Question

Two copper balls each weighing 10 g 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 $2.0\times {10}^{8}N$

$\text{Given:}$ Mass of balls, $m=10g$

Distance between the balls, $r=10cm=0.1m$

$\text{Step 1: No. of electrons transferred from one ball to another}$

As number of atoms/molecules in the Mass-number grams of a element is equal to Avogadro number,

Number of atoms in 63.5 g of copper = $6.022\times {10}^{23}$

$\Rightarrow$Number of atoms in 1 g of copper = $\frac{6.022\times {10}^{23}}{63.5}$

$\Rightarrow$Number of atoms in 10g of copper = $\frac{10}{63.5}\times 6.022\times {10}^{23}=0.948\times {10}^{23}$

For every ${10}^6$ atoms, one electron is transferred.

$\therefore$ Number of electrons transferred from 1 ball to another = $\left(n\right)=0.948\times {10}^{23}/{10}^6=0.948\times {10}^{17}$

$\text{Step 2: Charge acquired by the balls :}$

Charge acquired by balls is, $q=ne=1.6\times {10}^{-19}\times 0.948\times {10}^{17}=1.52\times {10}^{-2}C$

$\text{Step 3: Electrostatic force between the balls :}$

So electrostatic force between the balls is,

$F=\frac{q^2}{4\pi {ϵ}_0{r}^2}=9\times {10}^9\times \frac{(1.52\times {10}^{-2}{)}^2}{{0.1}^2}\approx 2\times {10}^{8}N$

Hence, Option C is correct.