A uniform nonconducting rod of mass ma and length l, with charge density `lambda` as shown in fig. is hanged at the midpoint at origin so that it can rotate in a horizontal plane without any friction. A uniform electric field E exists parallel to x axis in the entire region. Calculated the period of small oscillation of the rod .

Answer

`2pisqrt((m)/(3E lambda))`.

Explanation

`tau = int_(0)^(l//2) d tau_(1)+int _(0)^(l//2) d tau_(2)`
`=2int _(0)^(l//2)E lambdadx(x sin theta)=(E lambda l^(2))/(4) sin theta`

At `theta` is small , `tau~~(E lambda l^(2))/(4) theta`
`(ml^(2))/(12)(alpha)=-(E lambda l^(2)theta)/(4)` or `omega=sqrt((3 E lambda)/(m))`
`:. T=2(pi)sqrt((m)/(3E lambda))`.