Suggested+Solutions+Test+FyANVC07+Ch+16 19+Electricity

8/3/2019 Suggested+Solutions+Test+FyANVC07+Ch+16 19+Electricity

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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:

Enjoy it!

[email protected] for sale.Free to use for educational purposes. 1


2/14



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:


3/14



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]


4/14



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]


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]


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 =


5/14


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/]



6/14



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


7/14



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]


8/14



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]


9/14


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


NFBC 0.12=



10/14



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


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=


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.


11/14



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


12/14



( )( ) 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


13/14



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.


14/14


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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