Consider a long thin wire of uniform linear charge density λ. A point P is taken at a distance r from the midpoint C of the wire. The electric field at point P due to the wire is E.
Let E be the electric field at point P due to the wire
Consider a small length element dx on the wire with centre O, such that OC = x
Let q be the charge on this element, q = λdx
The electric field at A due to the element is
\(\begin{array}{l}dE = \frac{1}{4\pi \epsilon _{0}}\frac{\lambda dx}{OP^{2}}\end{array} \)
OP2 = r2 + x2
\(\begin{array}{l}dE = \frac{1}{4\pi \epsilon _{0}}\frac{\lambda dx}{(r^{2}+x^{2})}\end{array} \)
The electric field (dE) is resolved into two rectangular components, dEcosθ and dEsinθ. dEsinθ is the parallel component, and dEcosθ is the perpendicular component. When the whole wire is considered, the parallel component dEsinθ is cancelled. Only the perpendicular component dEcosθ affects point P.
\(\begin{array}{l}dE = \frac{1}{4\pi \epsilon _{0}}\frac{\lambda dx cos\theta }{(r^{2}+x^{2})}\end{array} \)
In the ΔOCP, x = r tanθ —–(2)
Differentiating equation (1) we get
dx = rsec2θdθ
r2 + x2 = r2 + r2 tan2θ
= r2 (1+ tan2θ) = r2 sec2θ
Substituting in (1)
\(\begin{array}{l}dE= \frac{\lambda rsec^{2}\theta d\theta }{4\pi \epsilon_{0}r^{2}sec^{2}\theta }cos\theta\end{array} \)
\(\begin{array}{l}dE= \frac{\lambda }{4\pi \epsilon_{0}r }cos\theta d\varepsilon \theta\end{array} \)
The wire is of infinite length. Therefore, A and B are infinite distances apart. Therefore, θ varies from -π/2 to +π/2. The electric field intensity at the point P due to the whole wire is
\(\begin{array}{l}E = \int_{-\pi /2}^{\pi /2}\frac{\lambda }{4\pi \epsilon _{0}r}cos\theta d\theta\end{array} \)
=
\(\begin{array}{l}\frac{\lambda}{4 \pi \in_{0} r}[\sin \theta]_{-\pi / 2}^{\pi / 2}=\frac{\lambda}{2 \pi \in_{0} r}\end{array} \)
