JEE Main201910 Jan 2019Evening ShiftPhysicsElectrostaticsActual
Charges - q and + q , located at A and B , respectively, constitute an electric dipole. Distance A B = 2 a , O is the mid point of the dipole and O P is perpendicular to A B . A charge Q is placed at P where O P = y and y ≫ 2 a . The charge Q experiences an electrostatic force F . If Q is now moved along the equatorial line to P ' such that O P ' = y 3 the force on Q will be close to y 3 ≪ 2 a
Options
- A27 F .
- BF 3 .
- C3 F .
- D9 F .
Correct answer
A. 27 F .
Step-by-step solution
Electric field intensity at the distance y along the perpendicular bisector is E 1 = K P y 3 So, force at distance y on Q , F = Q ( E 1 ) = Q K P y 3 ... ( 1 ) F ' = Q K P y 3 3 ... ( 2 ) Dividing 1 by 2 , we get, F ' F = 3 3 ∴   F ' = 27 F