question_answer

Charges -q and +q located at A and B. respectively, constitute an electric dipole. Distance \[AB=2a\], 0 is the mid-point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where \[OP=y\] and \[y\,\,>\,\,>\,\,2a\]. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P’ such that \[OP'=\left( \frac{y}{3} \right)\],  the force on Q will be close to- \[\left( \frac{y}{3}>>2a \right)\] [JEE Main 10-Jan-2019 Evening]

A) 9F  

B) 3F   

C) F/3  

D) 27F

Correct Answer: A

Solution :

[a]

\[\,\left| \overrightarrow{P} \right|=q.2a\]

\[F=QE\]

\[F\,\,=\,\,Q.\frac{2Kp}{{{y}^{3}}}\]

\[\frac{F}{F'}=\,\frac{Q.\frac{2Kp}{{{y}^{3}}}}{Q.\frac{2Kp}{{{(y/3)}^{3}}}}\]

\[\frac{F}{F'}=\,\frac{1}{27}\]

\[F'=27\,F\]