A) 0.25 m from the charge \[1\times {{10}^{-9}}C\]
B) 0.75 m from the charge \[9\times {{10}^{-9}}C\]
C) Both [a] and [b]
D) at all points on the line pining the charges
Correct Answer: C
Solution :
\[mg={{V}_{w}}{{\rho }_{w}}\,g+{{V}_{oil}}\,g\] Suppose at a .distance \[{{V}_{w}}\] from the charge of having minimum magnitude, the force on the test charge is zero \[{{V}_{oil}}\] \[{{\rho }_{w}}\] \[{{\rho }_{oil}}\] \[\Rightarrow \] \[m=(2\times 10\times 10\times 1)\,+(8\times 10\times 10\times 0.6)\] \[m=200+480=680\,g\] \[{{x}_{1}},\,{{x}_{2}},\,{{x}_{3}}\] \[{{k}_{1}},\,{{k}_{2}},\,{{k}_{3}},\,{{k}_{4}}....,\] \[{{x}_{5}}={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...\] \[\Rightarrow \] \[\frac{mg}{{{k}_{s}}}=mg\,\left( \frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{{{k}_{3}}}+... \right)\] \[\therefore \] This gives \[\frac{1}{{{k}_{s}}}-\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{{{k}_{3}}}+...\] or \[{{k}_{1}}=k,\] As \[{{k}_{2}}=2k,\] is length So, \[{{k}_{3}}=3k,...\] So, the required point is a distance of 0.25 m from the charge \[\frac{1}{{{k}_{s}}}=\frac{1}{k}+\frac{1}{2k}+\frac{1}{4k}+\frac{1}{8k}+...\] or 0.75 m from the charge\[\Rightarrow \].You need to login to perform this action.
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