Su'aalaha Tusaalaha ah ee Ciidanka Coulomb

9 Tusaalooyin oo ku saabsan Su'aalaha Ciidanka Coulomb

1. Saddex qaybood ayaa loo diyaariyey sida ku cad sawirka hoose. Xoogga Coulomb ee uu ku dhacay darajada B waa …. (k = 9 x 109 Nm2C-2, 1 μC = 10-6 C)
Tusaalaha Ciidanka Coulomb Su'aal 1aad

A. 09 x 10 1 N si loogu dallaco C

B. 09 x 10 1 N si loogu dallaco A

C. 18 x 10 1 N si loogu dallaco C

D. 18 x 10 1 N si loogu dallaco A

E. 36 x 10 1 N si loogu dallaco C

Dood

Waa la ogyahay in :

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

q B = 10 µC = 10 x 10 -6 = 10 -5 Coulombs

q C = 20 µC = 20 x 10 -6 = 2 x 10 -5 Coulombs

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

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

k = 9 x 10 9 Nm 2 C −2

La weydiiyay : Ciidanka Coulomb oo ay soo mareen dacwad B

Jawaab :

Waxaa jira laba awoodood oo Coulomb ah ama awoodo koronto oo ku shaqeeya dallacaadda B, kuwaas oo kala ah xoogga Coulomb ee u dhexeeya dallacaadaha A iyo B (F AB ) iyo xoogga Coulomb ee u dhexeeya dallacaadaha B iyo C (F BC ). Xoogga Coulomb ee uu la kulmay dallacaadda B waa natiijada F AB iyo F BC.

Xoogga Coulomb ee u dhexeeya dallacaadaha A iyo B:

Tusaalaha Ciidanka Coulomb Su'aal 2aad

Kharash A waa togan yahay, kharash B-na waa togan yahay, sidaa darteed F AB wuxuu u gudbayaa kharash C.

Xoogga Coulomb ee u dhexeeya dallacaadaha B iyo C:

Tusaalaha Ciidanka Coulomb Su'aal 3aad

Kharash B waa togan yahay, kharash C-na waa togan yahay, sidaa darteed F BC waxay u socotaa kharash A.

Ciidanka Coulomb oo uu khibrad u leeyahay darajada B:

F B = F BC – F AB = 180-90 = 90 N

Baaxadda xoogga Coulomb ee uu la kulmo charge B (F B ) waa 90 Newtons. Jihada F B waxay la mid tahay jihada F BC , taas oo ah charge A.

Jawaabta saxda ah waa B.

2. Cabbirka iyo jihada xoogga Coulomb ee ku jira dallacaadda B waa... ( k = 9 x 10 9 Nm 2 C −2 , 1 μC = 10 −6 C)

Tusaalaha Ciidanka Coulomb Su'aal 4aad

A. 2,5 k Q 2 r -2 dhanka bidix

B. 2,5 k Q 2 r -2 dhanka midig

C. 2 k Q 2 r -2 dhanka bidix

D. 2 k Q 2 r -2 dhanka midig

E. 1 k Q 2 r -2 dhanka bidix

Dood

Waa la ogyahay in :

Dalac A (q A ) = +Q

Dalac B (q B ) = -2Q

Dalacaadda C (q C ) = -Q

Masaafada u dhaxaysa khidmadaha A iyo B (r AB ) = r

Masaafada u dhaxaysa khidmadaha B iyo C (r BC ) = 2r

k = 9 x 10 9 Nm 2 C −2

La weydiiyay : baaxadda iyo jihada ciidanka Coulomb ee ku jira weerarka B

Jawaab :

Xoogga Coulomb ee u dhexeeya dallacaadda A iyo dallacaadda B:

Tusaalaha Ciidanka Coulomb Su'aal 5aad

Kharash A waa togan yahay, kharash B-na waa taban yahay, sidaa darteed jihada F AB waxay u jeeddaa kharash A.

Xoogga Coulomb ee u dhexeeya dallacaadda B iyo dallacaadda C:

Tusaalaha Ciidanka Coulomb Su'aal 6aad

Dalacaadda B waa taban, Dalacaadda C-na waa taban, sidaa darteed jihada F BC waxay u jeeddaa dalacaadda A

Xoogga ka dhashay ee ku shaqeynaya dacwadda 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

Jihada ciidanka Coulomb waxay u socotaa dhanka A ama dhanka bidix.

Jawaabta saxda ah waa A.

3. Laba dallac koronto ayaa si gaar ah loo dhigay sida ku cad jaantuska. Dallaca A waa 8 µC, xoogga soo jiidashada leh ee labada dallacna waa 45 N. Haddii dallaca A loo wareejiyo dhanka midig 1 cm iyo k = 9.10 9 Nm 2 .C -2 , markaa xoogga soo jiidashada leh ee labada dallacna waa…

A. 45 NTusaalaha Ciidanka Coulomb Su'aal 7aad

B. 60 N

C. 80 N

D. 90 N

E. 120 N

Dood

Waa la ogyahay in :

Dareeraha korontada ee A (q A ) = 8 µC = 8 x 10 -6 Coulombs

Xoogga korantada ee u dhexeeya labada dallacaad (F) = 45 Newton

Masaafada u dhaxaysa labada dallac (r AB ) = 4 cm = 0,04 mitir = 4 x 10 -2 mitir

Joogto ah (k) = 9 x 10 9 Nm 2 .C -2

Su'aal : Awoodda korantada ee u dhaxaysa labada dallac haddii dallac A loo wareejiyo dhanka midig 1 cm ama 0,01 mitir

Jawaab :

Marka hore xisaabi kharashka korontada ee B, ka dibna xisaabi xoogga korontada ee u dhexeeya labada dallac ee korontada, haddii dallacyada korontada ee A loo wareejiyo dhanka midig 1 cm.

Kharashka korontada ee B :

Qaaciddada Sharciga 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 )

Koronto laga helo 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 Coulombs

Xoogga korontada ee u dhexeeya dallacaadaha korantada A iyo B :

Haddii dallacaadda A loo wareejiyo dhanka midig 1 cm, masaafada u dhaxaysa labada dallac waxay noqonaysaa 3 cm = 0,03 mitir = 3 x 10 -2 mitir.

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

F = (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

Jawaabta saxda ah waa C.

4. Laba dallac koronto oo P iyo Q ah oo 10 cm u jira waxay la kulmaan xoog soo jiidasho leh oo ah 8 N. Haddii dallac Q loo wareejiyo 5 cm dhanka dallacsiinta P (1 µC = 10 -6 C iyo k = 9 x 10 9 Nm 2 .C -2 ), markaa xoogga korontada ee dhaca waa...

A. 8 NTusaalaha Ciidanka Coulomb Su'aal 8aad

B. 16 N

C. 32 N

D. 40 N

E. 56 N

Dood

Waa la ogyahay in :

Masaafada u dhaxaysa khidmadaha P iyo Q (r PQ ) = 10 cm = 0,1 m = 1 x 10 -1 m

Xoogga korontada ee u dhexeeya dallacaadaha P iyo Q (F) = 8 N

Dakhli koronto Q (q Q ) = 40 µC = 40 x 10 -6 C

Joogto ah (k) = 9 x 10 9 Nm 2 .C -2

Su'aal : Awoodda korantada ee u dhaxaysa dallacaadaha P iyo Q haddii dallacaadda Q loo wareejiyo 5 cm dhanka dallacaadda P

Jawaab :

Marka hore xisaabi lacagta korontada ee P, ka dibna xisaabi xoogga korontada ee u dhexeeya labada dallac ee korontada, haddii lacagta korontada ee Q loo wareejiyo 5 cm dhanka dallacda P.

Dalacaadda korontada 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

Awoodda korantada ee u dhaxaysa dallacaadaha korantada P iyo Q :

Haddii dallacaadda Q loo wareejiyo bidixda 5 cm, masaafada u dhaxaysa labada dallac waxay noqonaysaa 5 cm = 0,05 mitir = 5 x 10 -2 mitir.

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

Jawaabta saxda ah waa C.

5. Fiiri sawirka soo socda ee dallacaadaha korontada. Xoogga korontada ee uu la kulmo dallacaadda q B waa 8 N (1 µC = 10 -6 C) iyo (k = 9.10 9 Nm 2 .C -2 ). Haddii dallacaadda q B loo wareejiyo 4 cm laga bilaabo A, markaa xoogga korontada ee uu la kulmo q B hadda waa…

A. 2 NTusaalaha Ciidanka Coulomb Su'aal 9aad

B. 4 N

C. 6 N

D. 8 N

E. 10 N

Dood

Waa la ogyahay in :

Masaafada u dhaxaysa khidmadaha A iyo B (r AB ) = 2 cm = 0,02 m = 2 x 10 -2 m

Xoogga korontada ee u dhexeeya dallacaadaha A iyo B (F) = 8 N

Dalac koronto A (q A ) = 2 µC = 2 x 10 -6 C

Joogto ah (k) = 9 x 10 9 Nm 2 .C -2

Su'aal : Awoodda korantada ee u dhaxaysa dallacaadaha A iyo B haddii masaafada u dhaxaysa labada dallacaadood ay tahay 4 cm

Jawaab :

Marka hore xisaabi lacagta korantada B, ka dibna xisaabi xoogga korontada ee u dhexeeya labada dallac ee korontada haddii masaafada u dhaxaysa labada dallac ee korontada ay tahay 4 cm = 0,04 mitir = 4 x 10 -2 mitir.

Lacag koronto 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 Coulombs

Xoogga korontada ee u dhexeeya dallacaadaha A iyo 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

F = (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

Jawaabta saxda ah waa A.

6. Saddex dallacaad koronto ayaa la dhigay meelaha geeska saddexagalka ABC iyadoo dhererka dhinaca AB = BC = 20 cm iyo isla baaxadda dallacaad (q = 2µC) sida sawirka dhinaca (k = 9.10 9 Nm 2 .C -2 , 1 µ = 10 -6 ). Cabbirka xoogga korontada ee ka shaqeeya barta B waa….

A. 0,9√3 NTusaalaha Ciidanka Coulomb Su'aal 10aad

B. 0,9√2 N

C. 0,9 N

D. 0,81 N

E. 0,4 N

Dood

Waa la ogyahay in:

Dareeraha barta A (q A ) = 2 µC = 2 x 10 -6 Coulomb

Dareeraha barta B (q B ) = 2 µC = 2 x 10 -6 Coulomb

Dareeraha barta C (q C ) = 2 µC = 2 x 10 -6 Coulomb

Masaafada B ilaa C (r BC ) = 20 cm = 0,2 mitir = 2 x 10 -1 mitir

Masaafada B ilaa A (r BA ) = 20 cm = 0,2 mitir = 2 x 10 -1 mitir

k = 9.10 9 Nm 2 .C -2

Su'aal: Baaxadda awoodda korontada ee ka shaqeysa barta B

Jawaab:

Xoogga korantada ee u dhexeeya dallacaadaha dhibcaha B iyo 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

Dareeraha korontada ee dhibcaha B iyo C waa togan, sidaas darteed jihada xoogga korontada F BC waxay u jirtaa bidix, meel ka fog barta C.

Xoogga korontada ee u dhexeeya dallacaadaha dhibcaha B iyo 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

Dareeraha korontada ee dhibcaha B iyo A waa togan, sidaas darteed jihada xoogga korontada F BA waa hoos, meel ka fog barta A.

Labada awoodood ee korontada waxay sameeyaan xagal qumman, sidaas darteed awoodda korantada ee ka dhalata ee ka shaqeysa barta B waxaa lagu xisaabiyaa iyadoo la adeegsanayo qaacidada Pythagorean.

Tusaalaha Ciidanka Coulomb Su'aal 11aad

Jawaabta saxda ah waa B.

7. Fiiri sawirka soo socda!

Saddex dallacaad Q 1 , Q 2 , iyo Q 3 ayaa la dhigay cidhifyada saddexagal midig ABC. Dhererka AB = BC = 30 cm. Marka la eego k = 9.10 9 Nm 2 .C -2 iyo 1 µ = 10 -6 markaa xoogga Coulomb ee ku jira dallacaadda Q 1 waa….

A. 1 NTusaalaha Ciidanka Coulomb Su'aal 12aad

B. 5 N

C. 7 N

D. 10 N

E. 12 N

Dood

Waa la ogyahay in:

Dareeraha barta A (q A ) = 3 µC = 3 x 10 -6 Coulomb

Dareeraha barta B (q B ) = -10 µC = -10 x 10 -6 Coulomb

Dareeraha barta C (q C ) = 4 µC = 4 x 10 -6 Coulomb

Masaafada B ilaa C (r BC ) = 30 cm = 0,3 mitir = 3 x 10 -1 mitir

Masaafada B ilaa A (r BA ) = 30 cm = 0,3 mitir = 3 x 10 -1 mitir

k = 9.10 9 Nm 2 .C -2

Su'aal: Xoogga Coulomb ee natiijada ka soo baxda barta B

Jawaab:

Xoogga korantada ee u dhexeeya dallacaadaha dhibcaha B iyo 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

Dakhalka korantada ee barta B waa taban, dakhalka korontada ee barta C waa togan, sidaas darteed jihada xoogga korontada F BC waxay u socotaa dhanka midig ee barta C.

Xoogga korontada ee u dhexeeya dallacaadaha dhibcaha B iyo 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

Dareeraha korantada ee barta B waa taban, dareeraha korontada ee barta A waa togan, sidaas darteed jihada xoogga korontada F BA waxay kor ugu socotaa barta A.

Labada awoodood ee korontada waxay sameeyaan xagal qumman, sidaas darteed awoodda korantada ee ka dhalata ee ka shaqeysa barta B waxaa lagu xisaabiyaa iyadoo la adeegsanayo qaacidada Pythagorean.

Tusaalaha Ciidanka Coulomb Su'aal 13aad

Jawaabta saxda ah waa B.

8. Saddex dallacaad koronto ayaa la dhigay meelaha geeska saddexagalka ABC iyadoo dhererka dhinaca AB = BC = 20 cm iyo isla cabbirka dallacaadda (q = 2µC) sida sawirka dhinaca (k = 9.109 Nm2.C-2, 1 µ = 10-6) Baaxadda xoogga korontada Tusaalaha Ciidanka Coulomb Su'aal 14aadka shaqaynta barta B waa….

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

Dood

Waa la ogyahay in:
Kharashka ku jira barta A (q)A) = 2 µC = 2 x 10-6 Coulomb
Kharashka barta B (q)B) = 2 µC = 2 x 10-6 Coulomb
Kharashka ku jira barta C (q)C) = 2 µC = 2 x 10-6 Coulomb
Masaafada u dhaxaysa B ilaa C (r)BC) = 20 cm = 0,2 mitir = 2 x 10-1 meter
Masaafada u dhaxaysa B ilaa A (r)BA) = 20 cm = 0,2 mitir = 2 x 10-1 meter
k = 9.109 Nm2.C-2
La weydiiyay: Baaxadda xoogga korontada ee ka shaqeeya barta B
Jawaab:

Xoogga korantada ee u dhexeeya dallacaadaha dhibcaha B iyo C:
FBC = k (q)B)(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 Newton
Dareeraha korantada ee dhibcaha B iyo C waa togan, sidaas darteed jihada xoogga korontada F waaBC dhanka bidix, meel ka fog barta C.

Xoogga korontada ee u dhexeeya dallacaadaha dhibcaha B iyo A:
FBA = k (q)B)(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 Newton
Dareeraha korantada ee dhibcaha B iyo A waa togan, sidaas darteed jihada xoogga korontada F waaBA hoos, meel ka fog barta A.

Labada awoodood ee korontada waxay sameeyaan xagal qumman, sidaas darteed awoodda korantada ee ka dhalata ee ka shaqeysa barta B waxaa lagu xisaabiyaa iyadoo la adeegsanayo qaacidada Pythagorean.
Tusaalaha Ciidanka Coulomb Su'aal 15aadJawaabta saxda ah waa B.

9. Fiiri sawirka soo socda!
Saddex dacwadood Q1,Q2, iyo Q3 waa cidhifka saddexagalka midig ee ABC. Dhererka AB = BC = 30 cm. Waa la ogyahay in Tusaalaha Ciidanka Coulomb Su'aal 16aadk = 9.109 Nm2.C-2 iyo 1 µ = 10-6 ka dibna ciidanka Coulomb ee ka dhashay dacwadda Q1 waa….

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

Dood
Waa la ogyahay in:
Kharashka ku jira barta A (q)A) = 3 µC = 3 x 10-6 Coulomb
Kharashka barta B (q)B) = -10 µC = -10 x 10-6 Coulomb
Kharashka ku jira barta C (q)C) = 4 µC = 4 x 10-6 Coulomb
Masaafada u dhaxaysa B ilaa C (r)BC) = 30 cm = 0,3 mitir = 3 x 10-1 meter
Masaafada u dhaxaysa B ilaa A (r)BA) = 30 cm = 0,3 mitir = 3 x 10-1 meter
k = 9.109 Nm2.C-2
La weydiiyay: Xoogga Coulomb ee natiijada ka soo baxday barta B
Jawaab:
Xoogga korantada ee u dhexeeya dallacaadaha dhibcaha B iyo C:
FBC = k (q)B)(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 Newton
Dakhalka korantada ee barta B waa taban, dakhalka korontada ee barta C-na waa togan, sidaas darteed jihada xoogga korontada F waaBC dhanka midig ee barta C.

Xoogga korontada ee u dhexeeya dallacaadaha dhibcaha B iyo A:
FBA = k (q)B)(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 Newton
Dakhalka korantada ee barta B waa taban, dakhalka korontada ee barta A waa togan, sidaas darteed jihada xoogga korontada F waaBA ilaa barta A.
 
Labada awoodood ee korontada waxay sameeyaan xagal qumman, sidaas darteed awoodda korantada ee ka dhalata ee ka shaqeysa barta B waxaa lagu xisaabiyaa iyadoo la adeegsanayo qaacidada Pythagorean.
Tusaalaha Ciidanka Coulomb Su'aal 18aadJawaabta saxda ah waa B.

Isha su'aasha:

Su'aalaha Imtixaanka Qaranka ee Fiisigiska ee Dugsiga Sare/Dugsiga Sare ee Xirfadda

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