Question

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $\lambda$ without using Gauss's law.
[Hint: Use Coulombs law directly and evaluate the necessary integral.]

Answer 1

Take a long thin wire of uniform linear charge density $\lambda$.

Consider a point A at a perpendicular distance $l$ from the mid-point $O$ of the wire, as shown in the following figure.

Let $E$ be the electric field at point $A$ due to the wire, $XY$.

Consider a small length element $dx$ on the wire section with $OZ=x$

Let $q$ be the charge on this piece.

$\therefore q=\lambda dx$

Electric field due to the piece,

$dE=\frac{1}{4\pi {\in }_0}\frac{\lambda dx}{(AZ{)}^2}$

However, $AZ=\sqrt{(l^2+x^2)}$

$\therefore dE=\frac{\lambda dx}{4\pi {\in }_0(l^2+x^2)}$

The electric field is resolved into two rectangular components. $dEcos\theta$ is the perpendicular component and $dEsin\theta$ is the parallel component.

When the whole wire is considered, the component $dEsin\theta$ is cancelled.

Only the perpendicular component $dEcos\theta$ affects point A.

Hence, effective electric field at point A due to the element $dx$ is $d{E}_1$.

$\therefore d{E}_1=\frac{\lambda dxcos\theta }{4\pi {\in }_0(x^2+l^2)}$ ... (1)

In $△AZO$,

$tan\theta =\frac{x}{l}$

$x=ltan\theta$ ... (2)

On differentiating equation (2), we obtain

$dx=l{\sec }^2\left(\theta \right)d\theta$

$d{E}_1=\frac{\lambda cos\left(\theta \right)d\theta }{4\pi {\epsilon }_{o}l}$

$E_1={\int }_{-\pi /2}^{\pi /2}\frac{\lambda \cos \left(\theta \right)d\theta }{4\pi {\epsilon }_{o}l}$

Solving, $E_1=\frac{\lambda }{2\pi {\epsilon }_{o}l}$