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  • A dipole comprises of two charged particles of identical magnitude q and opposite in nature. The mass 'm' of the positive charged particle is half of the mass of the negative charged

A dipole comprises of two charged particles of identical magnitude q and opposite in nature. The mass 'm' of
the positive charged particle is half of the mass of the negative charged particle. The two charges are separated
by a distance 'L'. If the dipole is placed in a uniform electric field 'E '; in such a way that dipole axis makes a very small angle with the electric field, 'E '. The angular frequency of the oscillations of the dipole when released is given by :

Option: 1

\sqrt{\frac{3 \mathrm{qE}}{2 \mathrm{ml}}}


Option: 2

\sqrt{\frac{8 \mathrm{qE}}{ \mathrm{ml}}}


Option: 3

\sqrt{\frac{8 \mathrm{qE}}{3 \mathrm{ml}}}


Option: 4

\sqrt{\frac{4 \mathrm{qE}}{ \mathrm{ml}}}


Answers (1)

best_answer

In this case, since masses of both charges are not same, therefore, we need to find center of mass (COM), about
which dipole will oscillate and then we will find the moment of Inertia about this axis, to find torque & hence  \omega .
As we know, COM will divide length in the inverse ratio of the masses, therefore, COM will be at a distance of

\frac{\mathrm{L}}{3}   from 2m &   \frac{\mathrm{2L}}{3}   from m.

 

MI about this axis                                                                                                                             

I=2 m\left(\frac{L}{3}\right)^2+\left(\frac{2 L}{3}\right)^2

\text { Or I }=\frac{2 \mathrm{~mL}^2}{\mathrm{a}}+\frac{4 \mathrm{~mL}^2}{\mathrm{a}}=\frac{6 \mathrm{~mL}^2}{\mathrm{a}}=\frac{2 \mathrm{~mL}^2}{3}                        

If released, it will oscillate about centre of mass. For small '\theta '
\mathrm{T}=-\mathrm{PE}\times \theta

\begin{aligned} & \Rightarrow\left[2 \mathrm{~m} \frac{\mathrm{l}^2}{9}+\mathrm{m} \frac{4 \mathrm{l}^2}{9}\right] \alpha=-\mathrm{qLE} \cdot \theta \\ & \Rightarrow \frac{2 \mathrm{ml}^2}{3} \alpha=-\mathrm{qLE} . \theta \\ & \Rightarrow \alpha=-\frac{3 q E}{2 m L} \theta \\ & \omega=\sqrt{\frac{3 q E}{2 m L}} \\ & \end{aligned}

 

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mansi

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