solution+fya16 19nvc09+electricity

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Suggested solutions Test FyA16_19NVCO09 NV-College

© [email protected] ☺Free to use for educational purposes. Not for sale. page 1 of 21

Test on Electricity: FyA16_19NVCO09Directions

Warning: There are more than one version of the test.

Test time 8:10-11:30.

Resources Calculators, and “Formulas for the National Test in Mathematics Courses A &

B”, and the formula sheet for Physics A, FYANV-College. You may also use

your personalized blue-booklet. The booklet must have your name and no

calculations or solution to any problem are supposed to be on it .

The test: For most of problems short answers are not enough. They require:

• that you write down what you do, that you explain your train of thought,

• that you, when necessary, draw figures.

• When you solve problems graphically/numerically please indicate how

you have used your resources.

Problems 9, 10, 18, and 19 are larger problems which may take up to 90minutes to solve completely. These Problems are of the greatest importance

for the highest grade, MVG. It is important that you try to solve them. A

description of what I will consider when evaluating your work, is attached to the

problem. 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.

Score The maximum score is 58 points, 26 of them VG points and 5MVG¤ problems.

Mark levels The maximum number of points you can receive for each solution is indicated

after each problem. If a problem can give 2 “Pass”-points and 1 “Pass with

distinction”- point this is written (2/1). Some problems are marked with ¤ ,which means that they more than other problems offer opportunities to show

knowledge that can be related to the criteria for Pass with Special Distinction in

Assessment Criteria 2000.

Maximum score: 58: 32G/26VG/5MVG¤

G: 19 points

VG: 38 points, at least 9 VG points

MVG: 40 points, at least 19 VG points and 3 MVG-quality works (all

categories M1to M5 must be represented in the solutions.)

Enjoy it! BehzadSubject ES DC ES ES ES ES ES ES ES ES

Problem 1 2 3 4 5 6 7 8 9 10 SumG 2 2 3 2 2 3 3 1 1 1 20

VG 1 6 4 11

MVG ¤ ¤

G

VG

MVG

Subject DC DC ES DC DC DC ES ES ES ES 59

Problem 11 12 13 14 15 16 17 18a 18b 19 Sum Total

G 3 2 2 2 2 0 1 0 0 0 12

VG 3 1 4 4 4 16

MVG ¤ ¤ ¤

G

VG

MVG

32

27





1. An object can not have a charge of 1. An object can not have a charge of

i. i. C 19106.1 −× .

ii. C 19104.6 −× .

iii. C 19100.1 −× .

iv. C 2.3 .

Answer: Alternative: ______ [1/0]

Why? Explain. [1/0]

Suggested answer:

Answer: Alternative iii. An object can not have charge of C 19100.1 −× .

It is smaller than the charge of electron. Charge of electron is thesmallest known elementary charge. The other charges are integer

multiplication of the elementary charge C e19106.1 −×= .

2. What is total electrical energy used by a W 2000 hair dryer operating for min25 ?

a. J 610 .0.3 ×

b. J 4100.5 × .

c. J 80 .

d. J 33.1 .


Why? Show your calculations. [1/0]

Suggested answer: Answer: Alternative a. J 6100.3 ×=W [1/0]

t PW ⋅= , J W 6100.360252000 ×=××= [1/0]

3. A cloud is at a potential of MV 0.9 relative to the ground. A charge of C 45 is

transferred in a lightning stroke between the cloud and the ground. Find the energy

dissipated. [3/0]

Suggested solution:

Answer: In the lightning MJ E P

410= electrical

energy is transferred to heat and other forms of energies.

Data: C Q 45= , V MV V 6100.90.9 ⋅==

Den totala energin är bevarad: 21 E E =

MJ MJ J V Q E P 41040510405100.945 66 ≈=×=×⋅=⋅=




4. Which diagram best represents the electrostatic force between an alpha particle with a

charge of 2+ elementary charges and a positively charged nucleus as a function of

their distance of separation?

A: B:

Distance

E l e c t r o s t a t i c

F o r c e

Distance


F o r c e

C: D:

Distance


F o r c e

Distance


F o r c e



Suggested answer: Answer: Alternative A. [1/0]

The force is inversely proportional to the square of the distance

between the charges, i.e.:2

21

r

QQk F

⋅= . Both particles are positive so

2

1

r F ∝ . [1/0]






Base your answers to the problems 5-10 on the figure below which illustrates two point

charges Q and placed at a distance

Base your answers to the problems 5-10 on the figure below which illustrates two point

charges Q and placed at a distancemC 50.121 −= mC 50.121 −= mC Q 500.22 = mmd 00.20mC Q 500.22 = mmd 00.20= from

each other. The charges are pinned to the page of the paper, and therefore are stationary.

5. The total potential energy of the system is

a. MJ 06. .14

b. MJ 06. .14−

c. kJ 06. .14−

d. kJ 06. 14

e. J 06. 14−

f. None above. It is ___________



Suggested solutions Answer: Alternative b: MJ E PE 06.14−= . [1/0]

Data: , ,C Q3

1 1050.12 −×−= C Q3

1 1050.2 −×= mr 31000.20 −×=

The potential energy of the system is

( )( )( )

MJ J r

QQk E

PE 06.1410406.1

1000.20

1050.21050.12109 7

3

33921 −=×−=

×

××−⋅×==

−

−−

Answer: MJ E PE

06.14−= [1/0]

6. If the point C is located at the middle of the line joining the two charges, i.e. it is at

mm00. from both charges, the work done to bring the charge mC Q 0.

from infinity to the point C is:

10 100−=

a. MJ 0. 900−

b. MJ 0. 900

c. kJ 00.90

d. kJ 00.90 to the left.−

e. J 00.90 to the right.




A B C D G

mC Q 50.121 −= mC Q 500.22 =





Suggested solutions Answer: Alternative b: MJ W 0.900= . [1/0]

The work done by an external force to bring the charge

from infinity to the point C is equal to the change in the potential

energy of the system:

mC Q 0.100−=

( )[ ]( )

J r

QQk

r

QQk E W

PE 3

333921

1000.10

1050.21050.12100.100109

−

−−−

×

×+×−×−⋅×=+=Δ= [1/0]

Answer: W [1/0] MJ J 0.90010000.9 8 =×=

7. If a unit positive charge is moved from the point B to the point D , the electric

potential energy of the system

a. increases

b. decreases

c. first increases, thendecreases

d. first decreases, then increases

e. remains constant through out the journey.


Why? Show your calculations. You may assume the point B is at from the

charge , and the point is at from the charge on the line joining the

charges as illustrated above. [2/0]

mm000.5

1Q D mm000.5 2Q

Suggested solutions Answer: Alternative a. [1/0]

The electric potential energy of the system at point B is:

B B

PE r

QQk

r

QQk

d

QQk E

B

2

2

1

121 ++=

( )( )( )

( )( )

[ ]( )

J E BPE 3

39

3

39

3

339

1000.15

1050.21109

1000.5

1050.121109

1000.20

1050.21050.12109

−

−

−

−

−

−−

×

×⋅⋅×+

×

×−⋅⋅×+

×

××−⋅×=

GJ GJ GJ GJ MJ E BPE 00.2100.2101406.000.2106.14 −≈−−=−−=

We may notice that the contribution of the electric potential energy of the

original system of is much smaller than the other two terms and itmay be ignored in the further calculations. Besides, it is the change of thepotential energy we are concerned about, not the total amount of it.

21 &QQ

The electric potential energy of the system at point C is:

[ ]( )

GJ J GJ r

QQk

r

QQk

d

QQk E

C C

PE C 001.9

1000.10

1050.2110901406.0

3

39

2

2

1

121 −≈×

×⋅⋅×+−=++=

−

−

Similarly at point : D GJ r

QQk

r

QQk

d

QQk E

D D

PE D001.3

2

2

1

121 −≈++=

Therefore, the electric potential energy of the system increasescontinuously in moving a unit charge from the point B to the point C

BC D

mC Q 50.121 −= mC Q 500.22 =




8. Moving mC and mC QQ 50.121 −= 500.22 = closer to each other would cause their

electric potential energy to

a. increases

b. decreases

c. first increases, then decreases

d. first decreases, then increases

e. remains constant through out the journey.


Why? Explain. [0/1]

Suggested explanations: Answer: Alternative b: decreases. [1/0]

The potential energy of the system is

r

QQk E PE

21 ⋅= , where

is the distance between them.

mr 31000.20 −×=

r

QQk E PE

21 ⋅= is inversely proportional to r :

if r decrease the magnitude of r

QQk E PE

21 ⋅= will increases, but due to the

fact that charges are of different sign and the potential energy of thesystem is negative, decreasing the distance between the charges willresult in a decrease in the electric potential (i.e. it negatively increases.)

[0/1]

9. Where, if any (other than infinity), on the line joining the charges, may a charge Q be

placed so that the net force on it is zero.

a. A which is a point to the left of the larger charge.

b. B which is a point closer to the larger charge somewhere between the charges.

c. C which is a point at the middle of the line joining charges.

d. D which is a point closer to the smaller charge somewhere between the charges.

e. G which is a point to the right of the smaller charge.

None above.


Why? Explain qualitatively. [0/2]

Calculate exactly the position of the point relative to one of charges. Discuss if the charge

is in equilibrium, and reason if the equilibrium is stable or not. Why?Q

[0/4/M1, M2, M3, M5 ¤]

Suggested solutions Answer: Alternative e: at the point G . [1/0]

If the total electric field at a point is zero, the total electric force on anycharge would be also zero.






The electric field at any point between the charges is nonzero and itsdirection is to the left. This is due to the fact that, in between charges,the electric field due to both charges is to the left. On the other hand,even though both to the left of the larger charge as well as to the rightof the smaller charge, the direction of the electric fields are in the

opposite directions, the total electric field is zero only at a point closerto the smaller charge. [0/1]

The electric field at any point between the charges is nonzero and itsdirection is to the left. This is due to the fact that, in between charges,the electric field due to both charges is to the left. On the other hand,even though both to the left of the larger charge as well as to the rightof the smaller charge, the direction of the electric fields are in the

opposite directions, the total electric field is zero only at a point closerto the smaller charge. [0/1]

At G , the electric field due to the chargeAt G , the electric field due to the charge1 E mC Q 00.1211 E mC Q 00.121 −= is to the

left, and that of is to the right. Due to the fact that is

closer to the smaller charge, the electric field is zero at the point. Theexact position of the charge may be calculated as:

mC Q 000.52 = G

If the point is to the right of G m x mC Q 500.22 = , and the charge of the

particle is assumed Q , using Coulomb’s law as well as Newton’s law:

21 F F =

( )21

2

2

xd

QQk

x

QQk

+

//=

// [0/1]

Using 21 550.12 QmC Q −=−=

( ) ( ) ( )( ) xd x xd x

xd x xd

Q

x

Q

xd

Q

x

Q+=⋅⇔+=⋅⇔

+=⇔

+

/=

/⇔

+−= 55

515 22

22212

2

22

1

2

2

( )( )15

1555

−

=⇔=−⋅⇔=−⋅⇔+=⋅d

xd xd x x xd x

( ) ( )cm xm

m x

d x 618.1618.1

15

02000.0

15≈⇔≈

−=⇔

−= [0/1]

Answer: At cmd

x 618.115=

−= to the right of the smaller charge

mC Q 500.22 = the net force on any charge particle is zero, and therefore,

the charge particle is at equilibrium. [0/1]

The equilibrium is stable. This is due to the fact that the net force to the

left or to the right of the point is non-zero, but towards the equilibrium.Lets assume the charge is a positive unit charge. If the charge is displacedleftward from the point , both forces increases but increases faster

and the net force will drive the positive charge away from the charges,

and towards the equilibrium. On the other hand, if we displaced thepositive unit charge to the right of the equilibrium, both forces on itdecreases, but decreases faster and the total force will be towards

left, i.e. it drives the particle towards the equilibrium point G .

GGF 2

Q

GF 2

If we displace the positive unit charge to the left of the equilibrium point

G, for example if ,cm x 0.1=

d

G

mC Q 500.22 =

x

mC Q 00.121 −=





( ) ( )N

QQk F

11

2

39

22

2 1025.201.0

110500.2100.9

01.0×=

××⋅×==

−

to the right

( ) ( )N

QQk F

11

2

39

22

1 1025.103.0

110500.12100.9

03.0×=

××⋅×==

−

to the left.

Therefore, since the resultant force will be

toward the right, i.e. towards the equilibrium point G.1

1111

2 1025.11025.2 F N N F =×>×=

Similarly, if we displace the positive unit charge to the right of theequilibrium point G, for example if cm x 0.2= ,

( ) ( )N

QQk F

10

2

39

22

2 10625.502.0

110500.2100.9

02.0×=

××⋅×==

−

to the right

( ) ( )

N QQ

k F 10

2

39

22

1 1003.7

04.0

110500.12100.9

04.0

×=××

⋅×==−

to the left

Therefore, since the resultant force will

be toward the left, i.e. towards the equilibrium point G. [0/1]2

1010

1 10625.51003.7 F N N F =×>×=

MVG-Quality The student shows in general the

highest “MVG” quality if he/she:

Formulates and develops the

problem, uses general methods with

problem solving.

Investigates and analytically shows

that the electric field and therefore the

electric force is zero at the point G .

The student must reason why the

equilibrium is stable.

M1

Analyses and interprets the results,

concludes and evaluates if they are

reasonable.

Analyses, interprets, and evaluates

the results using vector properties of

electric field. Qualitatively discusses

why the equilibrium is stable.,

M2

Carries out mathematical proof, or

analyses mathematical-physical

reasoning.

Analytically finds( )15 −

=d

x based

on vector properties of E , and F

M3

The presentation is structured, and

mathematical language is correct. It is

easy to follow the solution.


mathematical-physical language is

correct. It is easy to follow the

solution. Treats electric field as a

vector, uses proper units, SF.

M5





10. Where, if any (other than infinity), on the line joining the charges, the electric

potential is zero?

10. Where, if any (other than infinity), on the line joining the charges, the electric

potential is zero?

a. a. A which is a point to the left of the larger charge.

b. B which is a point closer to the larger charge somewhere between the charges.

c. C which is a point at the middle of the line joining charges.

d. D which is a point closer to the smaller charge somewhere between the charges.

e. G which is a point to the right of the smaller charge.

f. At the point D as well as the point G .

g. At the point A as well as the point B .

h. At the point A as well as the point G .

i. None above.

Answer: Alternative: ______ [1/0]Why? Explain qualitatively, and calculate exactly the position of the point relative to one

of charges. [0/4/M1, M2, M3, M5 ¤]

Suggested solutions

Answer: Alternative f: At the point as well as the point G , the D

electric potential is zero. [1/0]

Due to the fact that charges are of different sign, and the electricpotential is scalar, and therefore the total electric potential is thealgebraic sum of the electric potential due to the charges, at a point

closer to the charge whose magnitude is smaller, the electric potentialmay be equal to zero. If the point G is to the right of m x mCQ 500.22 =

21 V V V +=

( )012 =

++

xd

Qk

x

Qk

( ) xd

Q

x

Q

+−= 12 [0/1]

We may use 21 550.12 QmC Q −=−=

( ) ( ) ( ) xd x xd

Q

x

Q

xd

Q

x

Q

+=⇔

+

⋅−−=⇔

+−=

515 2212

( ) xd x +=

51

xd x +=5

d x x =−5

d x =4

cmmmd

x 5000.000500.04

02000.0

4==== to the right of mC Q 500.22 =

d

G

mC Q 500.22 =

x

mC Q 00.121 −=





Answer: At the point G is cmd

x 5000.04== to the right of ,mC Q 500.22 =

the electric potential is zero. [0/1]

If the point is to the left of D m x mC Q 500.22 =

21 V V V +=

( )012 =

−+

xd

Qk

x

Qk

( ) xd

Q

x

Q

−−= 12 [0/1]

Using 21 550.12 QmC Q −=−=

( ) ( ) ( ) xd x xd

Q

x

Q

xd

Q

x

Q

−

=⇔

−

⋅−−=⇔

−

−=515 2212

( ) xd x −=

51

xd x −=5

d x x =+5

d x =6

cmmmd

x 3333.0003333.06

02000.0

6==== to the left of mC Q 500.22 =

Answer: At the point , D cmd

x 3333.06== to the left of Q mC 500.22 =

on the line joining the charges, the electric potential is zero. [0/1]

d

mC Q 500.22 =

x

D

mC Q 00.121 −=




MVG-Quality The student shows in general the

highest “MVG” quality if he/she:

Formulates and develops the

problem, uses general methods with

problem solving.

Investigates and analytically shows

that the electric potential is zero at

the point G as well as the point

. Find D both points is necessary

for the MVG-point.

M1

Analyses and interprets the results,

concludes and evaluates if they are

reasonable.

Analyses, interprets, and evaluates

the results using scalar properties of

electric potential.

M2

Carries out mathematical proof, or

analyses mathematical-physical

reasoning.

Mathematical derivation of

cmd

x 5000.04== to the right of

mC Q 500.22 = , as well as

that of cmd

x 3333.06== to the

left of mC Q 500.22 = based on the

scalar properties of electric

potential.

M3



easy to follow the solution.



easy to follow the solution. Units, and

significant figures must be correct.

M5






Answer questions 111-15 based on the circuit below where Ω= 0.201 R , ,

and .

Ω= 0.402 R

Ω= 00.43 R

11. The equivalent resistance of the circuit isa. Ω 0.80

b. Ω 0.40

c. Ω 0.20

d. Ω 0.10

e. Ω00.5




Suggested solutions Answer: Alternative d: Ω= 0.10eq R [1/0]

Data: Ω= 0.201 R , Ω= 0.402 R

The 3 resistors , , and are in parallel. Therefore the equivalent

resistance of them is:1 R 2 R 1 R

121 R

0.40

5

0.40

212

0.40

2

0.40

1

0.40

2

0.20

1

0.40

1

0.20

11111

121121

=++

=++=++=++=

R R R R

Ω==⇔= 00.85

0.40

0.40

51121

121

R R

. [1/0]

Similarly, The 2 resistors , and are in parallel Therefore the

equivalent resistance of them is:

3 R 3 R

22 R

Ω=⇔===+= 00.200.2

1

00.4

2211133

33333

R R R R R

. [1/0]

121

R and are in series, their equivalent resistance is33

R

Ω=+=+= 0.1000.200.833121 R R Req Answer: Alternative d: Ω= 0.10eq R

V 25

1 R

2 R

1 R A

1V

3 R

3 R

1 A





12. The Ammeter A shows:

a. A00.5

b. A50.2

c. A25.1

d. A 625.0

e. A3125 .0




Suggested solutions: Answer: Alternative b: A I 50.2= [1/0]

Ohm’s law: A RV I RI V

eq

50.20.100.25 ===⇔= [1/0]

13. The voltmeter 1V shows:

a. V 0.50

b. V 0.25

c. V 0.20

d. V 0.10

e. V 00.5



Suggested solutions: Answer: Alternative c: V V V V 0.2012121 === [1/0]

The 3 resistors , , and are in parallel. Therefore, the potential

difference between their terminal is:1 R 2 R 1 R

V I RV V V 0.2050.20.812112121 =×=⋅=== [1/0]





14. The Ammeter 1 A shows:

a. A 0.2

b. A 0.4

c. A 0.8

d. A 0.16

e. A00.1



Why? Show your calculations.

Suggested solutions:

First method: Answer: Alternative e: A I 00.11 = [1/0]

The 3 resistors , , and are in parallel. Therefore, the potential

difference between their terminal is:1 R 2 R 1 R

V I RV V V 0.2050.20.812112121 =×=⋅===

Ohm’s law: AV

R

V I I RV 00.1

0.20

0.20

1

11111 =

Ω==⇔= [1/0]

Second method: We may solve the problem logically. is divided

to three parts. More current passes through the line with lesser

resistance. Therefore

A50.2

5

2of it, i.e. passes through and A0.1 Ω= 0.201 R

5

1of it, i.e. passes through A5.0 Ω= 0.202 R .

15. The total power dissipated in the resistors is:

a. W 5.62

b. W 0.25

c. W 5.12

d. W 5. 2

e. W 25 .0




Suggested solutions: First method:

Answer: Alternative a: [1/0] W P 5.62=

W IV P 5.620.2550.2 =×== [1/0]





Second method: Using , and A I 50.2= Ω= 0.10eq R

( ) W I RP eq5.625.20.10

22 =⋅==

Third method: Using V , andV 0.25= Ω= 0.10eq R

( )( )

W P R

V P

eq

5.620.10

0.2522

==⇔= :

16. What is the total amount of energy stored in a V 12 , h A ⋅90 car battery when it is

fully charged? [0/3]

Suggested solution: Answer: MJ U 9.3= energy is stored in the fully

charged battery. [0/1]

V QU ⋅= ,t

Q I = , t I Q ⋅=

C s At I Q 55 1024.31024.3360090 ×=⋅×=×=⋅= [0/1]

MJ J J V QU 9.3109.31089.3121024.3 665 =×≈×=××=⋅= [0/1]

MJ U 9.3=

17. In the center of all active stars, including the Sun, the plasma of hydrogen, i.e. protons

are pressed against each other due to enormous gravitational pressure of the great mass

of the star. As these protons, having positive charge of C e19106.1 −×=+ repel each

other more and more as they get closer to each other, the plasma gets hotter and hotter until it passes one million degree, and suddenly fusions of protons take place; fusing

hydrogen to helium. Calculate the electrostatic force between two protons separated

by a distance of m A10100.10.1 −×=° ?

i. N 810 repulsive.3.2 +×

ii. N 810 repulsive.3.2 −×

iii. N 48 repulsive.103.2 −×

iv. N 19 repulsive.103.2 −×



Suggested solutions: Answer: Alternative ii: N F 8103.2 −×= repulsive.

[1/0]

( )( )

N r

QQk F

8

210

2199

2

21 103.2100.1

106.1100.9 −

−

−

×=×

××== [0/1]





18. An α-particle (nucleus of He: 2 protons eQ 2+=α

, 1910 ,60.1 −×=e

kgm271070.6 −×=

α ) is shot from far away with an initial velocity

smv /1050.6 6×=α

directly toward a gold nucleus (79 protons: eQ Au79+= ) .

a) Calculate the voltage necessary to accelerate an α-particle to this velocity

smv /1050.6 6×=α

. [0/4/M1¤M2¤M3¤M5]

b) Calculate the closest distance the α-particle gets to the nucleus. Assume that the gold

nucleus remains stationary. [0/4/M1¤M2¤M3¤M5]

Suggested solution:

Data: eQ 2+=α

, eQ Au79+= , kgm

271070.6 −×=α

, smv /1050.6 6×=α

Problem: ?=V , ?min =r

a) The voltage necessary to accelerate the α-particle to the velocity

smv /105.6 6×=α

may be found using the principle of the

conservation of energy:

e

mv

e

mvV mveV E E E E E E Pf Kf PiKi f i

4222

12

222

00

=×

=⇔=⇔/+=+/⇔=

[0/2/M1¤M2¤M3¤M5]

( ) MV V

e

mvV 442.01042.4

1060.14

1050.61070.6

4

5

19

26272

=×≈××

×××==

−

−

[0/1]

Answer: α-particles must be accelerated in a potential difference of

MV V 442.0= to acquire velocity sm /1050.6 6×=α

v . [0/1]

b) To find the closest distance the most energetic α-particles reach weuse the conservation of energy principle and fact that “the closestdistance” is associated with zero kinetic energy (i.e. α-particle

stops and may be reflected back):

2

21

2

212120 2

2

12

1

mv

QQk

mv

QQk r

r

QQk mv E E E E E Pf Kf Ki f i ==⇔=⇔+/=⇔=

[0/2/M1¤M2¤M3¤M5]

( )

( )

mmv

QQk r closest

13

2627

2199

2

21 1057.2

105.61070.6

1060.17922109

2 −

−

−

×=×××

××××××== [0/1]

Answer: The most energetic α-particles approaching the nucleus of

gold with the velocity smv /105.5 6×=α

may come as close as

mr closest

131057.2 −×≈ . [0/1]

Note that mr closest

131057.2 −×≈ is larger than the “radius” of the

nucleus of the gold which is considered to be of the order of .

This means if the original accelerating potential was larger so that

the alpha-particle would get as close as to the “nucleus” of

the gold, it would be captured by the nucleus of gold, and the

m1410−

m1410−





nuclear strong force would bind it to the other nucleons, 118 would

be converted to an isotope of TI, i.e.: 120

nuclear strong force would bind it to the other nucleons, 118 would

be converted to an isotope of TI, i.e.: 120

Au79

TI 81

19. A point charge of mass gr m 50.2 at the end of an insulating string of length

cm0. is observed to be in equilibrium in a known uniform horizontal electric field,

C kN / , when the pendulum has swung so it is cm00.2 high. If the field

points to the right, as illustrated in the figure below, determine the magnitude and sign

of the point charge. First draw the free-body-diagram. [0/4/M1¤M2¤M3¤M5¤]

19. A point charge of mass gr m 50.2 at the end of an insulating string of length

cm0. is observed to be in equilibrium in a known uniform horizontal electric field,

C kN / , when the pendulum has swung so it is cm00.2 high. If the field

points to the right, as illustrated in the figure below, determine the magnitude and sign

of the point charge. First draw the free-body-diagram. [0/4/M1¤M2¤M3¤M5¤]

=

75

E 0.25=

Au79

TI 81

=

75

E 0.25=

Suggested Solutions:Suggested Solutions:

The charge is negative, because the pendulum swings to the left, i.e. inthe opposite direction of the field.The charge is negative, because the pendulum swings to the left, i.e. inthe opposite direction of the field.

[Free-body-diagram: 0/1/M1¤M2¤M3¤M5][Free-body-diagram: 0/1/M1¤M2¤M3¤M5]

Using the equilibrium condition,we may conclude that the angle

produced by the pendulum andthe vertical line,

Using the equilibrium condition,we may conclude that the angle

produced by the pendulum andthe vertical line,

°≈°=⎟ ⎠

⎞⎜⎝

⎛ = − 3.1326.13

75

73cos 1

α , and

the weight of the object , mg and

the electric force, , may

be written as

QE F E =

α α tantan ⋅=⇔= E

mgQ

mg

QE [0/2/M1¤M2¤M3¤M5]

C C E

mgQ μ α 231.01031.226.13tan

25000

8.90025.0tan 7 =×=°×=⋅= −

The charge is [0/1] C C Q μ 54.0104.5 7 =×−= −

C kN E /0.25=

C kN E /0.25=

QE F E =

mg

α

cm00.2




Aim of the subject

The subject of Physics aims at providing such knowledge and skills which are needed for

further studies in the natural sciences and technology, but also for studies and activities in

other areas. The aim is that pupils should experience the joy and intellectual stimulation,arising from being able to understand and explain phenomena in the surrounding world.

The aim is also to contribute to the pupils' knowledge of the natural sciences so that they can

take part in public debate on issues related to the natural sciences. This covers analysing and

developing their views on questions, which are important for both the individual and society,

such as e.g. energy and environmental issues, as well as ethical issues related to physics,

technology and society.

The subject also aims at providing advanced knowledge of the role of physics in the

development of Man’s world view. Not only has our knowledge of the universe increased ---

Man has moved from the centre of the world to a planet on the fringes of one of manygalaxies in the cosmos --- but our knowledge of microcosms has also increased. In addition,

the subject aims at providing increased understanding that theories and models are human

conceptual constructions, which can be changed in the light of new experience.

Goals to aim for

The school in its teaching of physics should aim to ensure that pupils:

develop their knowledge of the central concepts of physics, quantities and basic models,

develop their ability to speak and write, as well as reflect over the phenomena of physics, its

models and concepts,

develop their ability to quantitatively and qualitatively describe, analyse and interpret the

phenomena and processes of physics in everyday reality, nature, society and vocational life,

develop their ability to propose, plan and carry out experiments to investigate different

phenomena, as well as describe and interpret what is happening when using the concepts and

models of physics,

develop their ability with the help of modern technical aids to compile and analyse data, aswell as simulate the phenomena and processes of physics,

acquire knowledge of the development of the history of ideas concerning physics, and how

this has influenced Man's world view and the development of society,

develop the ability to analyse and evaluate the role of physics in society.

Structure and nature of the subject

The area of physics stretches from the very largest to the very smallest, from theories about

the development of the universe to the properties of the very smallest subatomic particles.

Being able to describe the movement of an object and what causes this movement, to





understand what light is and to study its properties, as well as study electricity and magnetism

have for long been central areas in physics. The concept of energy emerged as a connecting

link between areas which earlier were regarded as different. Primarily because our knowledge

of matter and its properties has increased, new areas of physics have developed. In a broad

sense physics today deals with matter, radiation and different types of interaction.

Characteristic of physics, as well as the other natural science subjects, is that knowledge is

built up in an interaction, on the one hand between experimental observations, and on the

other models and theories. Experiments also play a central role in the school's teaching of

physics. By means of laboratory work, pupils train their skills in planning experiments, using

measuring instruments and analysing data. Setting up hypotheses and carrying out

experiments to investigate phenomena, test or review a model are important elements.

Knowledge is used to discuss and explain phenomena in daily reality, nature and society.

Developments in the computer area have brought about a situation where there is now access

to powerful software for the analysis and stimulation of processes in physics. Such tools are

of great help within the framework of a specific model for discussing the influence of different factors on the outcome of an experiment.

The subject of Physics covers three courses.

Physics A deals with movement, energy and heat, light and electricity as well as the structure

of subatomic particles. The course also provides an orientation to the development of ideas in

physics, as well as the problems connected with the supply of energy. The relationship

between quantities in physics is mainly studied qualitatively, but some mathematical

treatment is covered. The course requires prior knowledge in Mathematics corresponding to

Mathematics A. The course is common to the Natural Science and Technology Programmes.

Physics B deals with the areas of mechanics, electromagnetism, mechanical and

electromagnetic waves, as well as atomic and nuclear physics. The course provides an

orientation to the evolution of the universe. The course includes an in-depth treatment of one

or more areas chosen on the basis of the teachers’ and pupils’ interests. The requirement for a

mathematical treatment of physics in this course is higher than in Physics A. The course

builds on some of the knowledge from Mathematics D. The course is common to the Natural

science branch of the Natural Science Programme.

Physics --- extension provides broader or advanced knowledge in some areas of physics, for

example astrophysics, solid-state physics, particle physics or the historical development of ideas in physics. The course presupposes Physic A. The course is optional.





FY1201 - Physics A

Goals

Goals that pupils should have attained on completion of the course

Pupils should:

be able to participate in planning and carrying out simple experimental investigations, as well

as orally and in writing report and interpret results

be able to reason over quantities in physics, concepts and models, as well as within the

framework of these models carry out simple calculations

GP. be able to describe and analyse some everyday phenomena and processes with the help of concepts and models from physics (GF: General Physics)

A. have an outline knowledge of the structure of the universe (U) and the composition of

matter in smaller sub-particles (A: atom), as well as the fundamental forces linking together

the planetary system (GF: gravitational force), atoms (AF) and atomic nuclei (NF)

CM. have a knowledge of the forces (F) and moments of a force (TQ: Torque), as well as be

able to use these concepts to describe equilibrium states and linear movements (LM):

(M. Classical Mechanics)

L. have a knowledge of light, its reflection and diffraction, as well as some applications in

this area

EC. DC. have a knowledge of electrical fields, electrical voltage and current, as well as

electrical energy and output (EC: Electrostatics, DC: Direct Cureent )

HTP. have a knowledge of heat, temperature and pressure

EN. have a familiarity with principles of energy and its conversion, be familiar with the

concepts of energy quality, and be able to use knowledge about energy to discuss energy

issues in society

HF. have a familiarity with some of the events from the historical development of physics,

and its consequences for society.


http://www3.skolverket.se/ki03/front.aspx?sprak=EN&ar=1011&infotyp=5&skolform=21&id=3053&extraId=






Grading criteria

Criteria for Pass (G)

G1 Pupils use appropriate definitions from physics, quantities, concepts, and models todescribe phenomena and processes in physics.

G2 Pupils participate in and carry out experiments on the basis of instructions.

G3 Pupils carry out calculations concerning problems of a routine nature.

G4 Pupils show through examples how concepts of physics are used when describing

everyday contexts.

G5 Pupils give examples of how a knowledge of physics contributes to a scientific view of

the world.

G6 Pupils present their work and co-operate in interpreting results and formulating

conclusions.

Criteria for Pass with distinction (VG)

VG1 Pupils give an account of the meaning of quantities in physics, concepts and models and

apply this knowledge to interpret and forecast observations in the surrounding world,

and also to carry out calculations.

VG2 Pupils work together over the choice of method and design of laboratory experiments.

VG3 Pupils process and evaluate results on the basis of theories and hypotheses set up.

VG4 Pupils apply concepts from physics and relationships from everyday and scientific

contexts.

VG5 Pupils describe the development of physics, and how this has contributed to the

formation of a scientific view of the world.

Criteria for Pass with special distinction (MVG)

MVG1 Pupils apply scientific ways of working, plan and carry out investigatory tasks, both

theoretically and experimentally, as well as interpret results and evaluate the validity

and reasonableness of their conclusions.

MVG2 Pupils use concepts and models from physics in an analytical and insightful way.

MVG3 Pupils analyse and discuss approaches to problem solving, using knowledge from

different fields of physics.

solution+fya16 19nvc09+electricity

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