Question

23.87: An electric dipole in a uniform horizontal electric field is displaced slightly from its equilibrium position as shown, where theta is small. The separation of the charges is 2a, and each of the two particles has mass m. a)Assuming the dipole is released from this position (at rest), show that its angular orientation exhibits simple harmonic motion (this is the same as showing that d^2 theta/d t^2 + omega * theta = 0 where omega is a collection of constants; this may be done by writing the torque equivalent of Newton's 2^nd Law and rearranging) with a frequency f= 1/2 pi Squareroot qE/ma .b) Suppose the

Answer 1

Torque exerted by electrostatic force on the positive charge

$\large \tau = \vec{r} \times \vec{F}$

$\large \tau = -rF\sin \theta$

Here electrostatic force F = qE, r = a Taking centre of the dipole as axis of rotation

$\large \tau = -aqE\sin \theta$

Similarly torque on negative charge

$\large \tau = -aqE\sin \theta$

Net Torque

$\large \tau _{net}= -2aqE\sin \theta$

From Newton's Second law of rotational Motion

$\large \tau_{net} = I\alpha = I\frac{\mathrm{d^{2}} \theta }{\mathrm{d} t^{2}}$

Here Moment of inertia of the dipole is given by

$\large I = ma^{2}+ma^{2} = 2ma^{2}$

Using all the equations we get

$\large 2ma^{2}\frac{\mathrm{d^{2}} \theta }{\mathrm{d} t^{2}} = -2aqE\sin \theta$

$\large ma\frac{\mathrm{d^{2}} \theta }{\mathrm{d} t^{2}} = -qE\sin \theta$

$\large \frac{\mathrm{d^{2}} \theta }{\mathrm{d} t^{2}} = -\frac{qE}{ma}\sin \theta$

For small angular displacement $\large \sin \theta \approx \theta$

$\large \frac{\mathrm{d^{2}} \theta }{\mathrm{d} t^{2}} = -\frac{qE}{ma} \theta$

So we can say that

$\large \frac{\mathrm{d^{2}} \theta }{\mathrm{d} t^{2}} \propto \theta$

So Dipole will execute the SHM.

Now comparing with standred equation of SHM we get

$\large \omega ^{2} = \frac{qE}{ma}$

$\large \omega = \sqrt{\frac{qE}{ma} }$

We know that

$\large \omega = 2\pi f= \sqrt{\frac{qE}{ma} }$   So

$\large f= \frac{1}{2\pi }\sqrt{\frac{qE}{ma} }$