Physics 2 (MT) - Quiz 1a - Solution

7/31/2019 Physics 2 (MT) - Quiz 1a - Solution

1/5

Quiz 1 Solution

Swiss German University (SGU)SGU Campus - BSD EduTownBumi Serpong Damai 15339Island of Java Indonesia

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Subject Physics 2 Score:

Lecturer Alexander A. Iskandar, Ph.D.Program MechatronicsFaculty EngineeringSemester 1Date 15 March 2012

Time 50 minutes

Topic

Name :

Answer in the space provided.

Use the last page for scrap paper, but do not detached this page.

If needed, use the following values of gravitational acceleration g = 9.8 m/s2, vacuum permittivity

constant0 = 8.85 10-12 C2/N m2(or k = (40)

-1 = 9 109 Nm2/C2) and the electron charge e = 1.6

10-19 C.

1. Point charges of +6.0 C and4.0 C are placed on an x-axis, atx = 8.0 m andx = 16 m,

respectively. What charge must be placed at x = 24 m so that any charge placed at the

origin would experience no electric force. (25 points)

Let the charge on the origin is Q (positive). Then the charge q1 = +6.0 C will repel it and

the charge q2 =4.0 C will attract it. Because the problem is a 1D problem, we can usesign to denote direction, hence the force on charge Q from the three other charges is

0

2416

104

8

1062

3

2

6

2

6

321

Qq

kQ

kQ

k

FFFFQ

Solving for q3 we have

Cq66

22

26

22

2

3 1045102

4

1

6310

16

4

8

624


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Quiz 1 Solution


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2. Two tiny conducting balls of identical mass m and identical charge q hanging from

nonconducting threads of lengthL are in equilibrium. Assume that is small such that tan can be replaced with its approximate equal sin . (a) Draw all forces that act on each of

the ball. (b) Determinex in terms ofq, m,L, g and . (25 points)

Using Newtons Law, we can determine x. First draw the forces

that act on one of the charges

2

2

04

1sin

cos

x

qT

mgT

Combining the two equation of motion above

and using small angle approximation (tan =

sin =x/(2L)) we have

2

2

0

2

2

0 4

1

2sin

4

1tan

x

q

L

xmgmg

x

qmg

or,

mg

Lqx

0

2

3

2

Fe

T

mg


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Quiz 1 Solution


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3. Four charges (q1 = q2 = +5e, q3 = +12e and q4 =3e) are placed as shown in the figure, the

distance dis 5 m. Determine the magnitude and direction of the net electric field at pointP due to all the charges. (25 points)

By symmetry the charges q1 and q2 do not contribute to the electric

field at point P (they have equal magnitude but opposite direction),

hence we need to calculate only contributions from q3 and q4.

32

0

3

12

4

1qfromwayPat

d

eE

and,

42

0

443

41 qtowardsPat

deE

Hence, the total field at P is equal to

312

19

9

2

0

43

/6481025)4(

106.110945

4

45

4

1

qfromawayPatCN

d

eEEEP


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Quiz 1 Solution


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4. A non-conducting rod with a uniformly distributed charge +Q. The rod forms a half-circle

with radius R. What is the electric field at point P (the center of the circle). Hint : aninfinitesimal arc of angle dhas a length ofds =Rd. (25 points)

Consider an infinitesimal element on the half-circle, the vertical

components of the field will cancels, hence we need only to calculate

the horizontal components,

0

2

0

sin4

1

R

dqE

where,

R

QRddsdq

Then,

2

0

2

0

2

00

2

0

0

2

0

22)2(

4cos

4

sin4

R

Q

RR

R

R

R

dR

RE

with a direction to the right.


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Quiz 1 Solution


Tel. +62 21 3045 0045

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Physics 2 (MT) - Quiz 1a - Solution