$(a)\;mga\qquad(b)\;\sqrt{3} mga\qquad(c)\;\large\frac{mga}{\sqrt{3}}\qquad(d)\;None\;of\;these$

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Answer : (b) $\;\sqrt{3} mga$

Explanation :

Electrostatic energy at H=0 is:

$E_{e}=\large\frac{\sigma q a }{2 \in_{0}}$

$E_{g}=0$

At $\;H=\large\frac{a}{\sqrt{3}}\;, \quad\; E_{e}=\large\frac{\sigma q}{2 \in_{0}}\;[\sqrt{a^2+H^2}-H]$

$E_{e}=\large\frac{\sigma q}{2 \in_{0}}\;[\sqrt{a^2+\large\frac{a^2}{3}}-\large\frac{a}{\sqrt{3}}]$

$E_{e}=\large\frac{\sigma q}{2 \in_{0}} \;[\large\frac{2a}{\sqrt{3}}-\large\frac{a}{\sqrt{3}}]$

$E_{e}=\large\frac{\sigma q}{2 \in_{0}}\;\large\frac{a}{\sqrt{3}}$

$E_{g}=\large\frac{mga}{\sqrt{3}}$

$T.E=E_{e}+E_{g}$

$T.E=\large\frac{mga}{\sqrt{3}}+\large\frac{\sigma q a}{2 \in_{0} \sqrt{3}}$

$T.E=\large\frac{mga}{\sqrt{3}}+\large\frac{2mga}{\sqrt{3}}$

$T.E=\sqrt{3} mg\;.$

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