Download Page 1 Problem An electron is released from rest in a uniform

Chapter 16 Solutions
4
Problem 15
Problem Compare
/ the electric force holding the electron in orbit around the proton (2 ' )
) in the hydrogen nucleus with the gravitational force between the same
electron and proton. What is the ratio of these two forces?
Solution The magnitude of the electric force is given by Coulomb’s law
1 1
(
+ ,.- /
2 (
/
Since the magnitude of the charge on each of the electron and the proton in the hydrogen
1
1
' )
( is , and the distance between the two is 2 ,
the magnitude of the electric force is
9 = (
+ & ? ;
/
,.- / 9< ' ) = ( 4 On the other hand the gravitational force is given by Newton’s universal law of gravitation
+ ( 2 (
According to the table on the inside of the front cover, the mass of the electron is 6 & E@
, while that of the proton is ( C (
, and the distance
33; ) (78 (
is the same. Finally, Newton’s gravitational constant is 6
C
.
The gravitational force is therefore
+
6
C 3 ; ) ( 8
6E@ C / 6(
(A 6 &
9< ' ) =(
: ; The ratio of the electric force to the gravitational force is therefore
+ & ? ;
4 E3 + : ;5
4
This is an indication of how much more powerful the electric force is than the gravitational force.
Problem 27
Problem An electron is released from rest in a uniform electric field and accelerates to the
) 8 (
. What is the magnitude and direction of the electric field.
north at a rate of '
Solution The force on a charge in an electric field is
1
and according to Newton’s second law the acceleration the charge will undergo is
4
Chapter 16 Solutions
Putting these two equations together we get
1
5
1
E@ &
For an electron with mass 6 &
and charge
.
Putting these and the acceleration of the electron into our formula for we get for the
magnitude of /
6 & E@
A ) 8 (
; 8& '
C Because an electron has a negative charge, but the field refers to the motion of a positive
test charge, the direction of the field is opposite the direction of the acceleration, or in
other words south.
Problem 29
Problem Determine the direction and magnitude of the electric field at the point shown
in Fig. 16-40. The two charges are separated by a distance of " and the point is
a distance F out on the perpendicular bisector of the line joining them. Express your
1
answers in terms of , F , " and .
y
+Q
a
a 2+
x
x
2
θ
x
E−
P
Solution
E+
−Q
Figure 3: (Figure 16-40 in the textbook.)
It is instructive to look at Figure 3 in order to solve this problem. Because the positive
and negative charges are of the same magnitude and placed symmetrically about the
point where we are to find the field, the F components of both electric fields and
are of the same magnitude but opposite direction. However, the G components are
of the two electric fields are of the same magnitude and both in the negative G direction.
So we only need to calculate the y component of one field, and double it, to have the
entire field. Now, the formula for the magnitude of the electric field from a point charge
is
1
,.-0/ 2 ( (
(
In this case, the distance between either charge and is 2 " # F . The G component is given by , where is the angle marked in the figure. Since we have
a right triangle,
"
(
( " # F
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