Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  AIMS  >>  Class12  >>  Physics  >>  Electric Charges and Fields
0 votes

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.

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App