##### Electrostatic Field of Continuous Charge Distributions
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The electric field generated by a continuous charge density \(\rho({\bf r})\) in volume \({\cal V}\) can be easily calculated from Coulomb's law using the superposition principle. Since each infinitesimal volume element \(d\tau' = dx' dy' dz'\) contains a charge \(dq' = \rho({\bf r}') d\tau'\), we have

- Gr4 (2.8)

\[ {\bf E} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \rho({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} \tag{E_vcd}\label{E_vcd} \]

Similarly, if the charge is spread out over a two-dimensional surface \({\cal S}\) with (surface) charge density \(\sigma({\bf r})\), we have over an infinitesimal area \(da'\) a charge \(dq' = \sigma({\bf r}') da'\), so

- Gr4(2.7)

\[ {\bf E} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal S} da' \sigma({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} \tag{E_scd}\label{E_scd} \]

Finally, for a line path \({\cal P}\) with linear charge density \(\lambda({\bf r}')\),

- Gr (2.6)

\[ {\bf E} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal P} dl' \lambda({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} \tag{E_lcd}\label{E_lcd} \]

Example

Find the electric field a distance \(z\) above the midpoint of a straight line segment of length \(2L\) carrying a uniform line charge \(\lambda\).

Solution: placing the line on the \(x\) axis, we have

\begin{align*} {\bf r} &= z \hat{\bf z}, {\bf r}' = x' \hat{\bf x}, dl' = dx', |{\bf r} - {\bf r}'| = \sqrt{z^2 + {x'}^2}, \end{align*}

so we have

\begin{align*} {\bf E} = \frac{1}{4\pi \varepsilon_0} \int_{-L}^L dx' \lambda \frac{z \hat{\bf z} - x' \hat{\bf x}}{(z^2 + {x'}^2)^{3/2}} = \frac{\lambda}{4\pi \varepsilon_0} \left[ z \hat{\bf z} \int_{-L}^L dx \frac{1}{(z^2 + x^2)^{3/2}} - \hat{\bf x} \int_{-L}^L dx \frac{x}{(z^2 + x^2)^{3/2}} \right] \end{align*}

The second integral vanishes by symmetry, whereas the first can be done in many ways, most easily by observing that \(\frac{d}{dx} \left( \frac{x}{\sqrt{z^2 + x^2}} \right) = \frac{1}{\sqrt{z^2 + x^2}} - \frac{x^2}{(z^2 + x^2)^{3/2}} = \frac{z^2}{(z^2 + x^2)^{3/2}}\), leading to

You could alternately proceed by using changes of variables \(y = zx\) followed by \(y = \tanh \alpha\): \(\int_{-L}^L \frac{dx}{(z^2 + x^2)^{3/2}} = \frac{1}{z^2} \int_{-L/z}^{L/z} \frac{dy}{(1 + y^2)^{3/2}} = \frac{2}{z^2} \int_0^{\mbox{atanh}(L/z)} d\alpha \cosh \alpha = \frac{2}{z^2} \left. \sinh \alpha \right|_0^{\mbox{atanh}(L/z)} = \frac{2}{z^2} \left. \frac{\tanh \alpha}{\sqrt{1 + \tanh^2 \alpha}} \right|_0^{\mbox{atanh}(L/z)}= \frac{2}{z^2} \frac{L}{\sqrt{z^2 + L^2}}\).

\begin{align} {\bf E} = \frac{\lambda}{4\pi \varepsilon_0} z \hat{\bf z} \left. \left( \frac{x}{z^2\sqrt{z^2 + x^2}} \right) \right|_{-L}^L = \frac{1}{4\pi \varepsilon_0} \frac{2\lambda L}{z \sqrt{z^2 + L^2}} \hat{\bf z}. \end{align}

For large distances \(z \gg L\), this looks like the field of a point charge \(q\lambda L\):

\({\bf E} = \frac{1}{4\pi \varepsilon_0} \frac{2\lambda L}{z^2} \hat{\bf z}\),

whereas for short distances \(z \ll L\) the field looks like that of an infinite wire:

\({\bf E} = \frac{1}{4\pi \varepsilon_0} \frac{2\lambda}{z}\).

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42