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Constant Electric Field in a Vacuum

Problem 3.33

A very thin round plate of radius \(R\) carrying a uniform surface charge density \(\sigma\) is located in vacuum. Find the electric field potential and strength along the plate's axis as a function of a distance \(l\) from its centre. Investigate the obtained expression at \(l \rightarrow 0\) and \(l \gg R\).

Reveal Answer
 3.33. φ=σl2ε0(1+(R/l)2−1),E=σ2ε0(1−ll2+R2). When l→0, then φ=σR2ε0,E=σ2ε0; when l≫R, then φ≈q4πε0l,E≈q4πεnl2, where q=σπR2\begin{aligned} &\text { 3.33. } \varphi=\frac{\sigma l}{2 \varepsilon_{0}}\left(\sqrt{1+(R / l)^{2}}-1\right), E=\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{l}{\sqrt{l^{2}+R^{2}}}\right) . \text { When }\\ &l \rightarrow 0, \text { then } \varphi=\frac{\sigma R}{2 \varepsilon_{0}}, E=\frac{\sigma}{2 \varepsilon_{0}} ; \quad \text { when } \quad l \gg R, \quad \text { then } \varphi \approx \frac{q}{4 \pi \varepsilon_{0} l},\\ &E \approx \frac{q}{4 \pi \varepsilon_{n} l^{2}}, \text { where } q=\sigma \pi R^{2} \end{aligned}
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