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. $



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