Question

Two positively charged particles each of mass 1.7x10-27 kg, carrying a charge of 1.6x10-19c are kept at a certain distance in air. If each charge experiences a repulsive force equal to its weight, find the distance of separation between the charges. (a)

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

Answer 2

Given that,

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

Charge on particles, $q=1.6\times 10^{-19}\ C$

To find,

The distance of separation between the charges.

Solution,

It is mentioned in the question that each charge experiences a repulsive force equal to its weight. So, it can be written as :

$\dfrac{kq^2}{d^2}=mg$

Here,

d is separation between the charges

So,

$d=\sqrt{\dfrac{kq^2}{mg}} \\\\d=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.7\times 10^{-27}\times 9.8}} \\\\d=0.117\ m$

So, the separation between the charges is 0.117 m.