Electrostatics

JEE Main level, single correct, medium

253 of 311

Question 253 of 311

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

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Answer: (A) 3qE2ml\sqrt{\dfrac{3qE}{2ml}}

The data given is
l=l=distance between the charges
q=q=charge
m=m=mass of the positive charge
2m2m==mass of negative charge
The moment of inertia of the system is
I=(m×2mm+2m)(l)2=2ml23I=\left(\dfrac{m\times 2m}{m+2m}\right){\left(l\right)}^{2}=\dfrac{2m{l}^{2}}{3}
The angular frequency can be written as
ω=pEI...(i)\omega =\sqrt{\dfrac{pE}{I}}...(i)
Substituting the value of moment of inertia in the equation (i) we get
ω=pE2ml23=3pE2ml2\omega =\sqrt{\dfrac{pE}{\dfrac{2m{l}^{2}}{3}}}=\sqrt{\dfrac{3pE}{2m{l}^{2}}}
The value of the dipole moment is
p=qlp=ql
Hence, the angular frequency becomes
ω=3qE2ml\omega =\sqrt{\dfrac{3qE}{2ml}}
This question was given bonus by NTA as none of the options matched in the original paper.

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