TWO IDENTICLE CHARGED RODS OF LENGTH L AND TOTAL CHARGE Q ARE PLACED IN LINE IN SUCH A MANNER THAT THE NEAREST ENDS OF THE RODS ARE AT A DISTANCE D .FIND THE MAGNITUDE OF FORCE ACTING ON ONE ROD WITH RESPECT TO OTHER

            L              l----D------l             L         

---------------------                 ---------------------

l---x---ldxl                           l---t----ldtl         

Let us assume the two rods placed as above,

Consider the small element dx of first rod at a distance x from the point extreme left to it.

Also consider the line element dt on second rod at a distance t from the point extreme left to it.

Force on line element dt due to element dx is given by

dF = k dq₁ dq₂ / (L - x + D + t)²

here dq₁ = Qdx/L and dq₂ = Qdt/L

Thus dF = k Q² dx dt / L² (L - x + D + t)²

Now force due to element dx of rod 1 on the entire rod 2 is

dF₁₂ = ₀∫^L   k Q² dx dt / L² (L - x + D + t)²

or dF₁₂ = (k Q² / L² )dx ₀∫^L   dt / (L - x + D + t)²

Similarly force on rod 2 due to 1 is

F₁₂ = ₀∫^L  [ (k Q² / L² ]dx  ₀∫^L   dt / (L - x + D + t)²

Evaluate this integral to find the force on one rod due to another.

Consider elements dx and dy on rods 1 and 2, at distances x and y from the ends. The given line charges have uniform charge density.

Answer comes out to be

Unless I've made  mistakes while integrating