Charges - q and + q , located at A and B , respectively, constitute an electric dipole. Distance A B = 2 a ,…

Charges -q and +q, located at A and B, respectively, constitute an electric dipole. Distance AB=2aO is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP=y and y2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P' such that OP'=y3 the force on Q will be close toy32a
  1. 27F.
  2. F3.
  3. 3F.
  4. 9F.

Solution

Electric field intensity at the distance y along the perpendicular bisector is

E1=KPy3

So, force at distance y on Q,

F=Q(E1)=QKPy3 ...(1)

F'=QKPy33 ...(2)

Dividing 1 by 2, we get,

F'F=33

 F'=27F

Asked in: JEE Main 2019 (10 Jan Shift 2)

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