# Electric field inside two parallel lines of charge

At what force will a particle be subjected to inside two parallel line charges with the same length and charge density. I get that in the middle the forces will cancel, but will they cancel if the particle is closer to the right or to the left? My teacher says yes and explained using integral but I really didn't understand why they cancel, seems to me that the net E would point to the left (if particle is close to right rod).

• You use the terms "right" and "left". That doesn't help because we don't know the orientation of the line charges. If you draw them on a piece of paper lying flat on a desk, are the line charges running left and right, or toward and away from you? Are they effectively infinitely long or finite? Is the charge density uniform or non-uniform (you said "same" but I interpreted that as one line being the same as the other, not necessarily uniform). – Bill N Aug 28 '15 at 21:14
• To build on Bill N's comment - if the wires are in the X direction, at +- y offset in the Y direction, and centered in X, are you asking about offsets in X or Y? And can you confirm the wire is finite, and uniformly charged? – Floris Aug 28 '15 at 22:29
• Excuses for miss information. On a xy plane, the lines are x=1, x=6 and the particle is on (2,4), for example. Assuming both have uniform charges and have the same lenght, which we can say it extends from y=0 to y=10. Let me know if I miss something and thanks. – João Pedro Aug 29 '15 at 2:52
• what particle is that which you are putting in between? – Jolie Aug 29 '15 at 3:58
• Only matter the position of the particle, I want to know he eletric field there! But it could be a test particle positive, and the rods also positive. – João Pedro Aug 29 '15 at 17:52

As given on the hyperphysics site, the field due to a finite line charge can be calculated. For a point that is a distance $z$ from a line extending from $x=-a$ to $x=b$, with a charge density $\lambda$ per unit length, the field is
$$E_z = \frac{\lambda}{4\pi\epsilon_0 z}\left(\frac{b}{\sqrt{z^2+b^2}}+\frac{a}{\sqrt{z^2+a^2}}\right)$$
It follows that when the source is not centered, the field due to the more distant line source will be weaker ($z$ is greater). In the limit where the line is very short, the problem reduced to that of two point sources - that should make the problem intuitively easy to solve.