Two perpendicular charged infinite lines

AI Thread Summary
The discussion centers on calculating the force vector between two perpendicular infinite charged lines, each with a charge density of 10^-3, where one half is positively charged and the other half is negatively charged. The user attempts to derive the force using the equation for charge density and integrating over the lines but encounters convergence issues with the integral. There is confusion regarding the treatment of the opposite charges on each line and how this affects the calculations. The user seeks clarification on whether their approach is correct or if adjustments are needed to account for the charge distribution. The conversation highlights the complexities of integrating forces in electrostatics involving infinite charge distributions.
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Homework Statement


On a long dielectric line a charge with density ##10^{-3}## is applied one half with positive charge and the other half with negative charge. Perpendicular to the first line and 5 cm away from it we have another line with the same charge density and also half of it is positive while the other half is negative. Points on the line where the charge changes from positive to negative, are the closest. (meaning they are 5 cm apart). Determine the vector of force between the lines.


Homework Equations





The Attempt at a Solution



I hope at least some of the things written below are ok:

Density of charge is ##\rho =e/l## meaning ##de=\rho \frac{dx}{l}##.

Vector from the origin to the point on horizontal line ##\vec r=(x,0,h)## if ##h## is the distance between the lines. Vector from the origin to the point on vertical line is ##\vec r^{'}=(0,y,0)##

So finally
##dF=dedE=\frac{\rho ^2}{l^2}dxdy\frac{1}{4\pi \epsilon _0}\frac{(x,-y,h)}{(x^2+y^2+z^2)^{3/2}}##

But if I integrate this from ##-\infty ## to ##\infty ## the integral does not converge... Did I do anything wrong or is this simply the correct answer? :/
 
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e is usually the (negative) charge of a single electron, using it for an arbitrary charge is a bit confusing.

Where did you take into account that the two halves have opposite charge?
 
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