Giải Bài Tập Vật Lý Đại Cương 111111

HD gii bi tp Vt L C 2 Email: [email protected] Website: www.caotu28.blogspot.com

Page 1

Bin son: Cao Vn T Trng: H CNTT&TT Thi Nguyn

HNG DN GII BI TP VT L I CNG 2

( Lu : Ti liu cha c thm nh nn vn cn nhng ch cha chnh xc hon ton)

PHN I: IN T HC

Chng I: TRNG TNH IN VT DN TRONG TRNG TNH IN

1. L thuyt. 1.1 S nhim in ca cc vt

+ Nhim in do c xt: hai vt khng nhim in khi c xt vi nhau th c th lm

chng nhim in tri du nhau.

+ Nhim in do tip xc: cho thanh kim loi khng nhim in chm vo qu cu

nhim in th thanh kim loi nhim in cng du vi in tch ca qu cu. a thanh kim loi

ra xa qu cu th thanh kim loi vn cn nhim in.

+ Nhim in do hng ng: a thanh kim loi khng nhim in n gn qu cu

nhim in nhng khng chm vo qu cu, th hai u thanh kim loi s nhim in. u gn

qu cu hn nhim in tri du vi in tch ca qu cu, u xa hn nhim in cng du vi

in tch ca qu cu. a thanh kim loi ra xa qu cu th thanh kim loi tr v trng thi khng

nhim in nh lc u.

Gii thch hin tng nhim in:

- Do c xt hay tip xc m cc electron di chuyn t vt ny sang vt kia.

- Do hng ng m cc electron t do s di chuyn v mt pha ca vt (thc cht y l s

phn b li cc electron t do trong vt) lm cho pha d electron tch in m v pha ngc li thiu

electron nn tch in dng.

1.2 So snh s ging nhau v khc nhau:

Ging nhau:

T l vi bnh phng khong cch.

T l thun vi tch ln.

u c hng s t l.

Khc nhau:

nh lut Coulomb nh lut vn vt hp dn ca Newton

Va l lc y va l lc ht ty vo du ca in tch.

T l vi ln ca in tch.

Hng s t l l: 2

9

2

.9.10

N mk

C

Ph thuc vo s c mt ca vt khc v mi trng gia hai vt.

L lc ht.

T l vi tch ln khi lng ca hai vt.

Hng s t l l: 2 2

11

2

.6,67.10

N mG

kg

Khng ph thuc vo s c mt ca vt khc v mi trng gia hai vt.

1.3 nh lut Culng.


Page 2


+ ln ca lc tng tc gia hai in tch im ng yn t l thun vi tch cc ln ca

hai in tch v t l nghch vi bnh phng khong cch gia chng.

F = k.2

21

.

||

r

qq

; k = 9.10

9

2

2

C

Nm; l hng s in mi ca mi trng; trong chn khng (hay gn

ng l trong khng kh) th = 1.

+ Vc t lc tng tc gia hai in tch im:

C im t trn mi in tch;

C phng trng vi ng thng ni hai in tch;

C chiu: y nhau nu cng du, ht nhau nu tri du;

C ln: F = 2

21

9

.

||10.9

r

qq

.

+ Lc tng tc gia nhiu in tch im ln mt in tch im:

1.4 in trng

+ in trng l mi trng vt cht tn ti xung quanh cc in tch.

+ Tnh cht c bn ca in trng l n tc dng lc in ln in tch t trong n.

+ in trng tnh l in trng do cc in tch ng yn gy ra.

+ Vc t cng in trng gy bi mt in tch im:

C im t ti im ta xt;

C phng trng vi ng thng ni in tch vi im ta xt;

C chiu: hng ra xa in tch nu l in tch dng, hng v pha in tch nu l in tch m;

C ln: E = 2

9

.

||10.9

r

q

.

+ n v cng in trng l V/m.

+ Nguyn l chng cht in trng: .

+ Lc tc dng ca in trng ln in tch:

F = q .

+ ng sc in l ng c v trong in trng sao cho hng ca tip tuyn ti bt k im

no trn ng sc cng trng vi hng ca vc t cng in trng ti im .

+ Tnh cht ca ng sc:

- Ti mi im trong in trng ta c th v c mt ng sc in v ch mt m thi. Cc

ng sc in khng ct nhau.

- Cc ng sc in trng tnh l cc ng khng khp kn.

- Ni no cng in trng ln hn th cc ng sc in s c v mau hn (dy hn),

ni no cng in trng nh hn th cc ng sc in s c v tha hn.

+ Mt in trng m cng in trng ti mi im u bng nhau gi l in trng u.

in trng u c cc ng sc in song song v cch u nhau.

1.5 ---

.......

2. Bi tp.

Bi 1-1:

Tm tt:

nFFFF ...21

nEEEE

...21

E


Page 3


28 11

19

0,5.10 ( ) 5.10 ( )

1,6.10 ( )

H

e

r cm m

q C

Fh =? Hng dn:

Theo bi ra ta c:

2

9 19 28

2 211

. . 9.10 .(1,6.10 )9,216.10 ( )

5.10

e e

h

H

k q qF N

r

Bi 1-2:

Tm tt: 19

31

8 10

1,6.10

9,1.10

10 10

e

e C

m kg

r cm m

ve = ?

Hng dn:

Theo bi ra th electron chuyn ng xung quanh ht nhn theo qu o trn di tc dng ca lc

hng tm nn ta c: 2 31 2

21 2

10

9,1.10 .9,1.10 . ( )

10ht

mv vF v N

r

Theo nh lut Culong ta c:

29 192

8

22 2

9.10 . 1,6.102,304.10 ( )

10

ecl

kqF N

r

Khi cn bng th: 21 2 8 2 12 69,1.10 2,304.10 2,53.10 1,6.10 ( / )ht clF F v v v m s

Bi 1-3:

Tm tt: 7

0

0

3,2.10

60

10 0,1

q C

l cm m

m = ?

Hng dn:

Theo hnh v ta thy mi qu cu u chu tc dng ca ba lc: Trng lc (P), sc cng (T) v

lc y Culong (clF )

Do cc qu cu ging nhau nn, in tch ca mi qu cu nhn c l: 7

701 2

3,2.101,6.10 ( )

2 2

qq q C

m

htF

v

r

2

T

0

2

q 02

q clF

A I r B

P

Hnh 2


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Do c iu kin cn bng nn: 0clP F T . Khi ta c: tanclF

P 1sin

2

r

l

Vi: 1

2

Vy:

2

02 22

0 0

2 2

1

.

. .4tan4 4.(sin ) .

cl

k q

F k q k qrP P r P P

T ta c:

29 72

0

2 20 0

1

9.10 3,2.100,039

4. 2 sin .tan 4. 2.0,1.sin30 .tan30

kqP N

l

Nn: 30,039

3,99.1010

Pm kg

g

Bi 1-4:

Tm tt: 0

3

0

2

' 54

800 ( / )

2

kg m

' ?

Hng dn:

Tng t cch gii trn ta c: 2

0

2

0 1 1 14 .4(2 sin ) .tan

qP

l (*)

Trong mi trng du ha th mi qu cu s chu tc dng ca lc y Acsimet P. Nn:

2

022

0 2 2 2

''

4 .4(2 sin ) .tan 2

qP P

l

(**)

0 0' ' ' ( ')P P mg m g Vg Vg Vg

Kt hp (*) v (**) ta c:

Ta xt: 2 2 0 0

0 0 01 1 1

2 2 0 0

2 2 2

( ' ) ' '.sin .tan' 1.sin 30 .tan30

.sin .tan ' ' 2.sin 27 .tan 27 '

VgP P

P Vg

T ta suy ra: 3' 2557,54 ( / )kg m

Bi 1-5.

Tm tt: 3

0

2

800 ( / )

2

kg m

? '

Hnh 3a

Hnh 3b

O

2

'T

'F

A B F

1q 2q

P

O

2

T

A B F

1q 2q

P


Page 5


Hng dn:

Theo hnh 3a th mi qu cu u chiu tc dng cu Trng lc(P), Sc cng dy (T), Lc y Culong

(F). Theo nh lut Culoong p dng trong bi ton ny th:

2

012

0 1 1

sin4 .(2 sin ) 2

q rF

l l

Trong trng hp cn bng th: ( Theo hnh v) 0P F T .

1 1cos cos

P mgT mg

T T .

2 2

0 01 12 2

0 1 1 0 1 1

sin sin4 2 sin 4 2 sin

q qFT

T l T l

.

Vy:

2

02 2 2

0 1 1 0 011 2 2

1 0 1 1 0 1

1

4 2 sinsintan

cos 4 2 sin . 4 2 sin .

( 1) (1)

q

l T q q

mg l mg l mgT

Khi vo mi trng: du th qu cu s chu thm lc y Acsimet F. Nh vy F s gim i 2 ln.

Vy lc ny ng nhin sc cng cng phi l T.

Khi ta c iu kin cn bng: ' 0dP F T

'

1 1

'cos 'cos '

' 'd d

d

P F mg FT mg F

T T

2

01 2

0 2 1

sin' 4 2 sin '

qF

T l T

.

Vy:

2

02 2

0 2 1 011 ' 2 '

1 0 2 1

4 2 sin 'sintan (2)

cos 4 2 sin

'd d

q

l T q

mg F l mg F

T

Do ta c: 1 1tan (1) tan (2)

2 2

0 0

2 2 '

0 1 0 2 1

'2 2 22

2

2

4 2 sin . 4 2 sin

1 1 1 1 1 1 1 1

' ' '

'.

1

d

d

q q

l mg l mg F

mg mg mg m g V V Vmg F


Page 6


(Gii thch: Ta c

3

' 3

4.

3

4' . '

3d

m V r

F V g r g

)

Bi 1-6:

Tm tt: 9 2 5 2

3

9

4.10 / 4.10 /

1 10

1,6.10

C cm C m

m g kg

q C

= ?

Hnh 4

Hng dn:

Gi thit bi cho c th hin hnh 4.

Xt n iu kin cn bng khi ta c: 02

0

. . . .tan

2 .

k q qF E q q

P r P P P

Vy: 5 9

0

12 3

4.10 .1,6.10tan 0,3616 19,87

2.8,85.10 .10 .10

Bi 1-7:

Tm tt: 8

1

8

2

8

3

2

2

2

3,2.10

4,8.10

9,6.10

3 3.10

4 4.10

5 5.10

q C

q C

q C

AC cm m

AB cm m

BC cm m

FthA = ?

Hnh 5 Hng dn:

Lc in tng hp tc dng nn q1 ti A l thF . Vy: 21 31thF F F

+ Phng: Nm trn ng thng AC.

Ta c: 31F + Chiu: Hng ra xa im A.

+ ln:

9 8 8

1 3

31 22 2

9.10 3,2.10 . 9,6.10.0,03( )

( ) 3.10

k q qF N

AC

.

A

T

F

B

P

31F

A (+)

thF q1

21F

B C

q2 (+) q3 (-)


Page 7


+ Phng: Nm trn ng thng AB.

21F + Chiu: Hng t A n B.

+ ln:

9 8 8

1 2 3

21 22 2

9.10 3,2.10 .4,8.10.8,64.10 ( )

( ) 4.10

k q qF N

AC

Theo d kin bi cho ta d dng nhn thy: 2 2 2BC AC AB . Nn ABC vung ti A.

Do :

+ ln: 22 3

31 21 0,03 8,64.10 0,0312( )thF F F N .

+ Phng: hp vi AB mt gc : 0313

21

0,03tan 3,472 73,93

8,64.10

F

F

.

+ Chiu: Nh hnh v.

Bi 1-7: *( Ch )

Tm tt: 9

7

0

1,6.10

3,2.10

8 0,08

q C

Q C

r cm m

?tdF

y

O x

Hnh 6

Hng dn:

Chia na vng xuyn thnh nhng phn t dl mang in tch dQ.

Ti tm O v tnh cht i xng nn tng cc thnh phn: 2 0 (1)dF

Khi hnh chiu ca lc tng hp nn cc trc Ox, Oy l: 1 2

2 2

2 2

sin ; s (2)F F

co

dF dF

T (1) v (2) ta c:

9 73

1 22 120 0

1,6.10 .3,2.100,7197.10 ( )

4 4 .8,85.10 .1. 0,08

qQF F N

r

Bi 1-9: B, do bi khng r.

Bi 1-10: ---Cha hc---(B xung sau)

Bi 1-11: Hng dn:

r

q

q M 4q

Hnh 7

Gi im M trn ng thng ni 2 in tch q v 4q in trng do 2 in tch gy ra trit tiu:

1 2 0E E E

Vy E1 = E2 Trong :

E1 l cng in trng do q gy ra ti

2

2dF

q 1dF

dl

r0 dF

2


Page 8


im M cch 1 khong r.

E2 l cng in trng do 4q gy ra ti im M v cch M 1 khong ( l r ).

Nh vy E1 = E2 .

2

2 2 22 2 2

.4 1 4 1 43 0,3 0,0225 0

0,15

0,05 5

0,15 ( )

kq k q q qr r

r r rl r l r r

r m cm

r m loai

Vy im M cch in tch 1 khong r = 5 cm.

Bi 1-12: ( Tham kho cch gii trong sch bi tp trang 75)

Bi 1-13: ( Tham kho cch gii trong sch bi tp trang 76)

Bi 1-14:

Tm tt: 9 2

8 6

2.10 / .

3.10 / 3.10 /

C m

C cm C m

F = ?

Hng dn:

Do c cng thc: F = q.E (1)

Khi ta c: q l (2)

Vi: l mt in di ca dy. l l mt n v chiu di ca dy ( Ta ly: l = 1m)

M: 0

(3)2

E

T (1), (2) v (3) suy ra: 6 9

4

12

0

3.10 .1.2.10. 3,4.10 ( )

2 2.8,85.10 .1

lF q E N

( Vi 1 ly mi trng chn khng) { Nhng p n trong sch bi tp l 3,4 (N), mk khng hiu lm ???)

Bi 1-15:

Tm tt:

0E

1 2q qr l

1. 1 2q q cng du

2. 1 2q q khc du

' ?l

A A

q1 l M q2

l

Hnh 8

Hng dn:

Theo gi thit th: 0E

Theo hnh v ta thy vct cng in trng ti mt im M bt k l: 1 2E E E .

Vi 1 2,E E ln lt l cc vct cng in trng do 1 2q q gy ra.


Page 9


Nn: 1 2E E . Vy im M nm trn AA.( Hnh v)

Khi ta c: 1 2 (1)E E . iu cho ta:

1

2

'(2)

'

ql

l l q

1. Xt trng hp 1: q1, q2 cng du th t (1) ta c: 0 ' 1l

12 1 1 1 1 2 1

2

1

1 2

''. ' . . '. ' .

'

.'

qll q l l q l q l q l q q l q

l l q

l ql

q q

2. Xt trng hp 2: q1, q2 khc du th t (1) ta c: ' 0l l

12 1 1 1 2 1 1

2

1

2 1

''. ' . . '. ' .

'

.'

qll q l l q l q l q l q q l q

l l q

l ql

q q

Bi 1-16:---Cha hc---( B xung sau)

Bi 1-17:---

Bi 1-18:

Tm tt:

1

8

1 2 3

( )

3,2.10

1

0q A

q q q C

AB AC a m

F

a) FA = ? EA = ? b) Q =? V tr?

Hnh 9

(Cn tip....)

Giải Bài Tập Vật Lý Đại Cương 111111

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