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A particle of mass m and charge q is loc...

A particle of mass m and charge q is located midway between two fixed charged particles each having a charge q and a distance 2l apart. Prove that the motion of the particle will be SHM if it is displaced slightly along the line connecting them and released. Also find its time period.

A

`sqrt((pi^(3)Mepsilon_(0)d)/(Qq))`

B

`sqrt((pi^(2)Mepsilon_(0)d^(3)/(Qq))`

C

`sqrt((pi^(3)Mepsilon_(0)d^(3))/(Qq))`

D

`sqrt((pi^(3)Mepsilon_(0))/(Qqd^(3))`

Text Solution

Verified by Experts

The correct Answer is:
c


Let a charge q is displaced by x from the mean position as shown in the figure (ii).
The restoring force acting on the charge q to bring it in its mean position is
`F=-[(1)/(4piepsilon_(0))(Q)/((d-x)^(2))-(1)/(4piepsilon_(0))(Qq)/((d+x)^(2))]`
`=-(Qq)/(4piepsilon_(0))[(1)/((d-x)^(2))-(1)/((d+x)^(2))]`
`=-(Qq)/(4piepsilon_(0))[((d+x)^(2)-(d-x)^(2))/((d^(2)-x^(2))^(2))]=(Qq)/(4piepsilon_(0))[(4dx)/((d^(2)-x^(2))^(2))]`
Since `xltltd`
`thereforeF=-(Qq)/(4piepsilon_(0))[(4dx)/(d^(4))]=-(Qqx)/(piepsilon_(0)d^(3))`
Acceleration , `a=(F)/(M)=-(Qqx)/(piepsilon_(0)d^(3)M)` ...(i)
As `aprop-x`
The charge will execute simple harmonic motion . The standard equation of SHM is
`a=-omega^(2)x` (ii)
Comparing (i) and (ii) , we get
`omega^(2)=(Qq)/(piepsilon_(0)d^(3)M)oromegasqrt((Qq)/(piepsilon_(0)d^(3)M))`
Time period , `T=(2pi)/(omega)=2pisqrt((piepsilon_(0)Md^(3))/(Qq))=2sqrt((pi^(3)epsilon_(0)Md^(3))/(Qd))`
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