A) \[12:1\]
B) \[8:1\]
C) \[24:1\]
D) \[4:1\]
Correct Answer: C
Solution :
\[{{F}_{\operatorname{int}ial}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{12\times (-8)}{{{r}^{2}}};\,{{F}_{\operatorname{int}ial}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{96}{{{r}^{2}}}\]where r is the distance between them. When the charges are brought in contact, then \[{{q}_{1}}={{q}_{2}}=\frac{12-8}{2}=\frac{4}{2}=2\mu F\] \[\therefore \] \[{{F}_{final}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2\times 2}{{{r}^{2}}}=\frac{4}{{{r}^{2}}}\times \frac{1}{4\pi {{\varepsilon }_{0}}}\] \[\Rightarrow \] \[{{\left| F \right|}_{final}}=\frac{4}{{{r}^{2}}}\times \frac{1}{4\pi {{\varepsilon }_{0}}}\] \[\therefore \] \[\frac{{{\left| F \right|}_{final}}}{{{\left| F \right|}_{final}}}=\frac{96}{4}=24\]
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