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 ratio 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.

Options

A. `+,-`
B. `-,+`
C. `+,+`
D. `-,-`

Answer

A

Explanation

Electric field at (2) tends to ` -oo`, hence the charge at (2) should be negative. There is a neutral point to the right of charges. This is possible onlu when the charge at (1) should be positive. Hence `Q_(1)` is positive and `Q_(2)` is negative. At netural point, `vecE_(1)=vecE_(2)`.
`(1)/(4 pi epsilon_(0)) (Q_(1))/((a+l)^(2))=(1)/(4 pi epsilon_(0))(Q_2)/(a^(2))` or `(Q_(1))/(Q_(2))=((a+l)/(a))^(2)`
Electric field at any position to the right of charges is
`E=E_(1)-E_(2)=(1)/(4 pi epsilon_(0))(Q_(1))/((l+x)^(2))-(1)/(4 pi epsilon_(0))(Q_(2))/(x^(2))`
For the maximum value of E,
`(dE)/(dx)=0 or d/(dx) [(1)/(4 pi epsilon_(0))(Q)/((l+x)^(2))-(1)/(4 pi epsilon_(0))(Q_(2))/(x^(2))]=0`
`Q_(1) (-2)/((l+x)^(3))-Q_(2)(-2)/(x^(3))=0`
or `(Q_(1))/((l+x)^(3))=(Q_(2))/(x^(3))` or `((l+x)/(x))^(3)=(Q_(1))/(Q_(2))`
or, `(l+x)/(x)=((Q_(1))/(Q_(2)))^(1//3)` or `(l)/(x) = ((Q_(1))/(Q_(2)))^(1//3)-1`
or `x=(l)/(((Q_(1))/(Q_(2)))^(1//3)-1)`.