Question

Two mutually perpendicular long straight conductors carrying uniformly distributed charges of linear charges densities `lambda_(1)` and `lambda_(2)` are positon at a distance a from each other. How does the interaction between the rods depends on a ?

Answer 1

Correct Answer - `F=(lambda_(1)lambda_(2))/(pi epsilon_(0))`
Let us make two - dimensional view of the situation. Beacuase of symmetry we can say the force on wire 2 `(lambda_2)` will be the along the negative y-direction.

`dF=E_(x) dq cos theta`
`F=int df=int(lambda_(1))/(2 pi epsilon_(0)sqrt(a^(2)+x^(2)))(lambda_(2)dx)cos theta`
where `cos theta=(a)/(sqrt(a^(2)+x^(2)))`
`F=(lambda_(1)lambda_(2))/(2 pi epsilon_(0)) a int_(-oo)^(oo) (dx)/(a^(2)+x^(2))=(lambda_(1)lambda_(2)a)/(2 pi epsilon_(0))xx1/a["tan"^(-1)(x)/(a)]_(-oo)^(oo)`
`F=(lambda_(1)lambda_(2))/(2 pi epsilon_(0))`
It is independent of a.