IIT JEE , AIEEE Forum - Physics , Magnetism

prasen_prasen

now its time for the second proof. as i've already told u that the answer of this is not coinciding this is not the questionu wanted. but i'm givin u the derivation.dont blame me for not giving u the derivation for frequency of this as not just me but no one on the world cant do it coz it won't be in periodic motion.give u the reason l8r meanwhile have a glance below

in the second case let the particle be displaced perpendicular to the linear system by x(<in this case the force acting on the middle particle by both the particles are at the same distance as even after the displacement.now the distance between the particles becomes (b^2+x^2)^1/2
now the magnitude of the force acting acting on the particle displaced by any one of the other 2 is
f=q^2/[4(pie epsilon not){(b^2+x^2)^1/2}^2]
f=q^2/[4(pie epsilon not)(b^2+x^2)]
now let this be making an angle @ with the initial linear system. now the resultant force acting on the particle becomes
F=2f sin@
F=2*q^2/[4(pie epsilon not){(b^2+x^2)]*sin@
but when in the system sin@=x/(b^2+x^2)^1/2
F=q^2/[2(pie epsilon not)(b^2+x^2)]*x/(b^2+x^2)^1/2
F=q^2/[2(pie epsilon not)(b^2+x^2)^(3/2)]x
but x<F=q^2/[2(pie epsilon not)(b^2)^(3/2)]x
F=q^2/[2(pie epsilon not){(b)^3}]x

Ask & Discuss Questions with Community & Experts