COULOMB'S LAW SOLUTIONS

1 a Fe k9Fe K

29rs2KErIIZFe DOUBLES

b Fe k9 9

4 K4 Fe INCREASES BY A FACTOR of 4

c Fe k

qK9Ig Fe DECREASES BY A FACTOR of 4

d Fe kY4 KELI4 Fe INCREASES BY A FACTOR of 4

e Fe k fK919

Fe NO CHANGE

f Fe k92

Ir

16 K

16 Fe INCREASED BY A FACTOR OF 16

2 given Fe KIRI 1 1.00 10 b 1.00 10 Gq _t 1.00 10 GC 9.00 109 2

q 1.00 106 C

9,00 102 71.00M

Fe 9.00 103 N ACCRACTIVE

3 GWEN91 92 6.0 10 6C Fe KYIr 1.0 10 Gm g oo oq

6.0 10 6146.015921.0 10 6 2

Fe 3.2 10 N

4 given Fe _K 9r 0.50M KEFe 5 ON q Kq _qz q

9.00 109

1.2 10 5C 12µL

1ele5 a g TOTAL _l 0kg 9.11 10 31 kg 1 electron

1.756 10 C

1.8 10 C

b given Fe kE 1.756 10 1.756 1g qz 1.756 10 C 9.00 109 c

1.0M 2.776 1032 N

Fe 2.8 10 N REPULSIVE

c given Fg GTI6.67 10 Y

o

M Mz 1.0kg 1 o

1 Om G 67 10 N

Fg MUCH WEAKER

G ga 8OµC 9,3 6.0µC qc tl20µCFa G Fa to

GIVEN

qa 8.0 106 C Fe _k9

q8.0 10 b 6.0 10

6

G9,3 6.0 10 C 9.00 10 o zoo

2

r 0.300 M 4.8 N4.8N LEFT

GIVEN

913 6.0 10 GC Fe k99 c 12.0 10 6C

g.oowoayl6OHO.to O

6

O Goo 2

r 0.600M1.8N1.8 N RIGHT

FneT Fa Fa1.8 4.8

3 ON3.0 N LEFT

7 qB 6OHCf Fc g 12.0

30.0cm Ifaqa t80µC

GIVEN

qa 8.0 106 C Fe _k9

913 6.0 10 GC goa8.010966.0210

6

0.300 2

r 0.300 M 4.8 N4.8N SOUTH

GIVEN

913 6.0 10 GC Fe k99 c 12.0 10 6C

g.oowoayl6OHO.to O

6

O Goo 2

r 0.600 m 1.8N1.8N EAST

Five _Fa FT

Eti

5 IN

o tan Eatan fief210

5 IN Zf EAST OF SOUTH

8 fFek9 k9

BECAUSE THE TWO ELECTRONS WILL REPEL PLACING THEL SECOND ELECTRON DIRECTLY BELOW WILL CAUSE AN

UPWARDS FORCE ON THE FIRST ELECTRON

g meg Fine 1 0

Fy Fe OFe FgKate MegrtMeg

g

r kmI

C9.00x109JCI.GOXIO.to

C9Txo9.8

5 I m

5 Im BELOW

9 a given F Feme 9.11 10 leg me K9erIqe 1.60 10 C z

k9e9nqn9.6 10 9C r mer 2.0 10 Om V J Me

9.00 109111.60151719.61019 1

2.0 10 10 9.11 1031

2.8 106 If

b given Fe Feme 9.11 10 leg me I K9erIqe 1.60 10 C

yzme4i

qn 9.6 10 9C K9eqnr 2.0 10 OM T me4n2r

1 Korean

19llxio.SI 4n2 2.0x1o0

9.00 109 1.60 10 9 9.6 10 19

4.6 10 lbs

10 Fc Feme Keefe

2 K 932

Met

takeFz 0.707 V