Question

An electric dipole is made up of two particles, one having charge $+1\mu \text{C}$ and mass $1\text{kg}$ and the other with charge $-1\mu \text{C}$ and mass $1\text{kg}$ separated by distance $1\text{m}$. It is in equilibrium in a uniform electric field of $20\times {10}^3\text{V}/\text{m}$. If the dipole is deflected through an angle $2^{\circ }$, then time taken by it to come again at equilibrium position is

• A. $2\pi \text{s}$
• B. $2.5\pi \text{s}$
• C. $4\text{s}$
• D. $5\text{s}$

Answer 1

The correct option is B $2.5\pi \text{s}$

The moment of inertia of the system about an axis through the centre of the dipole is,
$I=2\left(m{d}^2\right)=2\times 1\times {\left(\frac{1}{2}\right)}^2=\frac{1}{2}{\text{kg-m}}^2$

Also, dipole moment,

$p=q\left(2r\right)={10}^{-6}\times 1={10}^{-6}\text{C-m}$

Given: $E=20\times {10}^3\text{V/m}$

As the system is rotated through a very small angle from equilibrium position, the resulting motion will be an oscillation.

Hence, the time period will be,

$T=2\pi \sqrt{\frac{I}{pE}}=2\pi \sqrt{\frac{1/2}{{10}^{-6}\times 20\times {10}^3}}$

$\Rightarrow T=10\pi$

So, time taken to come to equilibrium position from the extreme position directly is,

$t=\frac{T}{4}=\frac{10\pi }{4}=2.5\pi \text{s}$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->