2.13 Find the electric field a distance \(s\) from an infinitely long wire, which carries a uniform line charge \( \lambda \).

Using Gauss’ law

.

$$\left| \left| \vec{E} \right| \right| ( 2 \pi s l ) = \frac{Q_{enc}}{\epsilon_0} = \frac{ \lambda l }{\epsilon_0}$$

$$\left| \left| \vec{E} \right| \right| = \frac{ \lambda }{ 2 \pi \epsilon_0 s }$$

$$\vec{E} = \frac{ \lambda }{ 2 \pi \epsilon_0 s } \hat{r}$$

The same answer can be obtained computing directly from the definition for the electric field. Assume, without loss of generality, the infinite wire is incident with the x-axis. Furthermore we will compute the electric field at a point \( ( 0, z ) \).

$$ d\vec{ E } = \frac{1}{ 4 \pi \epsilon_0 } \frac{ dq }{ x^2 + z^2 } \ \hat{ r } $$

$$ d\vec{ E } = \frac{1}{ 4 \pi \epsilon_0 } \frac{ \lambda }{ x^2 + z^2 } \ dx \ \hat{ r } $$

$$ \hat{ r } = \frac{x}{ \sqrt{ x^2 + z^2} } \ \hat { i } + \frac{z}{ \sqrt{ x^2 + z^2} } \ \hat{ j } $$

$$ \vec{ E } = \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ – \infty }^{ \infty} \! \frac{ x }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx \ \hat{ i } + \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ – \infty }^{ \infty } \! \frac{ z }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx \ \hat{ j } $$

$$ \vec{ E } = \left( \lim_{ a \rightarrow – \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ a }^{ 0 } \! \frac{ x }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx + \lim_{ k \rightarrow \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ 0 }^{ k } \! \frac{ x }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx \right) \ \hat{ i } + \cdots \\
\cdots + \left( \lim_{ b \rightarrow – \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ b }^{ 0 } \! \frac{ z }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx + \lim_{ l \rightarrow \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ 0 }^{ l } \! \frac{ z }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx \right) \ \hat{ j } $$

$$ \vec{ E } = \lim_{ a \rightarrow – \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ a }^{ 0 } \! \frac{ x }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx $$

$$ v = x^2 + z^2 \hspace{30mm} \frac{1}{2} \ dv = x \ dx $$

$$ \lim_{ a \rightarrow – \infty } \frac{ \lambda }{ 8 \pi \epsilon_0 } \int_{ a^2 + z^2 }^{ z^2 } \ \frac{ 1 }{ v^{\frac{3}{2}} } \ dv \ \hat{ i } $$
$$ \lim_{ a \rightarrow – \infty } \frac{ \lambda }{ 8 \pi \epsilon_0 } \left( \left. – 2 v^{ – \frac{1}{2} } \right|_{ a^2 + z^2 }^{ z^2 } \right) $$
$$ \lim_{ a \rightarrow – \infty } \frac{ \lambda }{ 8 \pi \epsilon_0 } \left( \frac{ – 2 }{ \sqrt{ z^2 }} + \frac{2}{ \sqrt{ a^2 + z^2 } } \right) = – \frac{ \lambda }{ 4 z \pi \epsilon_0 } $$

$$ \lim_{ k \rightarrow \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ 0 }^{ k } \! \frac{ x }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx $$

$$ v = x^2 + z^2 \hspace{30mm} \frac{1}{2} \ dv = x \ dx $$

$$ \lim_{ k \rightarrow \infty } \frac{ \lambda }{ 8 \pi \epsilon_0 } \int_{ z^2 }^{ k^2 + z^2 } \ \frac{ 1 }{ v^{\frac{3}{2}} } \ dv \ \hat{ i } = \lim_{ k \rightarrow \infty } \frac{ \lambda }{ 8 \pi \epsilon_0 } \left( \left. – 2 v^{ – \frac{1}{2} } \right|_{ z^2 }^{ k^2 + z^2 } \right) \hat{i}$$

$$ \lim_{ k \rightarrow \infty } \frac{ \lambda }{ 8 \pi \epsilon_0 } \left( \frac{ – 2 }{ \sqrt{ l^2 + z^2 }} + \frac{2}{ \sqrt{ z^2 } } \right) \hat{i} = \frac{ \lambda }{ 4 z \pi \epsilon_0 } \hat{i} $$

Therefore the sum of the electric field components parallel to \( \hat{i} \) is zero. Now consider the components parallel to \( \hat{j} \).

$$ \lim_{ b \rightarrow – \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ b }^{ 0 } \! \frac{ z }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx \ \hat{ j } $$

$$ x = z \tan \theta \hspace{30mm} dx = z \sec^2 \theta \ d \theta $$

$$\lim_{ b \rightarrow – \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ \tan^{-1} \left( \frac{ b }{ z } \right) }^{ 0 } \! \frac{ z^2 \sec^2 \theta }{ z^3 \left( \tan^2 + 1 \right)^{\frac{3}{2}} } \ d \theta \ \hat{ j } = \lim_{ b \rightarrow \infty } \frac{ \lambda }{ 4 z \pi \epsilon_0 } \int_{ \tan^{-1} \left( \frac{ b }{ z } \right) }^{ 0 } \ \cos \theta \ d \theta \hat{j}$$

$$ \lim_{ b \rightarrow – \infty } \frac{ \lambda }{ 4 z \pi \epsilon_0 } \left( \left. \sin \theta \, \right|_{\tan^{-1} \left( \frac{ b }{ z } \right)}^{ 0 } \right) = \lim_{ b \rightarrow – \infty } \frac{ \lambda }{ 4 z \pi \epsilon_0 } \left( – \sin \left( \tan^{-1} \left( \frac{ b }{ z } \right) \right) \right) \hat{j} = \frac{ \lambda }{ 4 z \pi \epsilon_0 } \hat{j}$$

$$ \lim_{ b \rightarrow \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ 0 }^{ b } \! \frac{ z }{ \left( x^2 + z^2 \right)^{\frac{3}{2}} } \ dx \ \hat{ j } $$
$$ x = z \tan \theta \hspace{30mm} dx = z \sec^2 \theta \ d \theta $$

$$\lim_{ b \rightarrow \infty } \frac{ \lambda }{ 4 \pi \epsilon_0 } \int_{ 0 }^{ \tan^{-1} \left( \frac{ b }{ z } \right) } \! \frac{ z^2 \sec^2 \theta }{ z^3 \left( \tan^2 + 1 \right)^{\frac{3}{2}} } \ d \theta \ \hat{ j } = \lim_{ b \rightarrow \infty } \frac{ \lambda }{ 4 z \pi \epsilon_0 } \int_{ 0 }^{ \tan^{-1} \left( \frac{ b }{ z } \right)} \ \cos \theta \ d \theta \hat{j}$$

$$ \lim_{ b \rightarrow \infty } \frac{ \lambda }{ 4 z \pi \epsilon_0 } \left( \left. \sin \theta \right|_{0}^{ \tan^{-1} \left( \frac{ b }{ z } \right)} \right) \hat{j} = \lim_{ b \rightarrow \infty } \frac{ \lambda }{ 4 z \pi \epsilon_0 } \left( \sin \left( \tan^{-1} \left( \frac{ b }{ z } \right) \right) + 0 \right) \hat{j} = \frac{ \lambda }{ 4 z \pi \epsilon_0 } \hat{j}$$

Therefore

$$\vec{E} = \frac{ \lambda }{ 4 z \pi \epsilon_0 } \hat{j} + \frac{ \lambda }{ 4 z \pi \epsilon_0 } \hat{j} = \frac{ \lambda }{ 2 z \pi \epsilon_0 } \hat{j}$$