Find the electric field a distance z above the midpoint of a straight line segment of length 2L that carries a uniform line charge λ (Fig. 2.6).

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Accepted Answer

The simplest method is to chop the line into symmetrically placed pairs (at ±x) quote the result of Ex. 2.1 (with d/2 → x, q → λ dx), and integrate (x : 0 → L).But here’s a more general approach:[latex]^{3}[/latex]

[latex]r=z\hat{z} , \acute{r}=x\hat{x} , d\acute{l} =dx ; [/latex]

[latex]η=r-r^′=z\hat{z}-x\hat{x}, η=\sqrt{z^2+x^2}, \hat{η}=\frac{η}{η}=\frac{z\hat{z}-x\hat{x}}{\sqrt{z^2+x^2}}[/latex].

[latex]E=\frac{1}{4\pi \varepsilon _{0} }\int_{-L}^{L}{\frac{\lambda }{z^{2}+x^{2}} }\frac{z\hat{z}-x\hat{x}}{\sqrt{z^{2}+x^{2}} }dx[/latex]

[latex]=\frac{\lambda }{4\pi \varepsilon _{0} }\left[z\hat{z}\int_{-L}^{L}{\frac{1 }{(z^{2}+x^{2})^{3/2}}dx }-\hat{x}\int_{-L}^{L}{\frac{x }{(z^{2}+x^{2})^{3/2}}dx }\right] [/latex]

[latex]=\frac{\lambda }{4\pi \varepsilon _{0} }\left[z\hat{z}(\frac{x }{z^{2}\sqrt{z^{2}+x^{2}}} )\mid ^{L}_{-L}-\hat{x}(-\frac{1 }{\sqrt{z^{2}+x^{2}}} )\mid ^{L}_{-L} \right][/latex]

[latex]=\frac{1}{4\pi \varepsilon _{0}}\frac{2\lambda L}{z\sqrt{z^{2}+L^{2}} }\hat{z}.[/latex]

[latex]^{3}[/latex]Ordinarily I’ll put a prime on the source coordinates, but where no confusion can arise I’ll remove the prime to simplify the notation.

For points far from the line (z >>L),

[latex]E≅\frac{1}{4\pi \epsilon_0} \frac{2 \lambda L}{z^2}[/latex].

This makes sense: From far away the line looks like a point charge q = 2λL. In the limit L → ∞, on the other hand, we obtain the field of an infinite straight wire:

[latex]E=\frac{1}{4\pi \epsilon_0} \frac{2 \lambda }{z}[/latex]. (2.9)

Question 2.2: Find the electric field a distance z above the midpoint of a...

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