A) 1.0
B) 0.75
C) 0.5
D) 0.25
Correct Answer: D
Solution :
For equilibrium of charge q. \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{-O\times q}{{{(r/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-Q)(-Q)}{{{r}^{2}}}=0\] \[\Rightarrow \] \[4q=Q\] \[\Rightarrow \] \[a=\frac{Q}{4}=\frac{1}{4}\] \[(\because \,Q=1\,C)\] \[=0.25\,C\]You need to login to perform this action.
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