2 charged line segments are placed on the y axis. L which is between [itex]a \leq y \leq b[/itex], and L' [itex]-b \leq y \leq -a[/itex]. Find the electric field at the point (0,0,h). L has line charge density [itex]\rho[/itex] and L' has [itex]- \rho[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

I believe that the z component of of the electric field will be cancelled out by symmetry. The only component is the y component. In this particular example, the component will be in the negative y direction that is [itex] \vec E = - E_y \vec j[/itex]

I plan to find the electric field generated by just one of the line segments and multiplying by 2.

[tex]\displaystyle{

\vec E = 2 \frac {1}{4 \pi \epsilon} \int_{-b}^{-a} \frac { - \rho y \vec j}{(h^2+y^2)^{\frac {3}{2}}}} = \frac {- \rho}{2 \pi \epsilon} \bigg( \frac {1}{\sqrt{b^2+h^2}} - \frac {1}{\sqrt{a^2+b^2}}\bigg) \vec j

[/tex].

The thing that troubles me is that [tex]\frac {1}{\sqrt{b^2+h^2}} - \frac {1}{\sqrt{a^2+h^2}} < 0[/tex]. E actually points in the positive y direction. What is wrong here?

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# E field calculation

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