solutions+g level+complementary++fybch16 19,+20nvco08+electricity+and+magnetism+published

8/3/2019 Solutions+G Level+Complementary++FyBCh16 19,+20NVCO08+Electricity+and+Magnetism+Published

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Solution G-level Complementary Test FyBCh16-19-20NVC08 Electricity & Magnetic Field NV-College

Physics B: FyBNVC08G-level Complementary Test on Electricity and Magnetism Ch 16-19, & 20

Instructions:

The Test Warning! There are more than one version of the test.

At the end of each problem a maximum point which one may get for a correct solution

of the problem is given.

Tools Approved formula sheets, ruler, and graphic calculator. You may use your green

personalized summary-formula-booklet which has your name on it. This should be

submitted along with the test.

Time: 13:10-16:00

Solutions to the problems must be answered on the original test paper.

Limits: This tests gives at maximum 22 points. It is only a complementary test to the G-testgiven earlier this week. The passing scores or qualification for a higher grade is judged

individually based on the student’s performance in the other two tests.

Please answer in this paper and as clear as possible. Do not forget the UNITS!Enjoy it! Behzad

Name:

16-20 E C C C, V C RC C, Q B, I B, Q Q, E Q, V

P 1 2 3 4a 4b 4c 4d 5 6 7a 7b Sum

G 1 1 1 1 2 2 2 2 2 2 2 18

G

16-20 B, Q B, Q Ch 16-20 Grade

P 8a 8a Total

G 2 2 22

G

Information regarding G-Test:

Teacher’s Comments

Student’s Comments

© [email protected] ☺Free to use for educational purposes. Not for sale. 1/7





1. In the Bohr’s model of the hydrogen atom, an electron circles a proton at a distance

m1110− . Find the electric field that acts on the electron. Show your calculations in

the space available below. [1/0]

1. In the Bohr’s model of the hydrogen atom, an electron circles a proton at a distance

m1110− . Find the electric field that acts on the electron. Show your calculations in


3.5 ×3.5 ×

Solutions Answer: m

V

E 11

101.5 ×=

( ) m

V

r

Qk E

1122199

22

199

211

199

2101.510513.0

1009.28

106.1109

103.5

106.1109 ×=×=

×

×⋅×=

×

×⋅×== +−

−

−

−

−

2. Ten identical capacitors of capacitance F 0. are connected in series. Find the

equivalent capacitance of the combination. Show your calculations in the space

available below. [1/0]

20

Solutions Answer: F eqC μ 0.2=

F F

C C

C C C C C C C

eqeq

eqeq

μ 0.210

.20

10

101...

1111

321

==⇔=⇔=⇔+++=

3. We triple the distance between the plates of a parallel plate capacitor of capacitance C .

Find the capacitance of the new capacitor. Show your calculations in the space available

below.

Answer: 32

C

C =

32

1

3002

C

d

A

d

AC =⋅== ε ε





4. A parallel plate capacitor, a resistor and a battery are connected as illustrated in the

figure below. The two parallel plates of the capacitor are

4. A parallel plate capacitor, a resistor and a battery are connected as illustrated in the

figure below. The two parallel plates of the capacitor are mm500.0 apart. The plates

are connected through a Ωk .100 resistor to a battery. When the plates are fully

charged, the electric field between the plates is mkV / ..200

a) Find the electric potential difference between the plates. Show your calculations in


Suggested solution: Answer: kV 100.0V =

Data: mmmd 4100.550.0 −×== , mV mkV E /10.200/.200

3×==

d E V ⋅= ⇔ kV V d E V 100.0.1001000.510.200 43 ==×××=⋅= −

b) If the parallel capacitor has an area of 200 and a plate separation of .4 cm

mm500.0 . Find its capacitance . [2/0]

Suggested Solutions: Answer: pF C 08.7=

( ) pF F d

AC 08.71008.7

105

1041085.8 12

4

412

0 =×=

×

×⋅×== −

−

−−ε [2]

c) Find the time constant of the circuit, and explain its significance. Show your

calculations in the space available below. [2/0]

Suggested solution: Answer: ns.708=τ

Data: Ω×=Ω= 310.100.100 k R F 121008.7 −×=; C

RC =τ ⇔ ( )( ) nssF .70810.7081008.710.100 9123 =×=×Ω×= −−τ

ns.708=τ [2]

d) Find total charge on the positive plate of the capacitor, when the capacitor is fullycharged. Show your calculations in the space available below. [2/0]

Suggested solution:Answer: nC 708.0Q =

Data: ; pF F C 08.71008.7 12 =×= −V kV 10010.0V ==

V C Q ⋅= ⇔ ( )( ) nC C F V Q 708.01008.71008.7100 1012 =×=×= −−

nC C Q 708.01008.7 10 =×= − [1]

R





5. A straight wire carrying a5. A straight wire carrying a A0.25 current has a length cm0.20 between the poles of a

magnet at an angle °.53 as illustrated in the figure

below. The magnetic field is fairly uniform at T 00.4 .

The magnitude and direction of the force on the wire.

Find the magnitude and the direction of the magnetic

force on the wire. Show your calculations in the spaceavailable below. [2/0]

Suggested solution: Answer: N F 0.16=

Data: A I 0.25= ,

mcm 200.00.20 ==l , °= .53θ , T B 00.4=

Problem: ?=F

B I F r

lrr

×= ⇔ θ sin B I F l= ⇔ N F 0.16127sin00.4200.00.25 =×××= [1/0]

Answer: The total force on the wire is N F 0.16= out of the page of

the paper towards the reader using the right hand rule. [1/0]

6. A charged particle moves in a circular path of radius r in a magnetic field B . The

velocity of the charged particle is tripled. Compare the radius of the circular path of the

charged particle. Explain. Show your calculations in the space available below. [2/0]

Suggested solution: Answer: BQ

vmr

⋅⋅=

BQ

vmr

r

vm BvQ

r

vmF

BvQF

⋅⋅=⇔⋅=⋅/⋅⇔

⎪⎩

⎪⎨

⎧

⋅=

⋅⋅= /2

2

Answer: According to BQ

vmr

⋅⋅= , if the velocity of the charge particle is

tripled, radius of its orbit must also be tripled. [2/0]

A0.25

T 00.4

°.53





7. A charge C 7. A charge C 0. charge is placed cm0.16 from an identical C 50+ μ 0.50+ charge.

a) What is the electric field at the point midway between the charges? [2/0]

b) What is the electric potential at the point midway between the charges? [2/0]

Suggested solution:Data:

C QQQ 0.5021 +=≡=C Q 0.103 +=

a) The electric field at the midway between the charges is zero, thisis due to the fact that the electric field due to C Q 0.501 += at the

midway is to the right, and that of C Q 0.502 += is to the left. The

magnitude of the electric field due to each charge midwaybetween the charges is identical. But 1 E

ris to the right, while 2 E

ris

to the left. Therefore, the vector sum of them must be zero, i.e.

021 =+= E E E rrr

. [2/0]

b) The electric potential at the midway is:r

Qk

r

Qk

r

Qk V M 221 =+=

mcmcm

r 080.00.82

0.16===

V V M

66

9 10125.1

08.0

105´1092 ×=

×⋅××=

−

[2/0]

C Q 0.501 +=C Q μ 0.502 +=

cm0.16





8. An electron is shot into the region between the poles of a magnet. The initial velocity of

the electron is

8. An electron is shot into the region between the poles of a magnet. The initial velocity of

the electron is sm /1000.3 6×

perpendicular to the direction of the

magnetic field. The magnetic field is

uniform and its magnitude is T 500.0 as

illustrated in the figure.

Data: C eQ 191060.1 −×−=−= ,

kgme

311011.9 −×=

a) Find the direction and magnitude of the magnetic force on the electron.

Explain. Show your calculations in the space available below.

[2/0]

Suggested solution: Answer: N F 131040.2 −×=

Data: C eQ 191060.1 −×−=−= , sm /1000.3 6×=v , T B 500.0=

The electron will experience a magnetic force which is into thepage of the paper and away from the reader of these words.Therefore, the electron will be deflected into the page and awayfrom the reader. The magnitude of the force is:

BvQF rrr

×⋅= ⇔

N BvQF 13619 1040.2500.01000.31060.190sin −− ×=××××=⋅⋅⋅=

b) Describe the path of the electron in the uniform magnetic field, and find thenecessary information regarding the path. [2/0]

Suggested solution: Answer: The electron will experience acentripetal acceleration, and as a result it will rotate in a circular

path of radius m BQ

vmr μ 2.34=

⋅

⋅= , normal to the page of the

paper

Data: C Qe

191060.1−×−= , smv /1000.3 6×= , T B 500.0= ,

kgme

311011.9 −×=

BvQF

rrr

×⋅= ⇔

BQ

vmr vmr BQ

r

vm BvQ

r

vmF

BvQF

⋅

⋅=⇔⋅=⋅⋅⇔⋅=⋅⋅⇔

⎪⎩

⎪⎨

⎧

⋅=

°⋅⋅⋅=2

2

90sin

mm BQ

vmr μ 2.341016.34

500.01060.1

1000.31011.9 6

19

631

=×=××

×××=

⋅

⋅= −

−

−

The electron will experience a centripetal acceleration, and as a

result it will rotate in a circular path of radius m BQ

vmr μ 2.34=

⋅

⋅= ,

normal to the page of the paper

C eQ 19106.1 −×−=−= smv /100.3 6⋅=

T B 500.0=

solutions+g level+complementary++fybch16 19,+20nvco08+electricity+and+magnetism+published

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