JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • 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 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 :

    [c]
    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 [a] but answer should be .


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