question_answer

If two like charges of magnitude \[1\times {{10}^{-9}}C\] and \[9\times {{10}^{-9}}C\] are separated by a distance of 1 m, then the point on the line joining the charges, where the force experienced by a charge placed at that point is zero, is

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\[\Rightarrow \].