Question

The point charges A=+3nc and b=+1nc are placed 5cm a part in air. The work

done to move charge B towards A by 1cm is



a. 2.0 x 10 J
b 1.35x10 J
c. 2.7 x10 J
d. 1.21x10 J
​

Answer 1

Explanation:

2.0 x 10 J

b 1.35x10 J

c. 2.7 x10 J

d. 1.21x10 J

Answer 2

Answer :

your answer is = 1.35×10−⁷J

Step by step :

Initial distance between the charges QA and QB,  d = 5 cm =0.05 m

5 cm =0.05 mPotential energy of the system initially,  Ui=dkQAQB   KQaQb Where k = 9×10⁹

d

∴  Ui = 9×10⁹×3×10‐⁹×1×10-⁹ = 5.4×10-⁷J

0.05

Now the charge B is moved by 1cm towards A.

Now the charge B is moved by 1cm towards A.Thus new distance between the charges d′ = 5 − 1 = 4 cm = 0.04 m

Potential energy of the system initially,  Uf =

KQaQb

d

∴  Uf = 9×10⁹×3×10−⁹×1×10−⁹ = 6.75×10−⁷J

0.04

Thus work done  W = Uf − Ui

∴ W = (6.75−5.4)×10−⁷= 1.35×10−⁷J