# Electric field of linear charge distributed over finite wire

Charge is distributed with constant linear charge density $lambda$ on the line of finite length shown in the figure. Find E at P. With the aid of the distance $R2$ and $R1$, express E in terms of angles $alpha_ 2$ and $alpha_1$ shown. Find E for the special case for which $L2 = L1 = L$ and P is in the (xy) plane.

In the given figure $rho$ is the distance from point P to axis z ($rho$ is perpendicular to z). Assume the figure has been redrawn (for simplicity) in the plane (xz) like below:

We can write

$d E_z=(K*lambda*dz/R^2) *cos⁡(beta)=(Klambda*dz/R^2)*(z/R)$

$d E_x=(K*lambda*dz/R^2) *sin⁡(beta)=(Klambda*dz/R^2) *(x/R)$

On the components we have

-the $z$ component

$E_z=Klambdaint_{-L_1}^{L_2} (z*dz/R^3)=Klambdaint_{-L_1}^{L_2}[z*d x/(z^2+x^2)^3]=Klambda*frac{1}{sqrt{z^2+x^2}}|_{-L_1}^{L_2}=$

$=(Klambda/x)*left ( frac{x}{sqrt{x^2+L_2^2}}-frac{x}{sqrt{x^2+L_1^2}}right )=(Klambda/x)*[cos(alpha_2)-cos(alpha_1)]$

and the $x$ component

$E_x=(Klambda x)int_{-L_1}^{L_2}[dz/(z^2+x^2)^3]=Klambda*frac{1}{sqrt{z^2+x^2}}|_{-L_1}^{L_2}=$

$=(Klambda/x)*left ( frac{L_2}{sqrt{x^2+L_2^2}}+ frac{L_1}{sqrt{x^2+L_1^2}}right )=(Klambda/x)*[sin(alpha_2)+sin(alpha_1)]$

Where

$K=1/(4piepsilon_0)$

For $L1=L2$ one has $alpha_1=alpha_2$ so that $cos(..)-cos(..) =0$ and $sin(..)+sin(..) =2*sin(..)$

$E_z=0$ and

$E_x=2Klambda/x*L/sqrt{(x^2+L^2 )}$