The electrostatic force of repulsion between two positive ions


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The electrostatic force of repulsion between two positively charged ions carrying equal charges is 3.7*10^-9 N , when they are separated by distance of 5 Amg. How many electrons are missing from each ions?

Thanks! _/\_​

Answered by admin @



Given,

Force between the two charges = 3.7*10^{-9}N

Distance between the charges, r = 5 А° = 5*10^{-10}m

q1 = q2 =q

We know,

F = \frac{1}{4\pi\epsilon_0}\frac{q^{2} }{r^{2} }

on putting the values,

3.7*10^{-9} = 9*10^{9} *\frac{q^{2}}{(5*10^{-10})^{2}}

\frac{3.7*10^{-9}*25*10^{20}}{9*10^{9}} = q^{2}

On solving we get,

q^{2} = 10.28*10^{-38} \\\\q = 3.20*10^{-19} C

Thus, Charge = 3.20*10^{-19} C

Now we know,

Q = \pmne              [here charge is positive]

on putting value,

3.20*10^{-19} = n *1.6*10^{-19}\\\\n = \frac{3.20}{1.6}

n = 2

Thus , the number of missing electrons is 2.


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