Two pieces of copper, each weighing 0.01 kg are placed at a distance of 0.1 m from each other.
One electron from per 1000 atoms of one piece is transferred to other piece of copper. What will be the coulomb force between two piece after the transfer of electrons?
(Atomic weight of copper is 63.5 g/mole. Avagadro’s number = 6 x 10²³/gram mole.)

Solution

Number of atoms in each piece of copper

= \frac{6 \times 10^{23} \times 0.01 \times 10^{3}}{63.5} = 9.45 \times 10^{22}

_atoms

Number of electrons transferred =

\frac{1}{1000} \times 9.45 \times 10^{22}

n = 9.45 \times 10^{19}

∴ C h a r g e s o n t h e e a c h p i e c e a f t e r t r a n s f e r

q_{1} = q_{2} = \pm ne = \pm 9.45 \times 10^{19} \times 1.6 \times 10^{- 19} = \pm 15.12 C r = 0.1 m

Coulombic force acting between two piece after the transfer of electrons is,

F = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r^{2}} = 9 \times 10^{9} \frac{(15.12)^{2}}{(0.1)^{2}} = 2.06 \times 10^{14} N

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