Question

Derive the expression of ELECTROSTATIC FIELD INTENSITY for an infinitely long wire containing charge "q".

Show the solution without using Gauss's Law.

#Revision Q12​

Answer 1

$\underline{\underline{\sf{\maltese\:\:Question}}}$

• Derive the expression of Electrostatic Field Intensity for an infinitely long wire containing charge $\sf{q}$

$\underline{\underline{\sf{\maltese\:\:Answer}}}$

• Expression of Electrostatic Field Intensity for an infinitely long wire containing charge $\sf{q}$ = $\sf{\lambda/2 \pi \varepsilon_{0} d}$

$\underline{\underline{\sf{\maltese\:\:Explanation}}}$

Consider the figure below, (Refer to Attachment)

• Consider a small element on the wire of length $\sf{dx}$. Charge on the element is, $\sf{d q=\lambda d x}$

Electric field due to the small element is,

$\to\sf{\displaystyle d E=\frac{1}{4 \pi \varepsilon_{0}} \frac{d q}{r^{2}}}$

$\to\sf{\displaystyle d E=\frac{1}{4 \pi \varepsilon_{0}} \frac{\lambda d x}{d^{2}+x^{2}}}$

Total electric field at the given point is only due to perpendicular components since parallel components get cancelled. Therefore,

$\to\sf{\displaystyle E=\int_{-\infty}^{\infty} d E \cos \theta}$

$\to\sf{\displaystyle E=\int_{-\infty}^{\infty} \frac{1}{4 \pi \varepsilon_{0}} \frac{\lambda d x}{d^{2}+x^{2}} \cos \theta}$

$\to\sf{\displaystyle E=\int_{-\infty}^{\infty} \frac{1}{4 \pi \varepsilon_{0}} \frac{\lambda d x}{d^{2}+x^{2}}\left(\frac{d}{\sqrt{d^{2}+x^{2}}}\right)}$

$\to\sf{\displaystyle E=\frac{\lambda d}{4 \pi \varepsilon_{0}} \int_{-\infty}^{\infty} \frac{d x}{\left[d^{2}+x^{2}\right]^{\frac{3}{2}}}}$

$\rule{315}{1}$

First Let us Simplify $\sf{\displaystyle\int_{-\infty}^{\infty} \frac{d x}{\left[d^{2}+x^{2}\right]^{\frac{3}{2}}}}$

• Compute the Indefinite Integral

$---\longrightarrow$ Apply Trigonometric Substitution : $\sf{x = d\tan(u)}$

$-\longrightarrow$ So dx = $\sf{d\sec ^2\left(u\right)du}$

$\sf{\displaystyle=\int \frac{1}{\left(d^2+\left(d\tan \left(u\right)\right)^2\right)^{\frac{3}{2}}}d\sec ^2\left(u\right)du}$

$\sf{\displaystyle =\int \frac{1}{d^2\sec \left(u\right)}du}$

$---\longrightarrow$ Take the Constant Out : $\sf{\int a \cdot f(x) d x=a \cdot \int f(x) d x}$

$\sf{\displaystyle =\frac{1}{d^2}\cdot \int \frac{1}{\sec \left(u\right)}du}$

$---\longrightarrow$ Use the following identity : $\sf{1/\sec(x)=\cos(x)}$

$\sf{\displaystyle=\frac{1}{d^2}\cdot \int \cos \left(u\right)du}$

$---\longrightarrow$ Use the common integral : $\sf{\int \cos (u) d u=\sin (u)}$

$\sf{\displaystyle=\frac{1}{d^2}\sin \left(u\right)}$

$---\longrightarrow$ Substitute back $\sf{u=\arctan \left((1/d)x\right)}$

$\sf{\displaystyle=\frac{1}{d^2}\sin \left(\arctan \left(\frac{1}{d}x\right)\right)}$

$---\longrightarrow$ Use the following identity : $\sf{\displaystyle\sin \left(\arctan \left(x\right)\right)=\frac{x}{\left(1+x^2\right)^{\frac{1}{2}}}}$

$\sf{\displaystyle=\frac{1}{d^2}\left(\frac{\frac{1}{d}x}{\left(1+\left(\frac{1}{d}x\right)^2\right)^{\frac{1}{2}}}\right)}$

$\sf{\displaystyle=\frac{x}{d^2\left(d^2+x^2\right)^{\frac{1}{2}}}}$

$---\longrightarrow$ Add a Constant to the Solution

$\sf{\displaystyle=\frac{x}{d^2\left(d^2+x^2\right)^{\frac{1}{2}}}+C}$

• Compute the Boundaries

$\sf{\displaystyle\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)}$

$\bullet\:\:\sf{\displaystyle\lim _{x \rightarrow-\infty}\left(\frac{x}{d^{2}\left(d^{2}+x^{2}\right)^{\frac{1}{2}}}\right)=-\frac{1}{d^{2}}}$

$\bullet\:\:\sf{\displaystyle\lim _{x \rightarrow \infty}\left(\frac{x}{d^{2}\left(d^{2}+x^{2}\right)^{\frac{1}{2}}}\right)=\frac{1}{d^{2}}}$

$\sf{\displaystyle=\frac{1}{d^2}-\left(-\frac{1}{d^2}\right)}$

$\sf{\displaystyle=\frac{1}{d^2}+\frac{1}{d^2}}$

$\sf{\displaystyle=\frac{2}{d^2}}$

$\rule{315}{1}$

$\to\sf{\displaystyle E=\left(\frac{\lambda d}{4 \pi_{\varepsilon_{0}}}\right)\left(\frac{2}{d^{2}}\right)}$

$\to\sf{\displaystyle E=\frac{\lambda d}{4\pi _{\varepsilon_0}}\cdot \frac{2}{d^2}}$

$\to\sf{\displaystyle E=\frac{\lambda d\cdot \:2}{4\pi _{\varepsilon_0}d^2}}$

$\to\sf{\displaystyle E=\frac{\lambda d}{2\pi _{\varepsilon_0}d^2}}$

$\boxed{\boxed{\to\sf{\displaystyle E=\frac{\lambda}{2 \pi \varepsilon_{0} d}}}}$

Answer 2

EXPLANATION.

Electric field intensity for an infinitely long wire containing charge ''q''.

As we know that, the formula.

⇒ Eₓ = kλ/d[Sin∅₁ + Sin∅₂].

⇒ Ey = kλ/d[Cos∅₁ - Cos∅₂].

As we know that,

In infinitely long wire ∅ varies from ∞ to ∞.

⇒ ∅₁ = 90°

⇒ ∅₂ = 90°.

E(net) = [E₁ + E₂].

⇒ Eₓ = kλ/d [ Sin90° + Sin90°].

⇒ Eₓ = kλ/d [1 + 1].

⇒ Eₓ = 2kλ/d.

⇒ Ey = kλ/d [ Cos90° - Cos90°].

⇒ Ey = kλ/d [ 0 - 0 ].

⇒ Ey = 0.

⇒ E(net) = E₁ + E₂.

⇒ E(net) = 2kλ/d + 0.

⇒ E(net) = 2kλ/d.

As we know that,

⇒ k = 1/4πε°.

⇒ E(net) = 2λ/4πε°d.

⇒ E(net) = λ/2πε°d.