Saved Bookmarks
| 1. |
An electric dipole of momentum vecpis placed in a uniform electric field. The dipole is rotated through a very small angle thetafrom equilibrium and is released. It executes simple harmonic motion with frequency f=1/(2pi)sqrt((pE)/I) where, I= moment of interia of the dipole. |
|
Answer» Solution :A dipole MAKES an angle 9 with an electric field from its equilibrium position. So, torque ACTING on it, `vectau = VECP xx vecE` `therefore tau = pEsintheta` This torque rotates the dipole in clockwise direction `therefore tau =-pEsintheta` Here `theta` is very small. `therefore sintheta =theta` `therefore tau =-pEtheta` but, `tau = 1alpha` and `alpha =-omega^(2)theta` `1alpha = -pEtheta`, where `alpha`is the angular acceleration in S.H.M. `therefore 1(-omega^(2)theta) = -pEtheta` `therefore omega = sqrt((pE)/1)` `therefore 2pif = sqrt((pE)/1) (therefore omega = 2pif)` `therefore f=1/(2pi)sqrt((pE)/1)` |
|