Question

Q.46
Two point charges 100 uC and 5 u C are placed at
points A and B respectively with AB = 40 cm. The
work done by external force in displacing the charge
5uC from B to C, where BC = 30 cm , angle
ABC – I 1
--=9x10'Nm2/C2
ABC = 3 and 7ten -
(1) 9 J
(3) 251​

Answer 1

Answer:

Work done in displacing charge 5μc for B to C is

W = 5 × 10–6 (VC – VB)  (1)

Let Q1 = 100μc, Q2 = 5μc

VB = [(kQ1) / d(AB)] & VC = [(kQ1) / d(AC)]

∴ VB = [(k × 100 × 10–6)/(0.4)] & VC = [(k × 100 × 10–6)/(0.5)]

∴ VB = [k/(0.4)] × 10–4 & VC = [k/(0.5)] × 10–4 from (1),

W = 5 × 10–6 [{(k × 10–4)/(0.5)} – {(k × 10–4)/(0.4)}]

    = 5 × 10–6 × 9 × 109[2 × 10–4 – 2.5 × 10–4]

∴    W = 45 × 103 × 10–4[– 0.5]

           = – 45 × 10–1 × 5 × 10–1

          = – 225 × 10–2  

          = – 2.25J

          = – (9/4) J