Ngā Tauira Pātai mō te Tohanga Coulomb

9 Ngā Tauira o ngā Pātai mō te Kaha Coulomb

1. E toru ngā kawenga kua whakaritea pērā i te pikitia i raro nei. Ko te kaha Coulomb e pāngia ana e te utu B ko …. (k = 9 x 109 Nm2C-2, 1 μC = 10-6 C)
Tauira Pātai mō te Kaha Coulomb 1

A. 09 x 10 1 N hei utu C

B. 09 x 10 1 N hei utu i a A

C. 18 x 10 1 N hei utu C

D. 18 x 10 1 N hei utu i a A

E. 36 x 10 1 N hei utu C

Kōrero

E mōhiotia ana :

q A = 10 µC = 10 x 10 -6 C = 10 -5 Coulombs

q B = 10 µC = 10 x 10 -6 = 10 -5 Ngā Coulomb

q C = 20 µC = 20 x 10 -6 = 2 x 10 -5 Ngā Coulomb

r AB = 0,1 mita = 10 -1 mita

r BC = 0,1 mita = 10 -1 mita

k = 9 x 10 9 Nm 2 C −2

I pātaihia : Te kaha Coulomb i pāngia e te utunga B

Whakautu :

E rua ngā kaha Coulomb, arā, ngā kaha hiko e pā ana ki te utu B, arā, ko te kaha Coulomb i waenganui i ngā utu A me B (F AB ) me te kaha Coulomb i waenganui i ngā utu B me C (F BC ). Ko te kaha Coulomb e pāngia ana e te utu B ko te hua o F AB me F BC.

Te kaha Coulomb i waenganui i ngā utu A me B:

Tauira Pātai mō te Kaha Coulomb 2

He pai te utu A, ā, he pai te utu B, nō reira ka anga a F AB ki te utu C.

Te kaha Coulomb i waenganui i ngā utu B me C:

Tauira Pātai mō te Kaha Coulomb 3

He pai te utu B, ā, he pai te utu C, nō reira ka anga a F BC ki te utu A.

Te kaha Coulomb e pāngia ana e te utu B:

F B = F BC – F AB = 180-90 = 90 N

Ko te rahi o te kaha Coulomb e pāngia ana e te utu B (F B ) he 90 Newton. Ko te ahunga o F B he rite ki te ahunga o F BC , arā, ki te utu A.

Ko te whakautu tika ko B.

2. Ko te rahi me te ahunga o te kaha Coulomb i runga i te utu B ko... ( k = 9 x 10 9 Nm 2 C −2 , 1 μC = 10 −6 C)

Tauira Pātai mō te Kaha Coulomb 4

A. 2,5 k Q 2 r -2 ki te taha maui

B. 2,5 k Q 2 r -2 ki te taha matau

C. 2 k Q 2 r -2 ki te taha maui

D. 2 k Q 2 r -2 ki te taha matau

E. 1 k Q 2 r -2 ki te taha maui

Kōrero

E mōhiotia ana :

Utu A (q A ) = +Q

Utu B (q B ) = -2Q

Utu C (q C ) = -Q

Ko te tawhiti i waenganui i ngā utu A me B (r AB ) = r

Ko te tawhiti i waenganui i ngā utu B me C (r BC ) = 2r

k = 9 x 10 9 Nm 2 C −2

I pātaihia : te rahi me te ahunga o te kaha Coulomb i runga i te utu B

Whakautu :

Te kaha Coulomb i waenganui i te utu A me te utu B:

Tauira Pātai mō te Kaha Coulomb 5

He pai te utu A, he kino te utu B, nō reira ko te ahunga o F AB e anga atu ana ki te utu A.

Te kaha Coulomb i waenganui i te utu B me te utu C:

Tauira Pātai mō te Kaha Coulomb 6

He kino te utu B, ā, he kino te utu C, nō reira ko te ahunga o F BC e anga atu ana ki te utu A.

Ko te kaha hua e pā ana ki te utu B:

F = F AB + F BC = 2 k Q 2 /r 2 + 0,5 k Q 2 /r 2 = 2,5 k Q 2 /r 2 = 2,5 k Q 2 r -2

Ko te ahunga o te kaha Coulomb e anga ana ki te utunga A, ki maui rānei.

Ko te whakautu tika ko A.

3. E rua ngā utu hiko e whakanohoia motuhaketia ana e ai ki te pikitia. Ko te utu i A he 8 µC, ā, ko te kaha kukume e pā ana ki ngā utu e rua he 45 N. Mena ka nekehia te utu A ki te taha matau mā te 1 cm, ā, ko k = 9.10 9 Nm 2 .C -2 , ko te kaha kukume e pā ana ki ngā utu e rua ko…

A. 45 NTauira Pātai mō te Kaha Coulomb 7

B. 60 N

C. 80 N

D. 90 N

E. 120 N

Kōrero

E mōhiotia ana :

Utu hiko i A (q A ) = 8 µC = 8 x 10 -6 Coulombs

Ko te kaha hiko i waenganui i ngā utu e rua (F) = 45 Newton

Ko te tawhiti i waenganui i ngā utu e rua (r AB ) = 4 cm = 0,04 mita = 4 x 10 -2 mita

Pūmau (k) = 9 x 10 9 Nm 2 .C -2

Pātai : Ko te kaha hiko i waenganui i ngā utu e rua mēnā ka nekehia te utu A ki te taha matau mā te 1 cm, 0,01 mita rānei

Whakautu :

Tuatahi, tatauhia te utu hiko i B, kātahi ka tatau i te kaha hiko i waenganui i ngā utu hiko e rua, mēnā ka nekehia te utu hiko i A ki te taha matau mā te 1 cm.

Te utu hiko i B :

Tātai Te ture a Coulomb :

F = k (q A )(q B ) / r 2

F r 2 = k (q A )(q B )

q B = F r 2 / k (q A )

Utu hiko i B :

q B = (45)(4 x 10 -2 ) 2 / (9 x 10 9 )(8 x 10 -6 )

q B = (45)(16 x 10 -4 ) / 72 x 10 3

q B = (720 x 10 -4 ) / (72 x 10 3 )

q B = 10 x 10 -7 Ngā Kuromi

Te kaha hiko i waenganui i ngā utu hiko A me B :

Ki te nekehia te utu i A ki te taha matau mā te 1 cm, ka noho te tawhiti i waenganui i ngā utu e rua hei 3 cm = 0,03 mita = 3 x 10 -2 mita

F = k (q A )(q B ) / r 2

W = (9 x 10 9 )(8 x 10 -6 )(10 x 10 -7 ) / (3 x 10 -2 ) 2

F = (9 x 10 9 )(80 x 10 -13 ) / (9 x 10 -4 )

F = (1 x 10 9 )(80 x 10 -13 ) / (1 x 10 -4 )

F = (80 x 10 -4 ) / (1 x 10 -4 )

F = 80 Newton

Ko te whakautu tika ko C.

4. E rua ngā utu hiko P me Q, he 10 cm te tawhiti, ka pāngia e te kaha kukume o te 8 N. Mēnā ka nekehia te utu Q ki te 5 cm ki te utu P (1 µC = 10 -6 C me te k = 9 x 10 9 Nm 2 .C -2 ), ko te kaha hiko e puta ana ko...

A. 8 NTauira Pātai mō te Kaha Coulomb 8

B. 16 N

C. 32 N

D. 40 N

E. 56 N

Kōrero

E mōhiotia ana :

Ko te tawhiti i waenganui i ngā utu P me Q (r PQ ) = 10 cm = 0,1 m = 1 x 10 -1 m

Ko te kaha hiko i waenganui i ngā utu P me Q (F) = 8 N

Utu hiko Q (q Q ) = 40 µC = 40 x 10 -6 C

Pūmau (k) = 9 x 10 9 Nm 2 .C -2

Pātai : Ko te kaha hiko i waenganui i ngā utu P me Q mēnā ka nekehia te utu Q kia 5 cm te anga atu ki te utu P

Whakautu :

Tuatahi, tatauhia te utu hiko P, kātahi ka tatauhia te kaha hiko i waenganui i ngā utu hiko e rua, mēnā ka nekehia te utu hiko Q kia 5 cm te anga atu ki te utu P.

Utu hiko P :

q P = F r 2 / k (q Q )

q P = (8)(1 x 10 -1 ) 2 / (9 x 10 9 )(40 x 10 -6 )

q P = (8)(1 x 10 -2 ) / 360 x 10 3

q P = (8 x 10 -2 ) / (36 x 10 4 )

q P = (1 x 10 -2 ) / (4,5 x 10 4 )

q P = (1/4,5) x 10 -6 Coulomb

Te kaha hiko i waenganui i ngā utu hiko P me Q :

Ki te nekehia te utu i Q ki te taha maui mā te 5 cm, ka noho te tawhiti i waenganui i ngā utu e rua hei 5 cm = 0,05 mita = 5 x 10 -2 mita

F = k (q P )(q Q ) / r 2

F = (9 x 10 9 )( (1/4,5) x 10 -6 )(40 x 10 -6 ) / (5 x 10 -2 ) 2

F = (2 x 10 3 )(40 x 10 -6 ) / (25 x 10 -4 )

F = (80 x 10 -3 ) / (25 x 10 -4 )

F = 3,2 x 10 1

F = 32 Newton

Ko te whakautu tika ko C.

5. Tirohia te ahua e whai ake nei o ngā utu hiko. Ko te kaha hiko e pāngia ana e te utu q B ko 8 N (1 µC = 10 -6 C) me (k = 9.10 9 Nm 2 .C -2 ). Mena ka nekehia te utu q B ki te 4 cm mai i A, ko te kaha hiko e pāngia ana e q B ko…

A. 2 NTauira Pātai mō te Kaha Coulomb 9

B. 4 N

C. 6 N

D. 8 N

E. 10 N

Kōrero

E mōhiotia ana :

Ko te tawhiti i waenganui i ngā utu A me B (r AB ) = 2 cm = 0,02 m = 2 x 10 -2 m

Ko te kaha hiko i waenganui i ngā utu A me B (F) = 8 N

Utu hiko A (q A ) = 2 µC = 2 x 10 -6 C

Pūmau (k) = 9 x 10 9 Nm 2 .C -2

Pātai : Ko te kaha hiko i waenganui i ngā utu A me B mēnā he 4 cm te tawhiti i waenganui i ngā utu e rua.

Whakautu :

Tuatahi, tatauhia te utu hiko B, kātahi ka tatauhia te kaha hiko i waenganui i ngā utu hiko e rua mēnā ko te tawhiti i waenganui i ngā utu hiko e rua he 4 cm = 0,04 mita = 4 x 10 -2 mita.

Utu hiko B :

q B = F r 2 / k (q A )

q B = (8)(2 x 10 -2 ) 2 / (9 x 10 9 )(2 x 10 -6 )

q B = (8)(4 x 10 -4 ) / (18 x 10 3 )

q B = (32 x 10 -4 ) / (18 x 10 3 )

q B = (32/18) x 10 -7

q B = (16/9) x 10 -7 Ngā Kuoromu

Te kaha hiko i waenganui i ngā utu A me B :

F = k (q A )(q B ) / r 2

F = (9 x 10 9 )(2 x 10 -6 )( (16/9) x 10 -7 ) / (4 x 10 -2 ) 2

W = (18 x 10 3 )( (16/9) x 10 -7 ) / (16 x 10 -4 )

F = (2 x 10 3 )(16 x 10 -7 ) / (16 x 10 -4 )

F = (2 x 10 3 )(1 x 10 -7 ) / (1 x 10 -4 )

F = (2 x 10 -4 ) / (1 x 10 -4 )

F = 2 Newton

Ko te whakautu tika ko A.

6. E toru ngā utu hiko kei ngā kokonga o te tapatoru ABC, ko te roa o te taha AB = BC = 20 cm, ā, he rite tonu te rahi o te utu (q = 2µC) ki te pikitia kei te taha (k = 9.10 9 Nm 2 .C -2 , 1 µ = 10 -6 ). Ko te rahi o te kaha hiko e pā ana ki te pūwāhi B ko….

A. 0,9√3 NTauira Pātai mō te Kaha Coulomb 10

B. 0,9√2 N

C. 0,9 N

D. 0,81 N

E. 0,4 N

Kōrero

E mōhiotia ana:

Utu i te pūwāhi A (q A ) = 2 µC = 2 x 10 -6 Coulomb

Utu i te pūwāhi B (q B ) = 2 µC = 2 x 10 -6 Coulomb

Utu i te pūwāhi C (q C ) = 2 µC = 2 x 10 -6 Coulomb

Te tawhiti mai i B ki C (r BC ) = 20 cm = 0,2 mita = 2 x 10 -1 mita

Te tawhiti o B ki A (r BA ) = 20 cm = 0,2 mita = 2 x 10 -1 mita

k = 9.10 9 Nm 2 .C -2

Pātai: Te rahi o te kaha hiko e pā ana ki te pūwāhi B

Whakautu:

Ko te kaha hiko i waenganui i ngā utu i ngā pūwāhi B me C:

F BC = k (q B )(q C ) / r BC 2

F BC = (9 x 10 9 )(2 x 10 -6 )(2 x 10 -6 ) / (2 x 10 -1 ) 2

F BC = (9 x 10 9 )(4 x 10 -12 ) / (4 x 10 -2 )

F BC = (36 x 10 -3 ) / (4 x 10 -2 )

F BC = 9 x 10 -1

F BC = 0,9 Newton

He pai te utu hiko i ngā pūwāhi B me C, nō reira ko te ahunga o te kaha hiko F BC kei te taha mauī, atu i te pūwāhi C.

Ko te kaha hiko i waenganui i ngā utu i ngā pūwāhi B me A:

F BA = k (q B )(q A ) / r BA 2

F BA = (9 x 10 9 )(2 x 10 -6 )(2 x 10 -6 ) / (2 x 10 -1 ) 2

F BA = (9 x 10 9 )(4 x 10 -12 ) / (4 x 10 -2 )

F BA = (36 x 10 -3 ) / (4 x 10 -2 )

F BA = 9 x 10 -1

F BA = 0,9 Newton

He pai te utu hiko i ngā pūwāhi B me A, nō reira ko te ahunga o te kaha hiko F BA kei raro, kei tawhiti atu i te pūwāhi A.

Ka hanga he koki matau ngā kaha hiko e rua, nō reira ka tatauhia te kaha hiko e puta mai ana i te pūwāhi B mā te whakamahi i te tātai Pythagoras.

Tauira Pātai mō te Kaha Coulomb 11

Ko te whakautu tika ko B.

7. Tirohia te pikitia e whai ake nei!

E toru ngā utu Q1 , Q2 , me Q3 kei ngā pito o te tapatoru matau ABC. Ko te roa o AB = BC = 30 cm. Mēnā ko k = 9.10 9 Nm2 .C -2 me 1 µ = 10 -6 , ko te kaha Coulomb i puta mai i runga i te utu Q1 ko ….

A. 1 NTauira Pātai mō te Kaha Coulomb 12

B. 5 N

C. 7 N

D. 10 N

E. 12 N

Kōrero

E mōhiotia ana:

Utu i te pūwāhi A (q A ) = 3 µC = 3 x 10 -6 Coulomb

Utu i te pūwāhi B (q B ) = -10 µC = -10 x 10 -6 Coulomb

Utu i te pūwāhi C (q C ) = 4 µC = 4 x 10 -6 Coulomb

Te tawhiti mai i B ki C (r BC ) = 30 cm = 0,3 mita = 3 x 10 -1 mita

Te tawhiti o B ki A (r BA ) = 30 cm = 0,3 mita = 3 x 10 -1 mita

k = 9.10 9 Nm 2 .C -2

Pātai: Te kaha Coulomb hua i te pūwāhi B

Whakautu:

Ko te kaha hiko i waenganui i ngā utu i ngā pūwāhi B me C:

F BC = k (q B )(q C ) / r BC 2

F BC = (9 x 10 9 )(10 x 10 -6 )(4 x 10 -6 ) / (3 x 10 -1 ) 2

F BC = (9 x 10 9 )(40 x 10 -12 ) / (9 x 10 -2 )

F BC = (360 x 10 -3 ) / (9 x 10 -2 )

F BC = 40 x 10 -1

F BC = 4 Newton

He kino te utu hiko i te pūwāhi B, ā, he pai te utu hiko i te pūwāhi C, nō reira ko te ahunga o te kaha hiko F BC kei te taha matau ki te pūwāhi C.

Ko te kaha hiko i waenganui i ngā utu i ngā pūwāhi B me A:

F BA = k (q B )(q A ) / r BA 2

F BA = (9 x 10 9 )(10 x 10 -6 )(3 x 10 -6 ) / (3 x 10 -1 ) 2

F BA = (9 x 10 9 )(30 x 10 -12 ) / (9 x 10 -2 )

F BA = (270 x 10 -3 ) / (9 x 10 -2 )

F BA = 30 x 10 -1

F BA = 3 Newton

He kino te utu hiko i te pūwāhi B, ā, he pai te utu hiko i te pūwāhi A, nō reira ko te ahunga o te kaha hiko F BA kei runga, e anga atu ana ki te pūwāhi A.

Ka hanga he koki matau ngā kaha hiko e rua, nō reira ka tatauhia te kaha hiko e puta mai ana i te pūwāhi B mā te whakamahi i te tātai Pythagoras.

Tauira Pātai mō te Kaha Coulomb 13

Ko te whakautu tika ko B.

8. E toru ngā utu hiko kei ngā kokonga o te tapatoru ABC, ko te roa o te taha AB = BC = 20 cm, ā, ko te rahi o te utu (q = 2µC) rite tonu ki te pikitia kei te taha (k = 9.109 Nm2.C-2, 1 µ = 10-6Te rahi o te kaha hiko Tauira Pātai mō te Kaha Coulomb 14ko te mahi i te pūwāhi B ko….

  1. 0,9√3 N
  2. 0,9√2 N
  3. 0,9 N
  4. 0,81 N
  5. 0,4 N

Kōrero

E mōhiotia ana:
Ko te utu i te pūwāhi A (qA) = 2 µC = 2 x 10-6 Coulomb
Ko te utu i te pūwāhi B (qB) = 2 µC = 2 x 10-6 Coulomb
Ko te utu i te pūwāhi C (qC) = 2 µC = 2 x 10-6 Coulomb
Te tawhiti mai i B ki C (rBC) = 20 henimita = 0,2 mita = 2 x 10-1 mita
Te tawhiti mai i B ki A (rBA) = 20 henimita = 0,2 mita = 2 x 10-1 mita
k = 9.109 Nm2.C-2
I pātaihia: Te rahi o te kaha hiko e pā ana ki te pūwāhi B
Whakautu:

Ko te kaha hiko i waenganui i ngā utu i ngā pūwāhi B me C:
FBC = k (qB)(qC) / rBC2
FBC = (9 x 109)(2 x 10-6)(2 x 10-6) / (2 x 10-1)2
FBC = (9 x 109)(4 x 10-12) / (4 x 10-2)
FBC = (36 x 10-3) / (4 x 10-2)
FBC = 9x10-1
FBC = 0,9 Niutona
He pai te utu hiko i ngā pūwāhi B me C, nō reira ko te ahunga o te kaha hiko F koBC ki te taha maui, atu i te pūwāhi C.

Ko te kaha hiko i waenganui i ngā utu i ngā pūwāhi B me A:
FBA = k (qB)(qA) / rBA2
FBA = (9 x 109)(2 x 10-6)(2 x 10-6) / (2 x 10-1)2
FBA = (9 x 109)(4 x 10-12) / (4 x 10-2)
FBA = (36 x 10-3) / (4 x 10-2)
FBA = 9x10-1
FBA = 0,9 Niutona
He pai te utu hiko i ngā pūwāhi B me A, nō reira ko te ahunga o te kaha hiko F koBA ki raro, ki tawhiti atu i te pūwāhi A.

Ka hanga he koki matau ngā kaha hiko e rua, nō reira ka tatauhia te kaha hiko e puta mai ana i te pūwāhi B mā te whakamahi i te tātai Pythagoras.
Tauira Pātai mō te Kaha Coulomb 15Ko te whakautu tika ko B.

9. Tirohia te pikitia e whai ake nei!
E toru ngā utu Q1,Q2, me Q3 kei te pito o te tapatoru matau ABC. Ko te roa o AB = BC = 30 cm. E mōhiotia ana Tauira Pātai mō te Kaha Coulomb 16k = 9.109 Nm2.C-2 ā, 1 µ = 10-6 kātahi ka puta te kaha Coulomb i runga i te utu Q1 ko….

  1. 1 N
  2. 5 N
  3. 7 N
  4. 10 N
  5. 12 N

Kōrero
E mōhiotia ana:
Ko te utu i te pūwāhi A (qA) = 3 µC = 3 x 10-6 Coulomb
Ko te utu i te pūwāhi B (qB) = -10 µC = -10 x 10-6 Coulomb
Ko te utu i te pūwāhi C (qC) = 4 µC = 4 x 10-6 Coulomb
Te tawhiti mai i B ki C (rBC) = 30 henimita = 0,3 mita = 3 x 10-1 mita
Te tawhiti mai i B ki A (rBA) = 30 henimita = 0,3 mita = 3 x 10-1 mita
k = 9.109 Nm2.C-2
I pātaihia: Te kaha Coulomb hua i te pūwāhi B
Whakautu:
Ko te kaha hiko i waenganui i ngā utu i ngā pūwāhi B me C:
FBC = k (qB)(qC) / rBC2
FBC = (9 x 109)(10 x 10-6)(4 x 10-6) / (3 x 10-1)2
FBC = (9 x 109)(40 x 10-12) / (9 x 10-2)
FBC = (360 x 10-3) / (9 x 10-2)
FBC = 40x10-1
FBC = 4 Niutona
He kino te utu hiko i te pūwāhi B, ā, he pai te utu hiko i te pūwāhi C, nō reira ko te ahunga o te kaha hiko F koBC ki te taha matau ki te pūwāhi C.

Ko te kaha hiko i waenganui i ngā utu i ngā pūwāhi B me A:
FBA = k (qB)(qA) / rBA2
FBA = (9 x 109)(10 x 10-6)(3 x 10-6) / (3 x 10-1)2
FBA = (9 x 109)(30 x 10-12) / (9 x 10-2)
FBA = (270 x 10-3) / (9 x 10-2)
FBA = 30x10-1
FBA = 3 Niutona
He kino te utu hiko i te pūwāhi B, ā, he pai te utu hiko i te pūwāhi A, nō reira ko te ahunga o te kaha hiko F koBA ki runga ki te pūwāhi A.
 
Ka hanga he koki matau ngā kaha hiko e rua, nō reira ka tatauhia te kaha hiko e puta mai ana i te pūwāhi B mā te whakamahi i te tātai Pythagoras.
Tauira Pātai mō te Kaha Coulomb 18Ko te whakautu tika ko B.

Pūtake pātai:

Ngā Pātai Ahupūngao Whakamātautau ā-Motu mō te Kura Tuarua/Kura Tuarua Mahi-ā-ringa

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