question_answer

Two electric dipoles, A, B with respective dipole moments \[{{\overrightarrow{d}}_{A}}=-4qa\widehat{i}\] and \[{{\overrightarrow{d}}_{B}}=-2qa\widehat{i}\] are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is - [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 .