A) 0.44 m
B) 0.65 m
C) 0.556m
D) 0.350m
Correct Answer: C
Solution :
From the question\[{{F}_{1}}={{F}_{2}}\] \[5\times {{10}^{-11}}C-2.7\times {{10}^{-11}}C\] \[\Rightarrow \] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{5\times {{10}^{-11}}\times q}{{{(0.2+x)}^{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2.7\times {{10}^{-11}}\times q}{{{x}^{2}}}\] \[\Rightarrow \] \[x=0.556m\] from\[IInd\]charge.You need to login to perform this action.
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