A) 24 cm from \[+9e\]
B) 12 cm from \[+9e\]
C) 24 cm from \[+e\]
D) 12 cm from \[+e\]
Correct Answer: B
Solution :
Suppose q is placed at a distance x from +9e, then for equilibrium net force on it must be zero i.e. \[\left| {{F}_{1}} \right|\text{ }=\text{ }\left| {{F}_{2}} \right|\] Which gives \[{{x}_{1}}=\frac{x}{\sqrt{\frac{{{Q}_{2}}}{{{Q}_{1}}}}+1}=\frac{16}{\sqrt{\frac{e}{9e}}+1}=12\,cm\]You need to login to perform this action.
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