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Question

$ \mathrm{E}_{\mathrm{i}} =\frac{\mathrm{KQ}^{2}}{2 \mathrm{a}}+\frac{\mathrm{KQ}^{2}}{4 \mathrm{a}}+\frac{\mathrm{KQ}}{2 \mathrm{a}} \cdot \mathrm{Q}

Vedclass · Accepted Answer

$\frac{{K{Q^2}}}{{4a}}$

Vedclass · Answer

$ \mathrm{E}_{\mathrm{i}} =\frac{\mathrm{KQ}^{2}}{2 \mathrm{a}}+\frac{\mathrm{KQ}^{2}}{4 \mathrm{a}}+\frac{\mathrm{KQ}}{2 \mathrm{a}} \cdot \mathrm{Q}=\frac{\mathrm{KQ}^{2}}{\mathrm{a}}+\frac{\mathrm{KQ}^{2}}{4 \mathrm{a}} $

$=\frac{5 \mathrm{KQ}^{2}}{4 \mathrm{a}} $

$ \mathrm{E}_{f} =\frac{\mathrm{K}(2 \mathrm{Q})^{2}}{4 \mathrm{a}}=\frac{\mathrm{KQ}^{2}}{\mathrm{a}} \quad \mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}}=\mathrm{H}=\frac{\mathrm{KQ}^{2}}{4 \mathrm{a}} $

Figure shows a solid metal sphere of radius $‘a’$ surrounded by a concentric thin metal shell of radius $2a$ . Initially both are having charges $Q$ each. When the two are connected by a conducting wire as shown in the figure, then amount of heat produced in this process will be

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