Question : Two charges of 5 µC & 20 µC are placed 20 cm apart. Calculate the work done in bringing the charges closer to 15 cm apart.
Doubt by Deepanshi U.
Solution :
q1 = 5 µC
= 5 × 10-6 C
q2 = 20 µC
= 20 × 10-6 C
rf = 15 cm
= 15 × 10-2 m
ri = 20 cm
= 20 × 10-2 m
W = Uf - Ui
W = [kq1q2/rf] - [kq1q2/ri]
W = kq1q2 [1/rf - 1/ri]
W = 9×109× 5 × 10-6 × 20 × 10-6 [1/(15 × 10-2) - 1/(20 × 10-2)]
W = 9 × 10-1 × [100/15 - 100/20]
W = 0.9 ×100×[(1/15)-(1/20)]
W = 90 × [(20-15)/300]
W = 90 × 5/300
W = 90 × (1/60)
W = 90/60
W = 3/2
W = 1.5 J
