[JEE Main Online Paper (Held On 10-Jan-2019 Morning]
A) \[\frac{\sqrt{2}\,R}{\sqrt{2}+1}\]
B) \[\frac{\,R}{\sqrt{2}+1}\]
C) \[\frac{\,\sqrt{2}\,R}{\sqrt{2}-1}\]
D) \[\frac{\,\,R}{\sqrt{2}-1}\]
Correct Answer: C
Solution :
On the x axis in left of A potential due to A and B is positive. But potential due to A is higher than that of B, between A & B sign of potential due to A & B is opposite. So potential can be same only in right of B on x axis. 
\[-\frac{k2qa}{{{x}^{2}}}=-\frac{k4qa}{{{(R+x)}^{2}}}\] \[\frac{R+x}{x}\,\,=\,\,\sqrt{2}\] \[R+x=\,\,x\sqrt{2}\] \[x=\,\frac{R}{\sqrt{2}-1}\] distance from \[A=R+\frac{R}{\sqrt{2}-1}\,\,\Rightarrow \,\,\frac{\sqrt{2}R}{\sqrt{2}-1}\] NTA has given the answer but answer should be .
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