COMEDK Exam 2018 Solved Paper

Given, Q1=12µF=12×10−6F
Q2‌=−8µF
‌=−8×10−6F
Initial value of electrostatic force of attraction between both spheres,
F1‌=‌

1

4πε0

‌

Q1Q2

r2

‌=‌

1

4πε0

⋅‌

12×10−6×8×10−6

r2

F1‌=‌

1

4πε0

‌

96×10−12

r2

. . . (i)
When both charged spheres are kept in contact, then new charges on the spheres will be
Q1′=Q2′‌=‌

Q1+Q2

2

‌=‌

12+(−8)

2

=2µF=2×10−6F
∴ Final value of electrostatic force between the spheres,
F2‌=‌

1

4πε0

‌

Q1′×Q2′

r2

‌=‌

1

4πε0

⋅‌

2×10−6×2×10−6

r2

F2‌=‌

1

4πε0

⋅‌

4×10−12

r2

. . . (ii)
From Eqs. (i) and (ii), we get
∴‌‌

F1

F2

=‌

‌ 1 4πε0 ⋅ 96×10−12 r2

1 4πε0 ⋅‌ 4×10−12 r2

⇒‌‌‌

F1

F2

=‌

24

1

⇒F1:F2=24:1