A point charge q of mass m is suspended vertically by a string of length l . A point dipole of dipole moment p is now brought towards q from inifinty so that charge moves aways. The final equilibrium position of the system including the direction of the dipole , the angles and distance is shown in the figure below . If the work done in brining the dipole to this position is N `xx ` (mgh) . where g is the acceleration due to gravity , then the value of N is _______ . (Note that for three coplanar forcess keeping a point mass in equilibrium `(F)/(sin theta)` is the same for all forces, where F is any one of the forces and `theta` is the angle between the other two forces ).

Answer

2

Explanation

`W = q.v + mgh`
` = q .(kp)/(x^(2)) + mgh " "……..(1)`
For equilibrium
`(qE)/(sin beta) = sin ( 90 - (alpha)/(2))`
`overset(qE)(sin (180-alpha)) = overset(mg)(cos((alpha)/(2))) , overset(qE)(sin alpha) = overset(mg)(cso (alpha)/(2))`
`qE = mg , qE = mg 2 sin .(alpha)/(2)`
`2 sin .(alpha)/(2) cos .(alpha)/(2) cos .(alpha)/(2) `
`q((Kp)/(x^(2))) = mg x sin .(alpha)/(2) " "= mg sin((alpha)/(2))^(h) sin.(alpha)/(2)`
`q_(x^(2))^(kp) = mgh " ".....(ii)`
From (i) and (ii)
W = 2mgh
x= 2