Question

Two copper balls, each weighing 10g are kept in air 10 cm apart. If one electron from every $10^{6}$ atoms is transferred from one ball to the other, the coulomb force between them is (atomic weight of copper is 63.5)

• A. $2.0 \times 10^{10}$N
• B. $2.0 \times 10^{4}$N
• C. $2.0 \times 10^{8}$N
• D. $2.0 \times 10^{6}$N

Answer 1

Answer: (C)

$2.0 \times 10^{8}$N

Answer 2

Number of atoms in given mass

$= \frac{10}{63.5} \times 6.02 \times 10^{23}$

= 9.48 × 10²²

Transfer of electron between balls

$= \frac{9.48 \times 10^{22}}{10^{6}}$

= 9.48 × 10¹⁶

Hence magnitude of charge gained by each ball.

Q = 9.48 × 10¹⁶ × 1.6 × 10^–19 = 0.015 C

Force of attraction between the balls

$F = 9 \times 10^{9} \times \frac{(0.015)^{2}}{(0.1)^{2}} = 2 \times 10^{8}N.$