# Electric field due to a linear charge distribution

Let's say I have a uniformly charged wire with some constant linear charge density. Along a line parallel to the $$z$$-axis and passing through $$(0,y,0)$$ and asked to find an electric field at a point $$(p,q,r)$$. Then the vector should be $$p \textbf{i} +(q-y)\textbf{j}+r\textbf{k}$$. My books say the field has no $$z$$ component and it takes vector as $$p\textbf{i}+(q-y)\textbf{j}$$. Can someone explain this?

• Simplify the problem considering as infinite straight line with constant linear charge density $\:\lambda\:$ the $\:x-$axis. You have then rotational symmetry with respect to this axis. The method of solution depends on what are you studying : Coulomb Law or Gauss Law. Commented Aug 14, 2022 at 18:30
• Related : Why don't stationary charge feel force from a current carrying wire?, Figure-03. Commented Aug 14, 2022 at 18:32

Consider two elementary segment of length $$dl_{1}$$and$$dl_{2}$$. They carry the charge $$dq= \lambda dl_{1}= \lambda dl_{2}$$ This charges create at the M two electric fields:$$\overrightarrow{d E_{1} }$$ and $$\overrightarrow{d E_{2} }$$. The electric field intensity only depends on the charge and distance between source and point of observation. So $$d E_{1}=d E_{2}=dE= \frac{dq}{4 \pi \varepsilon _{0} r^{2} }$$ At the point of observation, the electric fields add up. The vertical components cancel out and one gets: $$\overrightarrow{dE} =2.cos( \theta )dE \overrightarrow{i}$$