Question

A dipole has two charges $+1 \,\mu \,C$ and $-1 \,\mu\, C$ 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 $2^{\circ}$ 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, $m=1 \,kg$
separation distance between charges, $2 \,d=1 \,m$
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
$p=q \times 2 \,d \,\,[\because 2 \,d=1 \,m ]$
$\therefore p=1 \times 10^{-6} \times 1=10^{-6} \,cm$
$\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,
$T=2 \pi \sqrt{\frac{I}{p E}}$
$\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 $