Solution:
Work done in displacing charge of 5 μC from B to C is
W=5×10-6(VC-VB) where
VB=9×109×100×10-60.4=94×106 V and VC=9×109×100×10-60.5=95×106 V So W=5×10-6×(95×106-94×106)=-94 J
Two point charges 100 μC and 5 μC are placed at points A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 μC from B to C, where BC = 30 cm, angle ABC =π2 and 14πε0=9×109Nm2/C2 ,is
Two point charges 100 μC and 5 μC are placed at points A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 μC from B to C, where BC = 30 cm, angle ABC =π2 and 14πε0=9×109Nm2/C2 ,is
