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

The correct option is C $2.0\times {10}^{8}N$


Number of atoms in given mass $=\frac{10}{63.5}\times 6.02\times {10}^{23}$
$=9.48\times {10}^{22}$

Transfer of electron between balls $=\frac{9.48\times {10}^{22}}{{10}^6}$
$=9.48\times {10}^{16}$
Hence magnitude of charge gained by each ball.
$Q=9.48\times {10}^{16}\times 1.6\times {10}^{-19}=0.015C$
Force of attraction between the balls
$F=9\times {10}^9\times \frac{(0.015{)}^2}{(0.1{)}^2}=2\times {10}^{8}N.$