A) \[x=\sqrt{2}\,a\]
B) \[x=\frac{a}{\sqrt{2}}\]
C) \[x=\frac{\sqrt{2}\,a}{\sqrt{2}-1}\]
D) \[x=\frac{\sqrt{2a}}{\sqrt{2}+1}\]
Correct Answer: C
Solution :
The given situation can be shown as Suppose the field vanishes at a distance \[x,\]we have \[\frac{kq}{{{x}^{2}}}=\frac{kq/2}{{{(x-a)}^{2}}}\] \[\Rightarrow \] \[2{{(x-a)}^{2}}={{x}^{2}}\] or \[\sqrt{2}(x-a)=x\] \[\Rightarrow \] \[\sqrt{2}x-\sqrt{2}a=x\] \[\Rightarrow \] \[\sqrt{2}x-x=\sqrt{2}a\] \[\Rightarrow \] \[(\sqrt{2}-1)x=\sqrt{2}a\] \[\Rightarrow \] \[x=\left( \frac{\sqrt{2}a}{\sqrt{2}-1} \right)\]You need to login to perform this action.
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