Electrostatics Question 201
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) [KCET 2002]
Options:
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
Show Answer
Answer:
Correct Answer: C
Solution:
Number of atoms in given mass
$ =\frac{10}{63.5}\times 6.02\times 10^{23} $ = 9.48 x 1022
Transfer of electron between balls
$ =\frac{9.48\times 10^{22}}{10^{6}} $ = 9.48 x 1016
Hence magnitude of charge gained by each ball. Q = 9.48 x 1016 x 1.6 x 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. $
BETA
