question_answer

A \[4\mu F\] capacitor is charged by a 200 V supply. After charging it is disconnected from the supply and is connected to another uncharged \[2\mu F\]capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer:

            Given, \[{{C}_{1}}=4\mu F=4\times {{10}^{-6}},\] \[{{V}_{1}}=200\,V\] Initial energy of first capacitor             \[{{U}_{1}}=\frac{1}{2}{{C}_{1}}V_{1}^{2}=\frac{1}{2}\times (4\times {{10}^{-6}})\times {{(200)}^{2}}\]             \[=8\times {{10}^{-2}}\,J\] When another uncharged capacitor \[{{C}_{2}}=2\mu F\] is connected across first capacitor, \[{{V}_{2}}=0\] Then common potential, \[V=\frac{{{q}_{1}}+{{q}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{{{C}_{1}}{{V}_{1}}+0}{{{C}_{1}}+{{C}_{2}}}\]                                     \[=\frac{4\times {{10}^{-6}}\times 200}{(4+2)\times {{10}^{-6}}}\]                                     \[=\frac{400}{3}V\] Final energy,      \[{{U}_{2}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}}){{V}^{2}}\]                         \[=\frac{1}{2}\times (4+2)\times {{10}^{-6}}\times {{\left( \frac{400}{3} \right)}^{2}}\]                         \[=\frac{16}{3}\times {{10}^{-2}}J\]                         \[=5.33\times {{10}^{-2}}J\] Energy loss \[\Delta \,U={{U}_{1}}-{{U}_{2}}=8\times {{10}^{-2}}-5.33\times {{10}^{-2}}\] \[=2.67\times {{10}^{-2}}\,J\]