A) \[-(3+\sqrt{8})\]only
B) \[-3+\sqrt{8}\]only
C) \[-(3+\sqrt{8})\,or\ (-3+\sqrt{8})\]
D) \[+\sqrt{3}\]
E) \[-\sqrt{8}\]
Correct Answer: C
Solution :
Since the two charges (spheres) attract, they will be opposite in sign, ie,\[{{q}_{1}}\]and\[-{{q}_{2}}\]. Force, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{{{q}_{1}}\times \,-{{q}_{2}}}{{{d}^{2}}}\] After touching, charge on each will be\[\frac{{{q}_{1}}-{{q}_{2}}}{2}\]. New force, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}{{\left( \frac{{{q}_{1}}-{{q}_{2}}}{2} \right)}^{2}}\times \frac{1}{{{d}^{2}}}\] Given \[|F|=|F|\] On solving by quadratic equations, we get \[\frac{{{q}_{1}}}{{{q}_{2}}}=-(3+\sqrt{8})\]or\[(-3+\sqrt{8})\]
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