NEET Physics Electrostatics & Capacitance Question Bank NEET PYQ - Electrostatics and Capacitance

  • question_answer
    A point Q lies on die perpendicular bisector of an electrical dipole of dipole moment p. If the distance of Q from the dipole is r (much larger than the size of the dipole) then electric field at Q is proportional to:               [AIPMT 1998]

    A) \[{{p}^{-1}}\] and \[{{r}^{2}}\]

    B) \[p\] and \[{{r}^{-2}}\]

    C) \[{{p}^{2}}\] and \[{{r}^{-3}}\]

    D) \[p\] and \[{{r}^{-3}}\]

    Correct Answer: D

    Solution :

    [d] Electric field due to a dipole at bisector or at a point on its broad side on position is given by
    \[E=\frac{1}{4\,\pi {{\varepsilon }_{0}}}.\frac{p}{{{r}^{3}}}\]
    or         \[E\propto \,\,\frac{p}{{{r}^{3}}}\]                                            ....(i)
    where r is the distance of that point from centre of dipole.
    So, from Eq. (i)
    \[E\propto \,p\]
    and                   \[E\propto \,{{r}^{-3}}\]
    Note:    The electric field due to a dipole at its end on position is twice the value at its broad side on position i.e.,
    \[{{E}_{end-on}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{2p}{{{r}^{3}}}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner