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Question

Near the surface of the nucleus of a lead atom, the electric field has a strength of 3.4×1021 V/m3.4 \times 10^{21} \mathrm{~V} / \mathrm{m}. What is the energy density in this field?

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We need to determine the energy density of the field E=3.41021VmE=3.4 \cdot 10 ^{21} \hspace{1mm}\frac{\text V}{\text m}.

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