Question

two positively charged particles each of mass 1.7 *10 power minus 27 kg and carrying a charge of 1.6 *10 power minus 19 coulomb are placed r distance apart if each one experience repulsive force equal to its weight find r

Answer 1

hey
here is ur answer
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Answer : 0.117m


Explanation : F=Rq^2/r^2=mg


=9×10^9×(1.6×10^−19)^2/r^2=1.7×10^−27×10

r=1.17×10^−1m.
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I hope this will help
#Prem✌✌

Answer 2

Answer:

Distance, r = 0.11 m

Explanation:

It is given that,

Mass of charged particles, $m=1.7\times 10^{-27}\ kg$

Charge, $q=1.6\times 10^{-19}\ kg$

Each one experience repulsive force equal to its weight i.e. mg

So, $k\dfrac{q_1q_2}{r^2}=mg$

$r=\sqrt{\dfrac{kq_1q_2}{mg}}$

$r=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.7\times 10^{-27}\times 9.8}}$

r = 0.11 m

Hence, this is the required solution.