Two point like charges `Q_(1)` and `Q_(2)` are positioned at point 1 and 2. The firld intensity to the right of the chaege `Q_(2)` on the line that passes through the two charges varies accoreing to a law that is represented schematically in fig. The field intensity is assumed to be positive if its direction coincides with the positive direction on the x-axis. The distance between the charges is l.

The ratio of the absolute values of the charges `|Q_(1)//Q_(2)|` is

Answer

(i) x=7 cm (ii) x'=21 cm (iii) x=28 cm

Explanation

(a) Electric field near B and the right of B is along the positive x-axis so the sign of `Q_(b)` should be negative. There is a neutral point at x=24 cm, so the sign of `Q_(a)` should be opposite of `Q_(b)`. Hence `Q_(a)` should be positive.
(b) At netural point (at x=24 cm),
`k(Q_(a))/(24^(2))=k(Q_(b))/(3^(2))` or `|(Q_(a))/(Q_(b))|=((24)/(3))^(2)=64`
(c ) Electric field to the right of charges (at x=x).
`E=k(Q_(a))/(x^(2))-k(Q_(b))/((x-21)^(2))`
For field to be maximum
`(dE)/(dx)=0`
or `(kQ_(a)(-2))/(x^(3))-(kQ_(b)(-2))/((x-21)^(3)) =0` or `x=28 cm`.