19no i think that it is stable when displaced along y-axis...because the perpendicular bisector of a dipole's axis is equi potential ! thus du/dx is necessarily zero !
2 point charges +q and -q are fixed at (a,0,0) and (-a,0,0). A third point charge -q is at the origin. State with reason whether equilibrium is stable, unstable or neutral if it's slightly displaced :
a) along X-axis
b) along Y-axis
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7 Answers
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106If it is displaced along x-axis (say +ve direction). then the +q will attract it more -q will repel it... So, it will keep moving in +ve direction.. (altough later on it will return back) if it is displaced a bit along -ve x axis then the net force will be towards +ve x-axis ie. towards initial eqbm position but it will overshoot that and then the situation will be like that in the first case .. so i think (unstable eqbm) (not really sure)
If it is displaced along y-axis (say +ve ir -ve direction) then the net force will always be towards right.. So, the charge will initially move towards right and NOT towards equilibrium position So unstable
debo.. see my exp for second part.. it cleary does NOT come back to initial position
19yes,,,i was thinking exactly that , when i got this logic !!! where is the flaw in my logic ?
106yes u are correct in saying that "perpendicular bisector of a dipole's axis is equi potential ! thus du/dx is necessarily zero !"
But when the charge is displaced along perpendicular bisector, the component of force along the displaced direction is indeed zero.. but there may be forces in other directions (here towards right).. So, if the charge was restricted to move along the perp. bisector only, then it would be neutral equilibrium... but not here as there is no restriction on its motion
19right ! agreed ! in fact, i missed something very basic ! a point charge can never stay in stable equilibrium in an electric field !!!!
1Hey guys,for a body to be in equilibrium,the force on it must be zero.Thus the origin cannot be an equilibrium position,because a force will act in the direction towards the +q charge of the dipole.
