
.. θ
..
x
...........................................
.
...
..
...
.
. ..
..
. ..
.. θ
. ..
...
..
...
..
..
Enet
.
..
....
.
.
...
.
..
..
....
.
..
..
..
..
..
..
.
.
.....
..
..
...... Ee
.....
..
.
..
.
.
y .
..
.. ..
.. .
..
. •
proton = 2 = 2 8.99 × 109 = 3.6 × 102 N/C . . θ . ..
..
. ..
..
. ..
..
. ..
..
. •
electron Ex = 2Ee cos θ = ..
.. e
4πε0 d2 cos θ = 2k
N · m2
C2 e
cos θ
d2
(1.6 × 10−19 C)
cos 60◦
(2.0 × 10−6 m)2 ...
View
Full
Document
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

Click to edit the document details