Question

Two point charge `Q_(a)` and `Q_(b)` are positional at point A and B. The field strength to the right of charge `Q_(b)` on the line that passes through 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 coincides 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 strength is maximum.

Answer 1

Correct Answer - A::B::C
Over charge `Q_2` field intensity is infinite along negatve `x` -axis. Therefore `Q_2` is negative.
Beyon `xgt(l+a)`, field intensity is positive. Therefore `Q_1` is positive
b. At `x=l+a`, field intensity is zero.
`:. (kQ_1)/((l+a)^2)=(kQ_2)/a^2` or `|Q_1/Q_2|=((l+a)/a)^2`
c. Intensity at distance `x` from charge `2` would be
`E=(kQ_1)/((x+l)^2-(kQ_2)/x^2`
For `E` to be maximum `(dE)/(dx)=0`
or `-(2kQ_1)/((x+l)^3)+(2kQ_2)/x^3=0`
or `(1+l/x)^3=Q_1/Q_2=((l+a)/a)^2`
or `1+l/x=((l+a)/a)^(2//3))`
or 1+l/x=((l+a)/a)^(2//3)`
or `x=l/(((l+a)/a)^(2//3)-1)`
or `b=l/(((l+1)/a)^(2//3)-1)`