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  • Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Q 1.30: Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law.

[Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Answers (1)

best_answer

Let AB be a long thin wire of uniform linear charge density λ.

Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.

 

The charge on a small length dx on the line AB is q which is given as q = λdx.

So, according to Coulomb’s law, the electric field at P due to this length dx is

$\begin{aligned} & d E^{\prime}=\frac{1}{4 \pi \epsilon} \frac{\lambda d x}{(P C)^2} \\ & \text { But } P C=\sqrt{h^2+x^2} \\ & \Rightarrow d E=\frac{1}{4 \pi \epsilon} \frac{\lambda d x}{\left(h^2+x^2\right)}\end{aligned}$

This electric field at P can be resolved into two components dEcosθ and dEsinθ. When the entire length AB is considered, then the dEsinθ components add up to zero due to symmetry. Hence, there is only a dEcosθ component.

So, the net electric field at P due to dx is

$$
\begin{aligned}
& \mathrm{dE}^{\prime}=\mathrm{dE} \cos \theta \\
& \Rightarrow d E^{\prime}=\frac{\lambda d x \cos \theta}{4 \pi e\left(h^2+x^2\right)} \\
& \text { In } \triangle \mathrm{POC}, \\
& \tan \theta=\frac{x}{h} \\
& \Rightarrow \mathrm{x}=\mathrm{h} \tan \theta
\end{aligned}
$$

Differentiating both sides w.r.t. θ,

$\frac{d x}{d \theta}=h \sec ^2 \theta$

⇒ dx = h sec2θ dθ …………………….(2)

Also, h2 + x2 = h2 + h2tan2θ

⇒ h2 + x2 = h2(1+ tan2θ)

⇒ h2 + x2 = h2 sec2θ ………….(3)

(Using the trigonometric identity, 1+ tan2θ = sec2θ)

Using equations (2) and (3) in equation (1),

$$
\begin{aligned}
d E^{\prime} & =\frac{\lambda \cos \theta \times h \sec ^2 \theta}{4 \pi \epsilon \times h^2 \sec ^2 \theta} d \theta \\
d E^{\prime} & =\frac{\lambda \cos \theta}{4 \pi \epsilon h} d \theta
\end{aligned}
$$


The wire extends from $\theta=-\frac{\pi}{2}$ to $\theta=\frac{\pi}{2}$ since it is very long.
Integrating both sides,

$$
\begin{aligned}
E^{\prime} & =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\lambda \cos \theta}{4 \pi \epsilon h} d \theta \\
E^{\prime} & =\frac{\lambda}{4 \pi \epsilon h} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta d \theta \\
E^{\prime} & =\frac{\lambda}{4 \pi \epsilon h}\left(\sin \frac{\pi}{2}-\sin \left(-\frac{\pi}{2}\right)\right) \\
E^{\prime} & =\frac{\lambda}{4 \pi \epsilon h} \times 2 \\
E^{\prime} & =\frac{\lambda}{2 \pi \epsilon h}
\end{aligned}
$$
 

This is the net electric field due to a long wire with linear charge density λ at a distance h from it.

In the question  linear charge density = E

therefore 

$E^{\prime}=\frac{E}{2 \pi \epsilon h}$

 

Posted by

HARSH KANKARIA

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