COMEDK Exam 2022 Solved Paper

Given, q1=−20µC=−2×10−5C
q2=60µC=6×10−5C
Force of attraction between the both charges, kept at a certain distance r
F1=

Kq1q2

r2

=

K×2×10−5×6×10−5

r2

=

12×10−10K

r2

When both charges are touched, then new charge on both spheres,
q1′=q2′=

q1+q2

2

=

−20+60

2

=20µC
=2×10−5C
∴ Force between the two spheres,
F2=

Kq1′.q2′

r2

=

K.2×10−5×2×10−5

r2

=

4×10−10

r2

K
∴

F1

F2

=

12×10−10K r2

4×10−10 r2

=

12

4

=

3

1

∴ F1:F2=3:1