The energy density u is plotted against the d | vedclass

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Question

${\rm{u}} = \frac{1}{2}{ \in _0}{{\rm{E}}^2} = \frac{1}{2}{ \in _0}{\left( {\frac{{\rm{q}}}{{4\pi { \in _0}{\rm{r}}}}} \right)^2} = \frac{{{{\rm{q}}^2

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Vedclass · Answer

${\rm{u}} = \frac{1}{2}{ \in _0}{{\rm{E}}^2} = \frac{1}{2}{ \in _0}{\left( {\frac{{\rm{q}}}{{4\pi { \in _0}{\rm{r}}}}} \right)^2} = \frac{{{{\rm{q}}^2}}}{{32{\pi ^2}{ \in _0}{{\rm{r}}^2}}}$

$ \Rightarrow \log {\rm{u}} = \log \left( {\frac{{{{\rm{q}}^2}}}{{32{\pi ^2}{ \in _0}{{\rm{r}}^2}}}} \right) = \log {\rm{k}} - 2\log {\rm{r}}$

The energy density $u$ is plotted against the distance $r$ from the centre of a spherical charge distribution on a $log$-$log$ scale. The slope of obtianed straight line is :

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