First slide

Electrostatic force

Question

Two copper balls, each weighing 10g are kept in air 10 cm apart. If one electron from every 10⁶ atoms is transferred from one ball to the other, the coulomb force between them is (atomic weight of copper is 63.5)

Easy

A

$2 .0 \times 10^{10} N$
B

$2 .0 \times 10^{4} N$
C

$2 .0 \times 10^{8} N$
D

$2 .0 \times 10^{6} N$

Solution

Number of atoms in given mass $= \frac{10}{63 .5} \times 6 .02 \times 10^{23}$
$= 9 .48 \times 10^{22}$
Transfer of electron between balls $= \frac{9 .48 \times 10^{22}}{10^{6}}$
$= 9 .48 \times 10^{16}$
Hence magnitude of charge gained by each ball.
$Q = 9 .48 \times 10^{16} \times 1 .6 \times 10^{- 19} = 0 .015 C$
Force of attraction between the balls
$F = 9 \times 10^{9} \times \frac{(0 .015)^{2}}{(0 .1)^{2}} = 2 \times 10^{8} N$

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