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Atomic number of copper is 29, its atomic weight is 63.5 gm/mole. If two pieces of copper having weight 10 gm are taken and on one of the pieces 1 electron per 1000 atom is transferred to the other piece and there after this these pieces are placed 10 cm apart, then what will be the coulomb force between them.

1 Answer

In 1 mole copper (63.5 gm) there are $6 \times 10^{23}$ atoms (NA = Avogadro number $= 6 \times 10^{23}$ atoms)
Number of atoms $=\large\frac{6 \times 10^{25} \times 10}{63.5}$$=9.45 \times 10^{22}$
For every 1000 atoms, 1 electron is transferred therefore total number of transferred electron is $=\large\frac{9.45 \times 10^{22}}{1000}$$=9.45 \times 10^{19}$
Therefore charge on one piece is $9.45 × 10^{19}\; e$ and on the other will be $(9.45 × 10^{19}e)$
Force when they are kept 10 cm apart
$F= \large\frac{1}{4 \pi \in_0} \frac{(9.45 \times 10^{19})^2e^2}{(10 \times 10^{-2})^2}$
$\quad= \large\frac{9 \times 10^9 \times (9.45 \times 10^{19} )^2 \times (1.6 \times 10^{-19})^2}{(10 \times 10^{-2})^2}$
$\qquad= 2 \times 10^{14}\;N$
Hence A is the correct answer.
answered May 29, 2014 by meena.p
 

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