JEE Mains 29-July-2022 Shift 2 Solved Paper

Let two identical spheres have charge q. And distance between them =r
∴ Force between the spheres (F)=‌

kq2

r2

Now when an identical uncharged sphere C comes in contact with A, charge q on sphere A get's divided equally to both sphere.
So, both sphere A and C have charge =‌

q

2

Now, C get's in contact with B. So their total charge (q+‌

q

2

) gets divided equally.
So, charge on both B and C is =‌

q+ q 2

2

=‌

3q

4

Now, C is place midpoint between A and B.Repulsion force between A and C,
FAC=‌

k( q 2 )( 3q 4 )

(‌ r 2 )2

=‌

4k×3q2

8r2

Repulsion force between B and C,
FBC=‌

k( 3q 4 )( 3q 4 )

(‌ r 2 )2

=‌

4k×9q2

16r2

∴ Net force on C,
Fnet=FBC−FAC
=‌

4k×9q2

16r2

−‌

4k×3q2

8r2

=‌

kq2

r2

(‌

9

4

−‌

3

2

)
=‌

kq2

r2

×‌

3

4

=‌

3

4

F[. as F=‌

kq2

r2

]