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Suggested solutions Test FyANVC07 Electrostatics, DC-Circuit Ch16-19 NV-College
CHAPTER TEST: Electrostatics and DC currents Ch 16-19
Physics A
FALL 2008: FyANVC07
Warning: There are more than one versions of the test.Instructions
Test period 90 minutes.Tools Formula sheet, your personal summery booklet and ruler, Calculator.
The test For most items a single answer is not enough. It is also expected
that you write down what you do
that you explain/motivate your reasoning
that you draw any necessary illustrations.
After every item is given the maximum mark your solution can receive.
(2/3) means that the item can give 2 g-points (Pass level) and 3 vg-points
(Pass with distinction level).
Items marked with (Problems 1f, 1e, 6 and 7) give you a possibility to
show MVG-quality (Pass with special distinction quality) This means that
you use generalised methods, models and reasoning, that you analyse
your results and account for a clear line of thought in a correct
mathematical language.
Try all of the problems. It can be relatively easy, even towards the end of
the test, to receive some points for partial solutions. A positive evaluation
can be given even for unfinished solutions.
Enjoy it!
Mark limitsMaximum score: TOTAL 63 out of which 31 vg-points, and 3 MVG points
G: 19 points
VG: 38 points/ at least 9 VG points:
MVG: 43 points/ at least 18 VG points; MVG-quality work
At the aspect assessment of your work with exercise 1e, 6 and 7 I will consider
the depth of understanding of physics you have demonstrated
how well you have carried through the task
how well you have explained your work and motivated your conclusions
how well you have accounted for your work.
Only the marked problems in the box below will be graded.
1a 1b 1c 1d 1e 1e 1e 2 3 4 5a 5b 5c 5d 5e2/2 2/2 2/1 2/1 1/0 0/2 0/3/ 2/1 2/0 1/2 1/2 2/0 1/2 2/0 1/1
6a 6b 6c 7 Total TotalVg
Total
1/3 2/4/ 3/0 2/4/ 63 31
Grade Name:
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1. Answer the questions 1a to 1e based on the figure below which is a simple DC-circuit: Switch S is closed in questions 1a to 1e.:
1A
2A
3A
S
2V
1V
3V
5V
5I
=
00
.5
3
R
=
00
.5
1
R
= 00.54
R
= 0.102
R
=
0.
10
5
R
VV 0.40=
Suggested Solutions: The voltmeters and Ammeters show thefollowing values illustrated in the figure below:
S
=
00
.5
3
R
=
00
.5
1
R
= 00.54
R
= 0.102
R
=
0.
10
5
R
VV 0.40=
V20
V10
V10
V5
A2
A2
A1
A1
1.a The equivalent resistance of the circuit is:
i. 0.40
ii. 0.20
iii. 0.10
iv. 0.5
Answer: Alternative: ii: = 0.20R [1/0]
Show details of you calculations: [1/2]
Suggested solutions:
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To prove our claim completely we may first calculate the equivalent resistance of the
whole circuit.
= 00.53R and = 00.54R are in series. Their equivalent resistance is
=+= 0.1000.500.534R [1/0]
= 0.1034R and = 0.105R are in parallel. Their equivalent resistance is
===+=+= 00.52
0.10
0.10
2
0.10
1
0.10
1111345
534345
RRRR
[0/1]
00.5 ,=1R = 0.102R and = 00.5345R are in series. Their equivalent
resistance is the equivalent resistance of the circuit:
=++=++= 0.200.5100.5321 RRRReq
Answer: = 0.20eqR [0/1]
1.b What does the voltmeter 3V read in the DC-circuit above?
i. V0.40
ii. V0.20
iii. V0.10
iv. V0.5
Answer: alternative iv: VIRV 00.5333 == [1/0]
Show the details of you calculations: [1/2]Suggested solution:
To calculate the voltages across = 00.53R shown by voltmeter 3V , and the current
passing through the resistor = 00.51R shown by ammeter 1A , we may use the
information obtained above, i.e. the equivalent resistance of the circuit is
= 0.20eqR . The total voltage supplied by the source is VV 0.40= . Therefore: , the
total current passing through the battery, as well as the resistor = 00.51R may be
calculated using Ohms law as:
AR
VI
eq00.20.20
0.40
=== [0/1]
This is the same current that passes through = 00.51R and = 0.102R .
Therefore the current read by the Ammeters 1A and 2A are:
AIII 00.221 === . [1/0]
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The current AIII 00.221 === is divided to two equal amounts and
AAI
I 00.12
00.2
25 === passes trough = 0.105R and the rest, i.e.
AI 00.13 = will pass through = 00.53R and = 00.54R . This is due to the
fact that, as illustrated above, the equivalent resistance of the combination=+= 0.1000.500.534R is identical to = 0.105R and therefore the incoming
current due to the symmetry of the problem is divided to two equal amounts. [0/1]
Therefore, the voltage across the resistor = 00.53R , shown by the voltmeter 3V
may be calculated using ohms law as:
VVIRV 00.500.100.5333 === Answer: VIR 00.5333 ==V
1.c What does the ammeter 1A read in the DC-circuit above?
i. A00.4
ii. A00.2
iii. A00.1
iv. A500.0
Answer: alternative ii: AIII 00.221 === [1/0]
Show details of you calculations: [1/1]
Suggested solution: See the solution of 1c.
1.d The power dissipated in the resistor of resistance = 0.105
R is:
i. W00.1
ii. W00.5
iii. W0.10
iv. W0.40
Answer: alternative iii: WP 0.105 = [1/0]
Show details of you calculations: [1/1]
Suggested solution:
Using the information obtained above, i.e.: AI 00.15 = we may calculate the power
dissipated in the resistors = 0.105R is:
( ) WIRP 0.1000.10.10 22555 === [1/1] Answer: WP 0.105 =
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1.e If the resistors are replaced by light bulbs of the same resistance. What happens whenthe switch S is opened? [1/0]
i. All remaining light bulbs will shine brighter.
ii. All remaining light bulbs will shine dimmer.
iii. Light bulb 5R will shine brighter but light bulb 1R and 2R will be dimmer.
iv. Light bulb 5R will be dimmer but light bulb 1R and 2R will shine brighter.
Answer: Alternative iii: Light bulb 5R will shine brighter but light bulb 1R and 2R will
be dimmer.
Explain conceptually in the space provided below your motivation for the choicemade above [0/2]
Prove your claim by performing calculations necessary in sufficient details and as
clear as possible. [0/3/]
Conceptual solutions:
When the switch S is opened, the resistors 3R and 4R are taken out of the circuit and the
system just consists of 1R , 2R and 5R . Due to the fact that the combination 4334 RRR +=
and 5R were initially connected in parallel, their equivalent resistance 345R was smaller
than 5R . Therefore, by removing 3R and 4R from the system the total resistance of the
circuit is increased. Thus less current passes through 1R and 2R and consequently, they will
be dimmer than that they were initially. But the story of 5R is different. It will shine
brighter. This is due to the fact that 5R has lost its parallel companion 4334 RRR += , and itdoes not need to share the current with its parallel neighbors anymore. Initially one half
of the main current 1I was passing through 5R and the other half was passing through 3R
as well as 4R . Now competition is gone an all current passes through the resistor 5R and it
is therefore brighter than it was initially before switching off the current. [0/2]
Calculation:.
Notice: A major part of the calculations give below is provided above. In order to get a
complete picture of the situation, they are repeated again:
With the switch S open, the equivalent resistance of the system is521
RRRReq
++=
==++=++= 0.250.250.100.1000.5521 RRRReq [0/1]
The current through the light bulbs is:
AR
VIIII
eq
newnewnew60.1
0.25
00.40521 ===== [0/1]
The power dissipated in each resistor (light bulb) is:
( ) WIRP newnew 8.1260.100.522
111 === , ( ) WIRP newnew 6.2560.10.1022
222 ===
( ) WIRP newnew 6.2560.10.10 22555 === [0/1/]
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Therefore the light bulb = 0.102R and = 0.105R shine twice as bright as
= 00.51R .
To prove our claim completely we may first calculate the equivalent resistance of the
whole circuit1
= 00.53R and = 00.54R are in series. Their equivalent resistance is
=+= 0.1000.500.534R
= 0.1034R and = 0.105R are in parallel. Their equivalent resistance is
===+=+= 00.52
0.10
0.10
2
0.10
1
0.10
1111345
534345
RRRR
= 00.51R , = 0.102R and = 00.5345R are in series. Their equivalent
resistance is the equivalent resistance of the circuit:
=++=++= 0.200.5100.5321 RRRReq
The current through the battery is AR
VI
eq
00.20.20
0.40===
This is the same current that passes through = 00.51R and = 0.102R .
Therefore the current read by the Ammeters 1A and 2A are: AIII 00.221 === .
Power dissipated in the resistors = 00.51R and = 0.102R are:
( ) WIRP 0.200.200.5 22111 === and ( ) WIRP 0.400.20.1022
222 ===
The current AIII 00.221 === is divided to two equal amounts and
AAI
I 00.12
00.2
25 === passes trough = 0.105R and the rest, i.e.
AI 00.13 = will pass through = 00.53R and = 00.54R . This is due to the
fact that, as illustrated above, the equivalent resistance of the combination
=+= 0.1000.500.534R is identical to = 0.105R and therefore the incoming
current due to the symmetry of the problem is divided to two equal amount.
Power dissipated in the resistors = 0.105R is: ( ) WIRP 0.1000.10.1022
555 ===
Conclusion:
nowdimmer0.20
8.12111
1
1RPP
WP
WPnew
new
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2. An object can not have a charge of [1/0]
i. C19106.1 .
ii. C10102.3 .
iii. C29
106.1
iv. nC4.6 .
Answer: Alternative iii. An object can not have charge of C29106.1 .
Why? Explain: [0/1]
Suggested answer:
It is smaller than the charge of electron. Charge of electron is the smallest known
elementary charge. The other charges are integer multiplication of the elementary charge
Ce19106.1 = . [0/1]
3. Calculate the electric potential needed to accelerate an electron to sm/100.5 7 . An
electron has a negative charge of C19106.1 and the mass kgme311011.9 = .
Ignore the relativistic effect.
i. MV1.7 .
ii. kV1.7 .
iii. V1.7 .
iv. kJ1.7 .
Answer: Alternative ii, i.e.: kVV 1.7= . [1/0]
Show the details of your calculations: [1/0]
Data: Ce19106.1 = ; kgme
311011.9 = ; smv /100.5 7= ; ?=V
22
2 2
1
2
1
2
1 mvQ
VmvQVmvE
QVEW
KE
PE
==
=
==
( )( )
kVVmvQ
V 1.77117106.12
100.51011.9
2
119
27312 =
==
Answer: kV1.7=V [1/0]
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4. Which circuit diagram shows voltmeter4. Which circuit diagram shows voltmeter V and ammeter A correctly positioned to
measure the current through the resistor 1R , and the potential difference between two
terminals of it? [1/0]
a b
c d
= 0.102R
1A
1V
=
00
.5
1
R
= 0.102
R
1V
=
00
.5
1
R
VV 0.40=
VV 0.40=
= 0.102R
1V
=
00
.5
1
RVV 0.40=
= 0.102
R
1V
= 00.51
R
VV 0.40=
1A
1A
1A
Why? Explain. What do the other alternatives voltmeter and ammeter shows and why?
[0/2]
Answer: Alternative a. Voltmeter must be connected in parallel, and ammeter must be
connected in series with the resistor 1R .
Alternative b: In this case the voltmeter is connected in series with the resistor. Due to
the fact that internal resistor of a decent voltmeter is in general very large very small
current passes through the circuit and none of the measuring devices show anything. In
case of a very defective voltmeter with a smaller internal resistance, the ammeter will
break down. The current will rush through the ammeter (which in general has a very
small resistance) causing its fuse to melt. [0/1]
Alternative c: In this case the voltmeter is connected in parallel with the resistor andreads a correct value of the potential difference between two terminals of the resistor
1R . But the ammeter is placed improperly and reads only an extremely small amount of
the current that passes through the voltmeter.
Alternative d: In this case the ammeter is placed properly and reads the current passing
through the resistor 1R . But the voltmeter is placed incorrectly and reads instead the
potential difference between two terminals of the resistor 2R . [0/1]
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5. Three negative collinear point charges of equal charge Q are placed on a line as
illustrated. The charge A repels charge B by N0.3 .
a. Find the direction and magnitude of the force that charge C applies on the chargeB . [1/2]
b. Find the magnitude and direction of the resultant force on the charge B . [2/0]
c. Find the magnitude and direction of the electric field at the pointD at cmrAD 0.1=
to the right of the charge A . [1/2]
d. Find the electric potential at the point D . [2/0]
e. Find the total potential energy of the system. [1/1]
NFAB 0.3=
A
B CD
cmrAD 0.1=
cmrAB 0.2= cmrBC 0.1=
Suggested solutions: Data: NFAB 0.3= , cmrAB 0.2= , cmrBC 0.1= ;
QQQQ CBA ===
a. Answer: The force that charge C applies on the charge B is
NFBC 0.12= to the left as illustrated below. [1/0]
NQ
kFAB 0.3)02.0( 2
2
== NQ
kQ
kFBC 0.120.34)02.0(
4)01.0(
2
2
2
2
==
== [0/2]
Alternative method: We may calculate the magnitude of each chargeusing the fact that NFAB 0.3= , cmrAB 0.2= , cmrBC 0.1= :
NNQ
FAB6
2
29 100.30.3
)02.0(109 === 219
9
262
103
4
109
)02.0(100.3CQ
=
=
CQ1910
3
4 = CQ 101065.3 = [0/2] CQ 10107.3
Therefore: NQkFBC 62
19
92
2
100.12)02.0(
10
3
4
109)01.0(
=
==
NFBC 0.12= [1/0]
NFAB 0.3=
A
B C
cmrAB 0.2= cmrBC 0.1=
NFBC 0.12=
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b. The resultant force on the charge B is NFB 0.9= towards the left.
NFFF BABCB 0.90.30.12 === to the left. [1/0]
A
B C
cmrAB 0.2= cmrBC 0.1=
NFB 0.9=[1/0]
c. Answer: The resultant electric field at D: CkNED /2.8 Towards the
right
A B
C
D
cmrAD 0.1=
cmrAB 0.2= cmrBC 0.1=
CkNED /2.8
To find the magnitude and direction of the electric field at the pointD ,we need to find how large each charge are. This may be achieved byusing the information about the force between the charge A and chargeB:2
nCCCQ
QNQ
FAB
5.361065.3103
4
103
4
109
104100.3100.3
)02.0(109
1019
19
9
4626
2
29
==
=
===
[0/1]
The direction of electric field of a negative charge is a vector towards it:
CkNCNEAD /86.32/86332)01.0(
103
4
1092
19
9 =
=
Towards the left (A)
CkNCNEBD
/86.32/86332)01.0(
103
4
1092
19
9 =
=
Towards the right (B)
CkNCNECD
/22.8/2168)02.0(
103
4
1092
19
9 =
=
Towards the right (C)
The resultant electric field CDBDADD EEEErrrr
++= is:
Answer: CkNED /2.8 Towards the right [1/1]
2 This is done above in 1a as an alternative solution. It is repeated here for the reason of clarity.
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d. Answer: The electric potential at the point D is VV 820D [1/0]
The electric potential is an scalar quantity which is just the algebraicsum of the electric potentials of each charge at the point D, i.e.:
VVVVVCDBDADD
02.0
103
4
10901.0
103
4
10901.0
103
4
109
19
9
19
9
19
9
=++= [1/0]
VVVVD 8206.821
02.0
103
4
1045
19
9 =
=
e. Answer: The total potential energy of the system is J710 EPE 2.2 =
[1/0]
01.002.003.0
222
QkQkQkEPE ++=
JkQkQEPE819922 1022
06.0
1110
3
4109
06.0
11
06.0
632 ===++
= [0/1]
6. A CQ 0.101 += point charge is
located at the origin of a coordinate
system. A second point charge of
CQ 0.52
+= is placed on the x-axis
at mmx 0.4= as illustrated in the
figure below.
a. Calculate the electric field at apoint A on y-axis at position
( )mm0. . [1/3]3,0
b. Where on the x-axis, can a third test charge CQ 0.13 += be placed so that it
experience no force? [2/4/]
c. Calculate the electric potential at this point. [3/0]Suggested solution:
Data: CQ 0.101 += , CQ 0.52 += at mmx 0.4= ; A ( )mm0.3,0 ; CQ 0.13 +=
a) Answer: The electric field at a point A on x-axis at position ( )mm0. is3,0
C
NEA
10101.1 at 97 with the positive x-axis. [1/0]
( )( ) C
N
r
QkEA
10
23
69
2
1
11 100.1
100.3
100.10109 =
==
away from the charge.C
NEA
10
1 100.1 =
mmx 0.4= , mmy 0.3= mmyxr 0.543 22222 =+=+= mr3
2 100.5=
CQ 0.52 +=
x
y
CQ 0.101 +=mmx 0.4=
( )mm0.3,0
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( )( ) C
N
r
QkEA
9
23
69
2
2
22 108.1
100.5
100.5109 =
==
;C
NEA
9
2 108.1 = away from the charge.
C
NEEE AAAx
5
4108.1
5
4cos
9
222 === [0/1]C
NEAx
9
2 1044.1 =
C
NEEE AAAy
5
3108.1
5
3sin
9
222 === C
NEAy
9
2 1008.1 =
C
N
C
NEEE AyAyAy
99
21 1008.11010 +=+= [0/1]C
NEAy
91008.11 =
Electric Field x-component y-component
C
NEA
9
1 1010= 01 =AxE
C
NEAy
9
1 1010=
C
NEA
9
2 108.1 = C
NEAx
9
2 1044.1 = C
NEAy
9
2 1008.1 =
C
NEA
91008.11 =
C
NEAx
91044.1 =
C
NEAy
91008.11 =
( ) ( )22AyAxA EEE +=
( ) ( ) CN
EA2929
1008.111044.1 +=
C
NEA
10101.1
=
Ax
Ay
E
E1
tan
=
= 4.97
1044.1
1008.11tan
9
91
Answer: 97 [0/1]
b) Answer: At ( )0,34.2 mm the electric field due to CQ 0.101 += , andCQ 0.52 += is zero and therefore no point charge at the point will experience
any force due to these charges. [1/0]
If the electric field at a point is zero, the electric force on a given charge at the point must
be zero. Therefore, we may look for a point on the x-axis so that electric field at the point
is zero. Due to the fact that the charges involved, i.e. CQ 0.101 = , CQ 0.52 += are
both positive, the electric field is zero somewhere between them, i.e. at a point where its
x-coordinate is mmx 0.30
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Therefore, the magnitude of the electric field due to the charge CQ 0.101 = must be
equal to the magnitude of the electric field of the charge CQ 0.52 += at ( )0,x .
( ) ( ) 12
004.0100.5
100.10
004.0
100.5100.102
2
6
6
2
6
2
6
2
2
2
2
1
1 =
=
=
/=/
x
x
xxr
Qk
r
Qk [0/1]
( )2
004.0=
x
x ( ) xxx 22004.0004.02 == 2004.02 =+ xx [0/1]
) 2004.021 =+x ( )
mmmx 34.2002343.021
2004.0=
+= [0/1/]
Check:( ) ( )
82.182.1004.0
100.5
002343.0
100.102
6
2
6
=
=
xQED
c) Answer: The electric potential at ( )0,34.2 mm is: MVV . V 6651056.6 7 ==
x
Qk
x
QkV
+=
004.0
21 [0/1]
( ) ( )VV
002343.0004.0
100.5109
002343.0
100.10109
69
69
+
=
[0/1]
VV71056.6 = [0/1]
Note that even though the electric field at the point is zero, the electricpotential of the point is quite large and is 65.6 MV.
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7. An electron moving at one percent of the speed of light to the left enters a uniformelectric field region where the electric field is horizontal. If the electron is to be brought
to rest in the space of cm0.2 calculate the magnitude and direction of the electric
field. Ignore relativistic effect. [2/4/]
Suggested solutions: Answer: CkNE 3.1 towards the left. [1/0]
Data: cmx 0.2= , smcv /1031.0 7== , Ce 19106.1 = , kgme31
1011.9 = .
smcv /1031.0 7==Er
kgme311011.9 =
Ce 19106.1 =
EeEQFrrr
==
[0/1]
The direction of the electric field must also be towards left, i.e. in thesame direction as the direction of the motion of the electron. This is dueto the fact that the charge of electron is negative and therefore to stopthen we must apply the electric in the same direction as its motion in
order to get a force in the opposite direction: EeEQFrrr
== [1/1]
According to Newtons second law of motion: amFr
r
= .
On the other hand the equation of motion of the electron may be written
as: 2022 vvax = .
Therefore, combining these three laws results in:
xe
mvE
vx
m
eEvax
m
eEaeEma
vvax
amF
EeEQF
=
=
=
==
=
=
==
22022
2
0
2
0
2
02
0
2
r
r
rrr
[0/2]
( )( ) C
kN
C
N
xe
mvE 3.11281
102106.12
1031011.9
2 219
26312
0 =
=
=
[0/1/]
Alternative method:
x
vvavvax
22
2
0
22
0
2 == ( ) 214
2
26
/1025.2102.2
1030sma =
=
[1/1]
NNmaF161431 100498.21025.21011.9 === [1/1]
CN
CN
QFE 1281
106.1100498.2
19
16
===
Answer:CkNE 3.1 to the left. [0/2/]
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