Two particles are placed in air seperated by a distance of 10 cm. 20 esu charge is distributed between them in such a way that the force of repulsion between them is maximum. What is the value of this maximum repulsive force in dyne?

Explanation

Suppose charges on the particles be q esu and `(20-q)` esu.
Force of repulsion between the two particles,
`F=(qxx(20-q))/((10^(2))` dyn
For maximum value of `F,(dF)/(dq)=0`
`:.(dF)/(dq)=1/100 d/(dq)(20q-q^(2))=0`
or `20-2q=0` or `q=10`
So one particle contains 10 esu of charge and the other `(20-10)` or 10 esu of charge, i.e. the total charge is equally sharted between them.
Maximum force `F_("max")=(10xx10)/((10^(2))=1` dyn