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For a vector potential $$\overrightarrow {\bf{A}} ,$$ the divergence of $$\overrightarrow {\bf{A}} $$ is $$\overrightarrow \nabla \cdot \overrightarrow {\bf{A}} = - \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{Q}{{{r^2}}},$$    where Q is a constant of appropriate dimension. The corresponding scalar potential $$\phi \left( {r,\,t} \right)$$  that makes $$\overrightarrow {\bf{A}} $$ and $$\phi $$ Lorentz gauge invariant is

For a vector potential $$\overrightarrow {\bf{A}} ,$$ the divergence of $$\overrightarrow {\bf{A}} $$ is $$\overrightarrow \nabla \cdot \overrightarrow {\bf{A}} = - \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{Q}{{{r^2}}},$$    where Q is a constant of appropriate dimension. The corresponding scalar potential $$\phi \left( {r,\,t} \right)$$  that makes $$\overrightarrow {\bf{A}} $$ and $$\phi $$ Lorentz gauge invariant is Correct Answer $$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Qt}}{{{r^2}}}$$

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