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 Online Paper (Held On 10-Jan-2019 Evening]

A) 9F   

B) 3F   

C) F/3                   

D)                  27F

Correct Answer: A

Solution :

\[\,\left| \overrightarrow{P} \right|=q.2a\] \[F=QE\] \[F\,\,=\,\,Q.\frac{2Kp}{{{y}^{3}}}\] \[\frac{F}{F'}=\,\frac{Q.\frac{2Kp}{{{y}^{3}}}}{Q.\frac{2Kp}{{{(y/3)}^{3}}}}\] \[\frac{F}{F'}=\,\frac{1}{27}\] \[F'=27\,F\]