# Suggested solution Part II V2

Ηλεκτρονική - Συσκευές

7 Οκτ 2013 (πριν από 3 χρόνια και 8 μήνες)

82 εμφανίσεις

FyANVC0
3 Electrostatics and DC Circuits Ch 16
-
19 [K6&7]

Part II V2

Name:……………………………

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:……………………………

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
)

-
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)

-
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:……………………………

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
)

-

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:……………………………

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:……………………………

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:……………………………

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: