Two charges each of magnitude Q are fixed at 2a distance apart. A third charge (-q of mass 'm') is placed at the mid point of the two charges; now -q charge is slightly displaced perpendicular to the line joining the charges then find its time period ?

(1)#2pisqrt((ma^3piepsilon_@)/(Qq))#
(2)#2pisqrt((2ma^3piepsilon_@)/(Qq))#
(3)#pisqrt((ma^3piepsilon_@)/(Qq))#
(4)#pisqrt((2ma^3piepsilon_@)/(Qq))#

1 Answer
Apr 10, 2018

(2)

Explanation:

For a small displacement #x# "perpendicular to the line joining the charges", the potential energy, #U(r) = -( Qq)/(4 pi epsilon_o r)#, of the displaced charge is now:

#U(x) = 2 times - ( Qq)/(4 pi epsilon_o sqrt(a^2 + x^2)) = - ( Qq)/(2 pi epsilon_o sqrt(a^2 + x^2))#

The force vector acting on the particle is the negative derivative wrt #x# of this conservative potential function. From symmetry we know that it will act "perpendicular to the line joining the charges".

# vec F = - (dU)/(dx) = -( Qq)/(2 pi epsilon_o) x/( (a^2 + x^2)^(3/2))#

By Taylor/Binomial expansion, for small #x# we have:

# x/( (a^2 + x^2)^(3/2)) = x/a^3 ( 1 + x^2/a^2)^(-3/2) = x/a^3 (1 - 3/2* x^2/a^2 + mathcal O (x^4))#

In otherwords #vec F# linearises as #vec F = -( Qq)/(2 a^3 pi epsilon_o ) x#, and it is this linearise restorative force that characterises the harmonic oscillator.

So from #m ddot x + ( Qq)/(2 a^3 pi epsilon_o ) x = 0#, we write this in the standard (#ddot x + omega^2 x = 0#) SHM pattern, ie # ddot x + ( Qq)/(2 m a^3 pi epsilon_o ) x = 0#.

Therefore;

#T = (2 pi )/omega = 2 pi sqrt( ( 2 m a^3 pi epsilon_o )/(Qq))#