Two point charge `Q_(a)` and `Q_(b)` are positional at point A and B. The field strenght to the right of charge `Q_(b)` on the line that passes therough the two charges varies according to a law represented schematically in fig. (without employing a definite scale). The field strength is assumed to be positive if its direction coincedes with the positive direction of the x-axis. The distance between the charges is `l=21 cm`.

(a) Find the sign of the charges.
(b) Find the ration between the absolute value of charge `Q_(a)` and `Q_(b)`.
(c ) Find the coordinate x of the point where the field strenght is maximum.

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`.