Download P23.2 P23.4 P23.11

The solution to the homework 1 of PHY114, edited by Xiaohua Liu
P23.2
(a) N=(
) (47
(b) # electrons added =
)= 2.
=
electrons added
Thus,
added)
=
2.38 electrons for every
already present
P23.4
contains
Ny(
)(10
)y2.3×
protons
With an excess of 1% electrons over protons, each person has a charge
q
C
So
N
This force is almost enough to lift a weight equal to that of the earth:
kg (9.8 m/
Nw
N
P23.11
(a) Let the third bead have charge Q and be located distance x from the left end of the
rod.
This bead will experience a net force given by
, where d =1.50 m
The net force will be zero if
, or
.
This gives an equilibrium position of the third bead of x=0.634d=0.634(1.50m)
=0.951m.
(b) Yes, if the third bead has positive charge. The equilibrium would be stable because
if charge Q were displaced either to the left or right on the rod, the new net force would
be opposite to the direction Q has been displaced, causing it to be pushed back to its
equilibrium position.
P23.14
Each charge exerts a force of magnitude
on the negative charge -Q: the top
charge exerts its force directed upward and to the left, and bottom charge exerts its force
directed downward and to the left, each at angle
, respectively, above and
below the x axis. The two positive charges together exert a net force:
or for
,
(a) The acceleration of the charge is equal to a negative constant times its displacement
from equilibrium, as in
, so we have Simple Harmonic Motion with
.
(b)
object with charge -Q.
(c)
P23.29
, where m is the mass of the
(a) From Example 23.6, the magnitude of the electric field produced by the rod is
N/C
(b) The charge is negative, so the electric field is directed towards its source, to the right.
P23.41
Field lines emerge from positive charge and enter negative charge.
(a) The number of field lines emerging from positive
proportional to their charges:
(b) From above,
is negative,
is positive.
and entering negative charge
is