BITSAT 2015 Paper

Initial potential energy of the system
=

1

4π∊0

[

q2

a

+

q2

a

+

q2

a

] =

1

4π∊0

(

3q2

a

)
= 9 × 109 (3×

(0.1)2

1

) = 27 × 107 J
Let charge at A is moved to mid-point O, Then final potential energy of the system
Uf =

1

4π∊0

[

2q2

(a∕2)

+

q2

a

]
= 5 ×

1

4π∊0

(

q2

a2

) = 45 × 107 J
Work done = Uf−Ui = 18 × 107 J
Also, energy supplied per sec = 1000 J (given)
Time required to move one of the mid-point of the line joining the other two
t =

18×107

1000

= 18 × 104 s = 50 h