Question

Two copper balls each weighing 10 g are kept in air 10 cm apart. If one electron from every ${{10}^{6}}$ atom is transferred from one ball to the other one, the coulomb force between them is (atomic weight of copper is 63.5):
$\begin{aligned} & \text{A}\text{. }2.0\times {{10}^{10}}N \\\ & \text{B}\text{. }2.0\times {{10}^{4}}N \\\ & \text{C}\text{. }2.0\times {{10}^{5}}N \\\ & \text{D}\text{. }2.0\times {{10}^{6}}N \\\ \end{aligned}$

Answer 1

Hint: Find the number of atoms in each copper ball. Then find the number of electrons transferred from one ball to another. Obtain the mathematical expression for coulomb force. Obtain the charge on each ball and put values on the equation to find the answer.

Complete step by step answer:
We have two copper balls of mass $m=10g$ each.
The balls are kept a distance of $d=10cm=0.1m$
Now, we need to find the number of atoms.
The number of atoms in a copper ball of mass 10 g is,
$n=\dfrac{m}{63.5}\times {{N}_{A}}$
Where, m is the mass of copper ball, 63.5 is the atomic weight of copper and ${{N}_{A}}$ is the Avogadro’s number with value ${{N}_{A}}=6.02\times {{10}^{23}}$
So,
$\begin{aligned} & n=\dfrac{10}{63.5}\times 6.02\times {{10}^{23}} \\\ & n=\dfrac{6.02}{63.5}\times {{10}^{24}} \\\ \end{aligned}$
Now, one electron is transferred from every ${{10}^{6}}$ atom from one ball to another. So, the number of electrons transferred will be,
$\begin{aligned} & {{n}_{e}}=\dfrac{1}{{{10}^{6}}}\times \dfrac{6.02}{63.5}\times {{10}^{24}} \\\ & {{n}_{e}}=\dfrac{6.02}{63.5}\times {{10}^{18}} \\\ \end{aligned}$
So, the ball to which the electron is transferred will be negatively charged and the ball from which the electron is transferred will be positively charged. Since the charge transfer is only between the two balls the charge on each ball will be equal but opposite.
Charge on each ball will be,
$\begin{aligned} & q=1.6\times {{10}^{19}}\times \dfrac{6.02}{63.5}\times {{10}^{18}} \\\ & q=1.516\times {{10}^{-2}}C \\\ \end{aligned}$
Now, the coulomb force on between two charges ${{q}_{1}}$ and ${{q}_{2}}$ at distance r apart is given by,
$F=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Putting the values on the above equation, the coulomb force between the balls is given by,
$\begin{aligned} & F=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{{{\left( 1.516\times {{10}^{-2}} \right)}^{2}}}{{{0.1}^{2}}} \\\ & \Rightarrow F=9\times {{10}^{9}}\times \dfrac{{{\left( 1.516\times {{10}^{-2}} \right)}^{2}}}{{{0.1}^{2}}} \\\ & \Rightarrow F=2.02\times {{10}^{8}}N \\\ & \Rightarrow F\approx 2.0\times {{10}^{8}}N \\\ \end{aligned}$
So, the correct option is (C).

Note:
$\dfrac{1}{4\pi {{\in }_{0}}}$ is the proportionality constant in the mathematical expression for coulomb’s force with value $9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$ . ${{\in }_{0}}$ is the permittivity of free space with value ${{\in }_{0}}=8.854\times {{10}^{-12}}F{{m}^{-1}}$ (Farad per meter). Coulomb force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.