FyANVC0
3 Electrostatics and DC Circuits Ch 16

19 [K6&7]
Part II V2
Name:……………………………
©
behzad.massoumzadeh@adm.huddinge.se
1
Part II
This part consists of generally long problems
.
Try all of the problems and answer
them
in detail
.
Total points: 64; G: 21; VG: 43 (min.
7 VG P); MVG:
min
47 (min 14 VG P & MVG quality)
Time period part I + II: 12:50

15:30
Good luck!
6.
Three
charge
s:
,
and
I are
placed at each
corner of a
n equilateral triangle of a side
.
Determine the magnitude and
direction of the force on
the
charge
.
(2/4
/¤
)
(1/1)
(1/1)
Force
Horizontal Component
Vertical Component
where
and
are unit vectors in the x and y direction.
(0/2/¤
)
Answer: Therefore, the force acting on the charge
due to the other charges
is
47
.5
N. This force make
s an angle of 65
degrees with the positive x axis.
FyANVC0
3 Electrostatics and DC Circuits Ch 16

19 [K6&7]
Part II V2
Name:……………………………
©
behzad.massoumzadeh@adm.huddinge.se
2
7.
An

particle (nucleus of He
: 2 protons
,
) is
shot from far
away
with an initial velocity
directly toward a g
old nucleus
(79
protons:
) .
a)
Calculate the voltage necessary to accelerate an

particle to this velocity
. (2/4)
b)
Calculate the closest distance the

particle gets to the nucleus.
Assume that the gold
nucleus remains stationary.
(2/4)
Suggested solution:
Data:
,
,
,
Problem:
,
a)
The voltage necessary to accelerate the

particle to the velocity
may be found using the principle of the conservation of energy:
(1/3)
(1/1
)
Answer:

particles must be accelerated in a potential difference of
to
acquire velocity
.
Second method:
b)
Calculate the closest distance the

particle gets to the nucleus. Assume that the gold
nucleus remains stationary.
(2/4)
To find the closest distance the most energetic

particles reach we use the
conservation of energy princi
ple and fact that “the closest distance” is associated
with zero kinetic energy (i.e.

particle stops and may be reflected back):
(1/3)
Answer: The most energetic

particles approaching the nucleu
s of gold with the
velocity
may come as close as
.
(1/1
)
Second method:
Third method:
FyANVC0
3 Electrostatics and DC Circuits Ch 16

19 [K6&7]
Part II V2
Name:……………………………
©
behzad.massoumzadeh@adm.huddinge.se
3
8.
Can a

diameter copper
(
)
wire have the same resistance as
a iron
wire
(
)
of the same length? Give numerical details.
(
2/
4)
Suggested solutions:
Data:
,
,
,
,
.
Problem:
Yes of course a

diameter copper wire may h
ave the same resistance as a platinum
wire of the same length but different cross

section area:
and
. Using the fact both of these wires have
identical lengths, the lengths can cancel out from both sides and w
e may obtain:
(1/3)
(1/1
)
Answer: A

diameter copper wire may have the same resistance as a

diameter
iron
wire of the same length.
FyANVC0
3 Electrostatics and DC Circuits Ch 16

19 [K6&7]
Part II V2
Name:……………………………
©
behzad.massoumzadeh@adm.huddinge.se
4
9.
A
and
charges are placed
apart. At what point
s
along the line joining them is.
a)
At what point(s) along the line joining them is the electric field zero?
(2/4)
b)
At what point(s) along the line joining them is the electric field zero?
(2/4)
c)
Analyze the pr
oblem in general
. i.e. Two charges
and
are placed
apart.
i.
At what points along the line joining them is the electric field zero?
(¤)
ii.
At what points along the line join
ing them is the electric potential zero?
(¤)
Suggested solutions:
Data:
,
,
,
,
Problem:
,
,
a)
Electric field is a vector quantity. Its direction is outward from a positive charge
and toward a negative charge. In
our arrangement, the electric field at a point to
the left of the positive charge may be zero. If the
direction to the right is
assumed positive, then the electric field due to
at a distance
to
the left of the charge is
positive (its direction is towards the charge
, i.e.
to the
right
). This field is neutrali
zed by a field due to
the positive
charge
, which is negative
, i.e. its direction is to the
left
(
away from
the
positive
charge). This points lies at
from the charge:
(1/3)
Answer: The total electric field is zero at a point
to the left of the
negative
charge
, on the line joining the charges,
as shown above.
(1/1
)
b)
On the other
hand the electric potential is a scalar quantity. Therefore, the point
may lie at a point
to the left of the positive charge on the line joining the two
charges. The potential is also at a point
to the left
of the negative charge
(
to the right of the positive charge) on the line joining the charges:
(1/2
)
FyANVC0
3 Electrostatics and DC Circuits Ch 16

19 [K6&7]
Part II V2
Name:……………………………
©
behzad.massoumzadeh@adm.huddinge.se
5
And:
(1/2
)
Answer: Therefore, the electric potential is zero at two points:
i.
it is zero
a point 0.04
m to the left of the positive cha
rge (and 0.1
4
m to
the left of the negative charge) on the line joining the charges.
ii.
It is zero on the line joining the charges, b
etween them and at a point
0.0775 m to the left of the positive charge (0.0225
m to the right of the
negative
charge.)
c)
Depending on the re
lative value (and signs) of the charges
and
the
problem may have different solutions. Let’s assume that
:
i.
If
the charges are of
different sign
, for example, if
is
positive and
is
negative (like the problem above).
The total
electric field is zero at a point
to the left of the positive charge
, on the line joining the charges, as shown
abov
e, where:
:
ii.
On the other hand, similar to the part b, the electric potential is zero
i.
at a point
to the left of the positive charge on the line joining the two
charg
es:
ii.
at a point
to the left of the negative charge (and
to the right of
the positive charge) on the line joining the two charges:
FyANVC0
3 Electrostatics and DC Circuits Ch 16

19 [K6&7]
Part II V2
Name:……………………………
©
behzad.massoumzadeh@adm.huddinge.se
6
iii.
If the charges
are of
the same sign
: both positive (or both negative) then
the electric potential is zero only at infinity. On the other hand the
electric field is zero at a point
to the left of the larger charge,
i.e.
(and
to the right of the smaller charge
) on the line joining the
charges:
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