11Opp side. Sorry forgot to mention that.
@ Ashish, but however u displace it, shouldn't a net force always act?
Two identical charges of magnitude +Q are fixed along the X-axis. A third charge −Q is placed midway between them at point P. Then small displacements of −Q are made in the directions indicated by the unit vectors. The −Q is stable with respect to displacement :-
a)\vec{i}\text{ and }\vec{j}
b)\vec{i}\text{ and }\vec{-j}
c)\vec{j}\text{ and }\vec{-k}
d)\vec-i}\text{ and }\vec{-k}
e)\text {Stable w.r.t to any displacement }
What do they mean by saying stable w.r.t to displacement?
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5 Answers
106i think they mean that it is stable when displaced along that direction
it should be (c)
11The two identical charges +Q are placed on the same side of origin on x-axis or opp side of origin on x-axis ???
1111@ Soumik, in the (c) option, net force always acts towards the =lbm position i.e is origin. So the net force will bring -Q towards its normal =lbm position. So stable =lbm.
11Acha - to ye baat hai!
So they are wanting us to get that displacement vector in which net force acts towards the equilibrium position!!!
Then it's ok. I mis-understood the qsn.
Thanks all of u.