Question

Two copper balls, each weighting $10$g are kept in air $10$cm apart. If one electron from every ${10}^6$ atom is transferred from one ball to the other, the coulomb force between them is (atomic weight of copper is $63.5$).

• A. $2\times {10}^{10}$N
• B. $2\times {10}^4$N
• C. $2\times {10}^8$N
• D. $2\times {10}^6$N

Answer 1

The correct option is C $2\times {10}^8$N

Given weight of copper ball $=10$ $g$

We know that, atoms contains in $63.5$ $g$ of copper $=6.022\times {10}^{23}$ atom

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

According to question from, ${10}^6$ atom we withdraw $1e^{-}$. Therefore, using urinary method

${10}^6$ atom $=1e^{-}$

$\Rightarrow 1$ atom $=\frac{1}{{10}^6}{e}^{-}$

$\Rightarrow \frac{6.022\times {10}^{24}}{63.5}atom=\frac{6.022\times {10}^{24}}{63.5\times {10}^6}{e}^{-}=9.483\times {10}^{16}{e}^{-}$

Charge on $1e^{-}=1.6\times {10}^{-19}$ $C$

Charge on $9.483\times {10}^{16}{e}^{-}=9.483\times {10}^{16}\times 1.6\times {10}^{-19}=1.517\times {10}^{-2}$ $C$

Now, from $1$ sphere that much of charge is withdrawn and to one it is added.

Therefore,

$Q_1=1.517\times {10}^{-2}$ $C$

$Q_2=-1.517\times {10}^{-2}$ $C$

$r=10\times {10}^{-2}$ $m={10}^{-1}$

$F^{\gamma }=\frac{K{Q}_1{Q}_2}{r^2}=\frac{9\times {10}^9\times 1.517\times {10}^{-2}\times (-1.517\times {10}^{-2})}{{10}^{-2}}=2.0711\times {10}^8$ $N$