Imibuzo emi-3 mayelana ne-Potential difference equation
1. Ishaja kagesi ihanjiswa ensimini kagesi efanayo enamandla angu-2√3 N ibanga elingama-20 cm. Uma isiqondiso samandla sise-engeli engu-30 o kuya ekufudukeni kweshaja kagesi, uyini umehluko emandleni kagesi angase abe khona ezindaweni zokuqala nezokugcina zeshaja kagesi.
Kwaziwa:
Amandla (F) = 2√3 N
Ibanga (ama) = 20 cm = 0.2 m
I-engeli (θ) = 30 o
Kufunwa: Umehluko wamandla kagesi
Isixazululo:
Umsebenzi wokudlulisa ishaja +q kusuka ku-a kuya ku-b ulingana nomehluko wamandla kagesi akhona kumaphuzu a-a no-b.
ΔEP = ΔW
Uma isiqondiso samandla u-F maqondana nesiqondiso sokudluliselwa kweshaja +q sibekwe nge-engeli u-θ, khona-ke umsebenzi wokudluliselwa kweshaja +q kusuka ku-a kuya ku-b uthi:
ΔW = F Δs cos θ = ( 2 √ 3 )(0,2)(cos 30) = (0,4 √ 3 )(0,5 √ 3 ) = (0,2)(3) = 0,6 Joule
2. Uma ukushaja kanye nomthamo wama-capacitor kwaziwa ukuthi kungu-5 µC kanye no-20 µF, ngokulandelana, nquma umehluko ongaba khona wama-capacitor.
Kwaziwa:
Ishaja kagesi (q) = 5 µC = 5 x 10 -6 C
Umthamo we-capacitor (C) = 20 µF = 20 x 10 -6 F
Okufunwayo: Umehluko wamandla we-capacitor (V)
Isixazululo:
Ifomula yomehluko wamandla we-capacitor, umthamo we-capacitor kanye nokushaja:
V = q / C = (5 x 10 -6 ) / 20 x 10 -6 = 5/20 = 0,25 Volts
3. Amashaja amabili QA = -4 µC kanye QB = 8 µC aqhelelene ngamasentimitha angu-16. Thola amandla kagesi endaweni ephakathi nendawo phakathi kwamashaja amabili.
Kwaziwa:
Ishaja kagesi 1 (Q A ) = 4 x 10 -6 C
Ishaja kagesi 2 (Q B ) = 8 x 10 -6 C
Ibanga (r) = 8 cm = 0,08 m
I-Coulomb constant (k) = 9 × 10 9 Nm 2 /C 2
Okufunwayo: Amandla kagesi ku-C (V C )
Isixazululo:
V A = k Q / r = (9 × 10 9 )(4 x 10 -6 ) / 0,08 = (36 × 10 3 ) / 0,08 = -450 Volts
V B = k Q / r = (9 × 10 9 )(8 x 10 -6 ) / 0,08 = (72 × 10 3 ) / 0,08 = 900 Volts
Amandla kagesi ku-C (V C ) = 900 – 450 = 450 Volts
