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)

A. 09 x 10¹ Utu N ki C

B. 09 x 10¹ N ki utanga A

C. 18 x 10¹ Utu N ki C

D. 18 x 10¹ N hei utu i a A

E. 36 x 10¹ Utu N ki C

Kōrero

E mōhiotia ana :

q_A = 10 µC = 10 x 10^-6 C=10^-5 Coulomb

q_B = 10 µC = 10 x 10^-6 = 10^-5 Coulomb

q_C = 20 µC = 20 x 10^-6 = 2x10^-5 Coulomb

r_AB = 0,1 mita = 10^-1 mita

r_BC = 0,1 mita = 10^-1 mita

k = 9 x 10⁹ Nm²C^-2

I pātaihia Te kaha Coulomb e pāngia ana e te utu B

Whakautu :

E rua Te kaha o Coulomb te kaha hiko rānei e pā ana ki te utu B, arā, 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_BCKo 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:

He tohu pai tō te utunga A, ā, he tohu pai tō te utunga B, nō reira ko F_AB ki te utu C.

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

He pai te utu B, ā, he pai te utu C, nō reira ko 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

Te rahi o te kaha Coulomb e pāngia ana e te utunga B (F_B) he 90 Newton. Ko te ahunga o F_B rite tonu ki te ahunga F_BC arā, e anga ana ki te utu A.

Ko te whakautu tika ko B.

2. PKo te rahi me te ahunga o te kaha Coulomb i runga i te utu B ko... (k = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C)

A. 2,5 k Q² r^-2 ki te taha maui

B. 2,5 k Q² r^-2 ki te taha matau

C. 2 k Q² r^-2 ki te taha maui

D. 2 k Q² r^-2 ki te taha matau

E. 1 k Q² 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

Te tawhiti i waenganui i ngā utu A me B (r_AB) = r

Te tawhiti i waenganui i ngā utu B me C (r_BC) = 2r

k = 9 x 10⁹ Nm²C^-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:

He pai te utu A, ā, he kino te utu B, nō reira ko te ahunga ko F_AB ki te utu A

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

He kino te utu B, ā, he kino te utu C, nō reira ko te ahunga o F_BC ki te utu A

Ko te kaha hua e pā ana ki te utu B:

F = F_AB +F_BC = 2 k Q²/r² + 0,5 k Q²/r² = 2,5 k Q²/r² = 2,5 k Q² 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 wehea 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. Mēnā ka nekehia te utu A ki te taha matau mā te 1 cm, ā, ko k = 9.10⁹ Nm².C^-2, kātahi ko te kaha kukume e pā ana ki ngā utu e rua ko...

A. 45 N

B. 60 N

C. 80 N

D. 90 N

E. 120 N

Kōrero

E mōhiotia ana :

Ko te utu hiko i A (q_A) = 8 µC = 8 x 10^-6 Coulomb

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 henimita = 0,04 mita = 4 x 10^-2 mita

Pūmau (k) = 9 x 10⁹ Nm².C^-2

I pātaihia 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.

Utu hiko i B :

Tātai Te ture a Coulomb :

F = k (q_A)(q_B) / r²

F r² = k (q_A)(q_B)

q_B = F r² /k (q_A)

Utu hiko i B :

q_B = (45)(4 x 10^-2)² / (9 x 10⁹)(8 x 10^-6)

q_B = (45)(16 x 10^-4) / 72 x 10³

q_B = (720 x 10^-4) / (72 x 10³)

q_B = 10x10^-7 Coulomb

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) / r²

F = (9 x 10⁹)(8 x 10^-6)(10 x 10^-7) / (3 x 10^-2)²

F = (9 x 10⁹)(80 x 10^-13) / (9 x 10^-4)

F = (1 x 10⁹)(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 kia 5 cm ki te utu P (1 µC = 10^-6 C me k = 9 x 10⁹ Nm².C^-2), kātahi ko te kaha hiko e puta ana ko...

A. 8 N

B. 16 N

C. 32 N

D. 40 N

E. 56 N

Kōrero

E mōhiotia ana :

Te tawhiti i waenganui i ngā utu P me Q (r_PQ) = 10 henimita = 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 10⁹ Nm².C^-2

I pātaihia 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 r² /k (q_Q)

q_P = (8)(1 x 10^-1)² / (9 x 10⁹)(40 x 10^-6)

q_P = (8)(1 x 10^-2) / 360 x 10³

q_P = (8 x 10^-2) / (36 x 10⁴)

q_P = (1 x 10^-2) / (4,5 x 10⁴)

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 mauī 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) / r²

F = (9 x 10⁹)( (1/4,5) x 10^-6)(40 x 10^-6) / (5 x 10^-2)²

F = (2 x 10³)(40 x 10^-6) / (25 x 10^-4)

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

F = 3,2 x 10¹

F = 32 Newton

Ko te whakautu tika ko C.

5. Tirohia te ahua o te utu hiko e whai ake nei. Ko te kaha hiko e pāngia ana e te utu q_B ko 8 N (1 µC = 10^-6 C) me (k = 9.10⁹ Nm².C^-2). Mēnā ko te utu q_B nekehia ki te 4 cm mai i A, kātahi ko te kaha hiko e pāngia ana ko q_B inaianei ko…

A. 2 N

B. 4 N

C. 6 N

D. 8 N

E. 10 N

Kōrero

E mōhiotia ana :

Te tawhiti i waenganui i ngā utu A me B (r_AB) = 2 henimita = 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 10⁹ Nm².C^-2

I pātaihia 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 r² /k (q_A)

q_B = (8)(2 x 10^-2)² / (9 x 10⁹)(2 x 10^-6)

q_B = (8)(4 x 10^-4) / (18 x 10³)

q_B = (32 x 10^-4) / (18 x 10³)

q_B = (32/18) x 10^-7

q_B = (16/9) x 10^-7 Coulomb

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

F = k (q_A)(q_B) / r²

F = (9 x 10⁹)(2 x 10^-6)( (16/9) x 10^-7) / (4 x 10^-2)²

F = (18 x 10³)( (16/9) x 10^-7) / (16 x 10^-4)

F = (2 x 10³)(16 x 10^-7) / (16 x 10^-4)

F = (2 x 10³)(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 te rahi o te utu (q = 2µC) ki te pikitia kei te taha (k = 9.10⁹ Nm².C^-2, 1 µ = 10^-6Ko te rahi o te kaha hiko e pā ana ki te pūwāhi B ko….

A. 0,9√3 N

B. 0,9√2 N

C. 0,9 N

D. 0,81 N

E. 0,4 N

Kōrero

E mōhiotia ana:

Ko te utu i te pūwāhi A (q_A) = 2 µC = 2 x 10^-6 Coulomb

Ko te utu i te pūwāhi B (q_B) = 2 µC = 2 x 10^-6 Coulomb

Ko te utu i te pūwāhi C (q_C) = 2 µC = 2 x 10^-6 Coulomb

Te tawhiti mai i B ki C (r_BC) = 20 henimita = 0,2 mita = 2 x 10^-1 mita

Te tawhiti mai i B ki A (r_BA) = 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:

F_BC = k (q_B)(q_C) / r_BC²

F_BC = (9 x 10⁹)(2 x 10^-6)(2 x 10^-6) / (2 x 10^-1)²

F_BC = (9 x 10⁹)(4 x 10^-12) / (4 x 10^-2)

F_BC = (36 x 10^-3) / (4 x 10^-2)

F_BC = 9x10^-1

F_BC = 0,9 Niutona

He pai te utu hiko i ngā pūwāhi B me C, nō reira ko te ahunga o te kaha hiko F ko_BC ki te taha maui, 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_BA²

F_BA = (9 x 10⁹)(2 x 10^-6)(2 x 10^-6) / (2 x 10^-1)²

F_BA = (9 x 10⁹)(4 x 10^-12) / (4 x 10^-2)

F_BA = (36 x 10^-3) / (4 x 10^-2)

F_BA = 9x10^-1

F_BA = 0,9 Niutona

He pai te utu hiko i ngā pūwāhi B me A, nō reira ko te ahunga o te kaha hiko F ko_BA 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.

Ko te whakautu tika ko B.

7. 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 ko k = 9.10⁹ Nm².C^-2 ā, 1 µ = 10^-6 kātahi ka puta te kaha Coulomb i runga i te utu Q₁ ko….

A. 1 N

B. 5 N

C. 7 N

D. 10 N

E. 12 N

Kōrero

E mōhiotia ana:

Ko te utu i te pūwāhi A (q_A) = 3 µC = 3 x 10^-6 Coulomb

Ko te utu i te pūwāhi B (q_B) = -10 µC = -10 x 10^-6 Coulomb

Ko te utu i te pūwāhi C (q_C) = 4 µC = 4 x 10^-6 Coulomb

Te tawhiti mai i B ki C (r_BC) = 30 henimita = 0,3 mita = 3 x 10^-1 mita

Te tawhiti mai i B ki A (r_BA) = 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:

F_BC = k (q_B)(q_C) / r_BC²

F_BC = (9 x 10⁹)(10 x 10^-6)(4 x 10^-6) / (3 x 10^-1)²

F_BC = (9 x 10⁹)(40 x 10^-12) / (9 x 10^-2)

F_BC = (360 x 10^-3) / (9 x 10^-2)

F_BC = 40x10^-1

F_BC = 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 ko_BC ki 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_BA²

F_BA = (9 x 10⁹)(10 x 10^-6)(3 x 10^-6) / (3 x 10^-1)²

F_BA = (9 x 10⁹)(30 x 10^-12) / (9 x 10^-2)

F_BA = (270 x 10^-3) / (9 x 10^-2)

F_BA = 30x10^-1

F_BA = 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 ko_BA 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.

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 ko 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 (q_A) = 2 µC = 2 x 10^-6 Coulomb
Ko te utu i te pūwāhi B (q_B) = 2 µC = 2 x 10^-6 Coulomb
Ko te utu i te pūwāhi C (q_C) = 2 µC = 2 x 10^-6 Coulomb
Te tawhiti mai i B ki C (r_BC) = 20 henimita = 0,2 mita = 2 x 10^-1 mita
Te tawhiti mai i B ki A (r_BA) = 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:
F_BC = k (q_B)(q_C) / r_BC²
F_BC = (9 x 10⁹)(2 x 10^-6)(2 x 10^-6) / (2 x 10^-1)²
F_BC = (9 x 10⁹)(4 x 10^-12) / (4 x 10^-2)
F_BC = (36 x 10^-3) / (4 x 10^-2)
F_BC = 9x10^-1
F_BC = 0,9 Niutona
He pai te utu hiko i ngā pūwāhi B me C, nō reira ko te ahunga o te kaha hiko F ko_BC ki te taha maui, 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_BA²
F_BA = (9 x 10⁹)(2 x 10^-6)(2 x 10^-6) / (2 x 10^-1)²
F_BA = (9 x 10⁹)(4 x 10^-12) / (4 x 10^-2)
F_BA = (36 x 10^-3) / (4 x 10^-2)
F_BA = 9x10^-1
F_BA = 0,9 Niutona
He pai te utu hiko i ngā pūwāhi B me A, nō reira ko te ahunga o te kaha hiko F ko_BA 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.
Ko 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 k = 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 (q_A) = 3 µC = 3 x 10^-6 Coulomb
Ko te utu i te pūwāhi B (q_B) = -10 µC = -10 x 10^-6 Coulomb
Ko te utu i te pūwāhi C (q_C) = 4 µC = 4 x 10^-6 Coulomb
Te tawhiti mai i B ki C (r_BC) = 30 henimita = 0,3 mita = 3 x 10^-1 mita
Te tawhiti mai i B ki A (r_BA) = 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:
F_BC = k (q_B)(q_C) / r_BC²
F_BC = (9 x 10⁹)(10 x 10^-6)(4 x 10^-6) / (3 x 10^-1)²
F_BC = (9 x 10⁹)(40 x 10^-12) / (9 x 10^-2)
F_BC = (360 x 10^-3) / (9 x 10^-2)
F_BC = 40x10^-1
F_BC = 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 ko_BC ki 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_BA²
F_BA = (9 x 10⁹)(10 x 10^-6)(3 x 10^-6) / (3 x 10^-1)²
F_BA = (9 x 10⁹)(30 x 10^-12) / (9 x 10^-2)
F_BA = (270 x 10^-3) / (9 x 10^-2)
F_BA = 30x10^-1
F_BA = 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 ko_BA 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.
Ko te whakautu tika ko B.

Pūtake pātai:

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