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

Options

A. `(l)/(((l+x_(1))/(x_(1)))^((2)/(3))-1)`
B. `(l)/(((l+x_(1))/(x_(1)))^((1)/(3))-1)`
C. `(l)/(((l+2x_(1))/(x_(1)))^((2)/(3))-1)`
D. `(l)/(((l+2x_(1))/(x_(1)))^((1)/(3))-1)`

Answer

A

Explanation

`because` Electric field near point b is `- infty`
`:. 'b'` shluld be negative electric field at `x_(1)` is O which possible only if 'a' and 'b' are of opposite sign
`:. 'a'` is positive
Charge b is negative and charge a is positive

E at `A = 0 (A = ) E =2)`
`rArr(|Q_(a)|)/((l+x_(1))^(2))=(|Q_(b)|)/((x_(1))^(2))`
`|(Q_(a))/(Q_(b))| =((l+x_(1))/(x_(1)))^(2)=(1+(l)/(x_(1)))^(2)`
W at a general X
`(K|Q_(a)|)/((l+x^(2)))-(K|Q_(b)|)/((x^(2)))`
`= K|Q_(a)|{(1)/((l+x)^(2))-|(Q_(b))/(Q_(a))|(1)/(x^(2))}`
If E is a maximum, `(dE)/(dx) = 0`
`rArr (-2)/((l+x)^(3))+((x_(1))/(l+x_(1)))^(2)(2)/(x^(3))=0`
`(l+x)^(2)=x^(3)((l+x_(1))/(x_(1)))^(2)`
`l+x = x((l+ x_(1))/(x_(1)))^((2)/(3))`
`x_(2)=(l)/(((l+x_(1))/(x_(1)))^((2)/(3))-1)`
`x_(2)=(l)/(((l+x_(1))/(x_(1)))^((2)/(3))-1),|Q_(a)/(Q_(b))|=(1+(l)/(x_(1)))^(2)`
Charge b is negative and charge a is positive.