Question

Q4. An electric dipole is made up of two particles having charges + luc, mass 1
kg and other with charge - 1pc and mass 1 kg separated by distance 1m. It is in
equilibrium in a uniform electric field of 20 x 103 V m-1. If the dipole is
deflected through angle 2° , time taken by it to come again in equilibrium is​

Answer 1

Given That,

First charge = 1 μC

Mass of first charge = 1 kg

Second charge = -1 μC

Mass of second charge = 1 kg

Distance = 1 m

Electric field $E=20\times10^{3}\ V/m$

Angle = 2°

We need to calculate the torque

Using formula of torque

$\tau=-2qlE\sin\theta$

For small θ,

$\tau=2qEl\theta$....(I)

We know that

$\tau=I\alpha$

Put the value of torque in equation (I)

$I\alpha=2qEl\theta$

$\alpha=\dfrac{2qEl\theta}{I}$

We know that,

The angular velocity is

$\omega^2=\dfrac{2qEl}{I_{0}}$

$\omega=\sqrt{\dfrac{2qEl}{I_{0}}}$

$\dfrac{2\pi}{T}=\sqrt{\dfrac{2qEl}{I_{0}}}$

$T=2\pi\sqrt{\dfrac{I_{0}}{2qEl}}}$

Here, Angular displacement is very small 2°.

So we can applied S.H.M

We need to calculate the time taken by it to come again in equilibrium

Using formula of time

$t=\dfrac{T}{4}$

Put the value of T

$t=\dfrac{2\pi}{4}\sqrt{\dfrac{I_{0}}{2qEl}}$

$t=\dfrac{\pi}{2}\sqrt{\dfrac{2ml^2}{2qEl}}$

$t=\dfrac{\pi}{2}\sqrt{\dfrac{ml}{qE}}$

Put the value into the formula

$t=\dfrac{\pi}{2}\sqrt{\dfrac{1\times0.5}{1\times10^{-6}\times20\times10^{3}}}$[/tex]t= 11.1\ sec[/tex]

Hence, The time taken by it to come again in equilibrium is 11.1 sec.

Answer 2

2.5pi

Explanation:

distance of Rcm is taken from centre of dipole