A) no change in the energy of the system
B) an increase in the energy of the system if \[{{Q}_{1}}{{R}_{2}}\ne {{Q}_{2}}{{R}_{1}}\]
C) always a decrease in the energy of the system
D) a decrease in the energy of the system if \[{{Q}_{1}}{{R}_{2}}={{Q}_{2}}{{R}_{1}}\]
Correct Answer: D
Solution :
Potential \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}}{{{R}_{1}}}\] and \[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{2}}}{{{R}_{2}}}\] If \[{{V}_{1}}={{V}_{2}},\] then \[\frac{{{Q}_{1}}}{{{R}_{1}}}=\frac{{{Q}_{2}}}{{{R}_{2}}}\] or \[{{Q}_{1}}{{R}_{1}}={{Q}_{2}}{{R}_{1}}\] In this case, there will be no flow of charges. However, if the given spheres have unequal potentials then the charges shall flow from higher potential to lower potential. Flow of charges shall be accompanied by loss of energy.
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