1. Laba dhibcood oo dacwad ah, QA = +8 μC iyo QB = -5 μC, waxaa kala soocaya masaafo r = 10 cm. Waa maxay baaxadda xoogga korontadaJoogtada ah k = 8.988 x 109 Nm2C-2 = 9x109 Nm2C-2.
La yaqaan:
Dalac A (q A ) = +8 μC = +8 x 10 -6 C
Dalacaadda B (q B ) = -5 μC = -5 x 10 -6 C
k = 9 x 10 9 Nm 2 C −2
Masaafada u dhaxaysa khidmadaha A iyo B (r AB ) = 10 cm = 0.1 m
La Doonayo : Cabbirka xoogga korontada
Xalka:
Qaacidda sharciga Coulomb :
Baaxadda xoogga korontada:
2. Laba walxood oo la dallacay sida ku cad jaantuska hoose. Q P = +10 μC iyo Q q = +20 μC waxaa kala soocaya masaafo r = 10 cm. Waa maxay baaxadda xoogga korontada ku shaqeeya.
La yaqaan:
Dalacsiinta P (QP ) = +10 μC = +10 x 10 -6 C
Dalacsiinta Q (Q Q ) = +20 μC = +20 x 10 -6 C
k = 9 x 10 9 Nm 2 C −2
Masaafada u dhaxaysa khidmadaha P iyo Q (r PQ ) = 12 cm = 0.12 m = 12 x 10 -2 m
La Doonayo: Cabbirka xoogga korontada
Xalka:
3. Saddex walxood oo la dallacay ayaa loo habeeyay xariiq sida ku cad jaantuska hoose. Dalacsiinta A = -5 μC, dalacsiinta B = +10 μC iyo dalacsiinta C = -12 μC. Xisaabi xoogga elektaroonigga saafiga ah ee ku jira walxaha B sababtoo ah labada dalacsiinta kale.
La yaqaan:
Dalac A (q A ) = -5 μC = -5 x 10 -6 C
Dalacaadda B (q B ) = +10 μC = +10 x 10 -6 C
Dalacaadda C (q C ) = -12 μC = -12 x 10 -6 C
k = 9 x 10 9 Nm 2 C −2
Masaafada u dhaxaysa walxaha A iyo B (r AB ) = 6 cm = 0.06 m = 6 x 10 -2 m
Masaafada u dhaxaysa walxaha B iyo C (r BC ) = 4 cm = 0.04 m = 4 x 10 -2 m
La Doonayo: Cabbirka iyo jihada xoogga elektrostaatigga saafiga ah ee ku yaal walxaha B
Xalka:
Xoogga saafiga ah ee ku jira walxaha B waa wadarta vector-ka ee xoogga F BA ee ku jira walxaha B ee ku jira walxaha A iyo xoogga F BC ee ku jira walxaha B ee ku jira walxaha C.
Xoogga F BA ee ku kacay walxaha B iyadoo loo marayo walxaha A:
Jihada xoogga korontada ku shaqeeya waxay tilmaamaysaa walxaha A (tilmaan bidix).
Xoogga F BC ee ku kacay walxaha B iyadoo la adeegsanayo walxaha A:
Jihada xoogga korontada ku shaqeeya waxay tilmaamaysaa walxaha C (tilmaan dhanka midig).
Xoogga elektaroonigga saafiga ah ee ku jira walxaha B :
F B = F AB – F BC = 675 N – 125 N = 550 Newtons.
Jihada xoogga elektrostaatigga saafiga ah ee ku yaal walxaha B waxay tilmaamaysaa walxaha C (waxay tilmaamaysaa dhinaca midig).
4. +Q 1 = 10 μC, +Q 2 = 50 μC iyo Q 3 waa la kala saaraa sida ku cad sawirka hoose. Waa maxay dallacaadda elektrostaatigga ah ee ku jirta walaxda 3 haddii xoogga elektrostaatigga ah ee saafiga ah ee walaxda 2 uu eber yahay.
La yaqaan:
Dalacaadda 1 (q 1 ) = +10 μC = +10 x 10 -6 C
Dalacaadda 2 (q 2 ) = +50 μC = +50 x 10 -6 C
Masaafada u dhaxaysa kharashyada 1 iyo 2 (r 12 ) = 2 cm = 0.02 m = 2 x 10 -2 m
Masaafada u dhaxaysa dallacaadda 2 iyo dallacaadda 3 (r 23 ) = 6 cm = 0.06 m = 6 x 10 -2 m
Xoogga elektaroonigga saafiga ah ee ku jira walxaha 2 (F2 ) = 0
La Rabay : charge 3 (q 3 )
Xalka:
Xoogga saafiga ah ee ku jira walaxda 2 waa wadarta vektor-ka ee xoogga F 21 ee ku shaqeeya walaxda 2 iyadoo la adeegsanayo walaxda 1 iyo xoogga F 23 ee ku shaqeeya walaxda 2 iyadoo la adeegsanayo walaxda 3.
Xoogga F 21 ee ku shaqeeya walxaha 2 iyadoo loo marayo walxaha 1:
Jihada xoogga korontada ku shaqeeya waxay tilmaamaysaa walxaha 3 (tilmaan dhanka midig).
Ciidanka F23 lagu sameeyay walxaha 2 iyadoo la adeegsanayo walxaha 3: 
Jihada xoogga korontada ku shaqeeya waxay tilmaamaysaa walxaha 1 (tilmaan bidix).
Xoogga elektaroonigga saafiga ah ee ku jira walxaha 2 = 0:
