Question

A dipole has two charges $+1\mu C$ and $-1\mu C$ and each of mass $1kg$. The separation between the charges is $1m$. An electric field $20\times 10^{3}V{m}^{-1}$ is applied on the dipole. If the dipole is deflected through $2^{\circ }$ from the equilibrium position, then the time taken by it to come to equilibrium position again is

• A. $2.5\pi s$
• B. $5\pi s$
• C. $15\pi s$
• D. $\pi s$

Answer 1

Answer: (A)

Given, mass of each charge, $m=1kg$
separation distance between charges, $2d=1m$
electric field, $E=20\times 10^{3}V/m$ and
deflection angle, $\theta =2^{\circ }$
Now, dipole moment $p=$ magnitude of any one charge $\times$ separation distance between the two charges
$p=q\times 2d[\because 2d=1m]$
$\therefore p=1\times 10^{-6}\times 1=10^{-6}cm$
$\therefore$ Moment of inertia about the separation distance is given as
$\therefore I=2\left(m{d}^2\right)=2\times 1\times (1/2{)}^2$
Moment of inertia, $I=1/2$
Time-period of oscillation is given as,
$T=2\pi \sqrt{\frac{I}{pE}}$
$\therefore$ Here, $I=$ moment of inertia
$p=$ dipole moment
and $E=$ electric field
$T=2\pi \sqrt{\frac{1/2}{10^{-6}\times 20\times 10^3}}=2\pi \sqrt{\frac{1000}{20\times 2}}=2\pi \times \frac{10}{2}$
$T=10\pi$
Then, the time taken to come to equilibrium position again is
$t=\frac{T}{4}=\frac{10\pi }{4}=2.5\pi s$