Question

A solid conducting sphere of radius 10 centimetre is enclosed by thin metallic shell of radius 20 centimetre a charge Q is equal to 20 microcoulomb is given to the inner sphere if the inner sphere is connected to the shell by a conducting wire find the heat generated in this process

Answer 1

Dear Student

Capacitance of spherical capacitors can be given as
$$\begin{gathered} C_1=\frac{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{ab}}{b-a}=\frac{4\times 3.14\times 8.85\times {10}^{-12}\times 10\times {10}^{-2}\times 20\times {10}^{-2}}{10\times {10}^{-2}}=2223.12\times {10}^{-14}F \\ {C}_2=4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{b}=4\times 3.14\times 8.85\times {10}^{-12}\times 20\times {10}^{-2}=2223.12\times {10}^{-14}\mathrm{F} \\ \mathrm{Total}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{system}\mathrm{before}\mathrm{connecting}\mathrm{with}\mathrm{wire}=\mathrm{energy}\mathrm{of}\mathrm{capacitor}1+\mathrm{energy}\mathrm{of}\mathrm{capacitor}2 \\ \mathrm{Total}\mathrm{energy}=\frac{{\mathrm{Q}}^2}{2{\mathrm{C}}_1}+\frac{{\mathrm{Q}}^2}{2{\mathrm{C}}_2}=\frac{400\times {10}^{-12}\mathrm{C}}{2\times 2223.12\times {10}^{-14}\mathrm{F}}+\frac{400\times {10}^{-12}\mathrm{C}}{2\times 2223.12\times {10}^{-14}\mathrm{F}} \\ \mathrm{Total}\mathrm{energy}=8.996\mathrm{J}+8.996\mathrm{J}=17.993\mathrm{J} \\ \mathrm{After}\mathrm{connecting}\mathrm{the}\mathrm{wire}\mathrm{all}\mathrm{charges}\mathrm{is}\mathrm{on}\mathrm{the}\mathrm{surface}\mathrm{of}\mathrm{outer}\mathrm{sphere},\mathrm{no}\mathrm{charge}\mathrm{is}\mathrm{on}\mathrm{inner}\mathrm{sphere}.\mathrm{So}\mathrm{total}\mathrm{energy}\mathrm{now}\mathrm{is} \\ \mathrm{Total}\mathrm{energy}\mathrm{Final}=0+\frac{{\mathrm{Q}}^2}{2{\mathrm{C}}_2}=0+\frac{400\times {10}^{-12}\mathrm{C}}{2\times 2223.12\times {10}^{-14}\mathrm{F}} \\ \mathrm{Total}\mathrm{energy}\mathrm{Final}=8.996\mathrm{J} \\ ∆\mathrm{Total}\mathrm{energy}=\mathrm{Total}\mathrm{energy}\mathrm{initial}-\mathrm{total}\mathrm{energy}\mathrm{final}=8.996\mathrm{J}-17.993\mathrm{J}=-9\mathrm{J} \\ \mathrm{Thus}\mathrm{heat}\mathrm{produced}\mathrm{is}+9\mathrm{J} \end{gathered}$$

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