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By van Elst H.

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Chapter 1), which is of fundamental importance to applications in electrostatics (and beyond). 8) Let us use this relation to obtain the magnitude of the electric field strength ✳ of a point charge ✄ at rest. By the isotropy of Euclidian space, ✳ cannot have ✗ ✄✜✢ ✄ ✡ . Now a preferred direction for a point charge. Thus, we have ✳❝✝ introducing a “Gaußian sphere”, of radius ✄ and with outward-pointing unit normal ✝ ✄ ✡ , so that ✄ lies at its centre, we find ✌ ✡ ✁ ✁ ✂ ✄✂✆☎ ❀ ✘✆✘ ✳ ✁ ✑✡ ✂☞✌ ✝✏✎✵✝ from which we obtain ✁ ✍ ✛✁ ✞ ✞ ✗ ✄✜✢ ✄ ✄ ✗ ✄✜✢❪✝ ✍ ✄ ✄ ☛✡ ✝ ✞ ✄ ✍ ✙ ◗☞☛✍✌✂ ✝ ✢ ✝ ✆✶✝ ✝ ✄ ✞ ✷ ✷ which is the result we were looking for.

5) ] for ✳ , we implicitly introduced the idealised concept of a test point charge ✄ [ or ✄ ] that does itself feel ✳ but that does not contribute to ✩ the generation of ✳ . That is, we have assumed that ✄ itself is not part of a ✩ given distribution of electric charges. To illustrate these results, let us give two numerical examples. The magnitude of the electric field strength generated by a single electron at rest at ✪❍ distance is ✮ ✳ ☎✆✮ ✎ ✩ ✿ ✝ ☞☛ ✙✕✙ ✣❲ ✦✎★✧ ✫ ✻✾❀ ✽✒❁ ❃ ✿ . Likewise, the magnitude of the electric field strength generated by a charge of ✂✁ at rest at ✪❍ distance is ✮✳ ❃ ✮ ✎ ✝ ★☛✍✛✕✛✞✣✡ ✦✎✕✫ ✻❋❀ ✽ ❁ ❃ ✿ .

5. 21) Thus, ✔●✘✮ ✱✼✟❤✱ ✮ can be interpreted as being proportional to the scalar potential of a unit point charge. , ✲ . , there are no boundary surfaces involved), this solution of Poisson’s equation will rarely be useful in actual applications. 33). ❛ Consider a volume ✔ ✲ with bounding surface . 24) is zero. 17) (for a given charge distribution ❬ ), as the surface integrals on the right-hand ❛ side still contain the (so far) unknown values on of both ✖ itself and its ❜▲✢✍✖✥✝ ✟ . So we have to find ways of obtaining normal derivative ✗ additional information on what values these quantities can possibly take.

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