A point charge `Q_(1)=-125mu C` is fixed at the center of an insulated disc of mass 1 kg. The disc rests on a rough horizontal plane. Another charge `Q_(2)=125mu C` is fixed verically above the center of the disc at a height h=1m. After the disc is displaced slightly in the horizontall direction (friction is sufficient to prevent slipping), find the period of oscillation of disc.

Answer

`2pi (h/Q)sqrt((6 pi epsilon_(0)Mh))`.

Explanation

Let the radius of the disc be R. If the disc is displaced x, The corresponding angular dispalcement `theta=x//R`.
The restoring torque `tau` about the point of contact of disc with ground.
`tau_(p)=(F sin theta)R`
`(F sin theta)R=1alpha=[(MR)^(2)/(2)+MR^(2)]alpha`
where `F=(Q^(2)x)/(4 pi epsilon_(0)(h^(2)+x^(2)))`
and `sin theta=(R )/(sqrt((h^(2)+x^(2)))`
Hence `(Q^(2)xR)/(4 pi epsilon_(0)(h^(2)+x^(2))^(3//2))=[(MR^(2))/(2)+MR^(2)]alpha`
or `alpha=(Q^(2)x)/(6 pi epsilon_(0)MR(h^(2)+x^(2))^(3//2))`
for `x ltlt h, alpha~=-(Q^(2)x)/(6 pi epsilon_(0)MRh^(3))`
`alpha=(a)/(R ) ~=-(Q^(2)x)/(6 pi epsilon_(0)MRh^(3))`
Negative sign is being introduced because angular acceleration and angular displacement are opposite to each other. Thus, ltbr. `a=(Q^(2)x)/(6 pi epsilon_(0)MRh^(3))`
Hence, `omega=sqrt((Q^(2))/(6 pi epsilon_(0)MRh^(3)))`
`T=2 pi sqrt((6 pi epsilon_(0)MRh^(3))/(Q^2)) =2pi (h/Q)sqrt((6 pi epsilon_(0)Mh))`.