A dipole has two charges `+1muC " and "-1muC` and each of mass 1 kg. The separation between the charges is 1 m. An electric field `20 times 10^(3)Vm^(-1)` is applied on the dipole. If the dipole is deflected through `2^(@)` from the equilibrium position, then the time taken by it to come to equilibrium position again is

Options

A. `2.5 pis`
B. `5 pi s`
C. `15pis`
D. `pis`

Answer

A

Explanation

Given, mass of each charge, m = 1 kg
separation distance between charges, 2d = 1m
electric field, `E=20 times 10^(3)V//m` and
deflection angle, `theta=2^(@)`
Now, dipole moment p = magnitude of any one charge `times` separation distance between the two charges
`" "p=q times 2d " "[because 2d=1m]`
`therefore " "p=1 times 10^(-6) times 1=10^(-6)cm " ...(i)"`
`therefore` Moment of inertia about the separation distance is given as
`therefore " "I=2(md^(2))=2 times 1 times (1//2)^(2)`
Moment of inertia, `I=1//2 " ...(ii)"`
Time period of oscillation is given as,
`" "T=2pisqrt(I/(pE))`
`therefore` Here,
`" "` I = moment of inertia
`" "`p = dipole moment
and E = electric field
`T=2pisqrt((1//2)/(10^(-6) times 20 times 10^(3)))=2pisqrt(1000/(20 times 2))=2pi times 10/2`
`T=10pi`
Then, the time taken to come to equilibrium position again is
`" "t=T/4=(10pi)/4=2.5pis`