Question 140: Electrostatics & Capacitance

Question

A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field $\overrightarrow{\mathrm{E}}_1=\mathrm{E}_0 \hat{x}$. If another electric field $\overrightarrow{\mathrm{E}}_2=2 \mathrm{E}_0(\hat{y}+\hat{z})$ is introduced to the system, what will be the percentage change in the frequency of the oscillation (approximate)?

Accepted Answer

For a rigid electric dipole in a uniform electric field, if it is given a small angular displacement, it performs SHM with angular frequency
$$ \omega=\sqrt{\frac{pE}{I}} $$
where
$p$ = dipole moment
$E$ = magnitude of effective electric field
$I$ = moment of inertia of the dipole
So, frequency depends on $\sqrt{E}$.
Initially, the field is
$$ \vec E_1=E_0 \hat x $$
Hence its magnitude is
$$ E_1=E_0 $$
So initial frequency is
$$ f_1 \propto \sqrt{E_0} $$
Now another field is added:
$$ \vec E_2=2E_0(\hat y+\hat z) $$
Let us find its components:
along $y$: $2E_0$
along $z$: $2E_0$
So the resultant field is
$$ \vec E=\vec E_1+\vec E_2=E_0\hat x+2E_0\hat y+2E_0\hat z $$
Its magnitude is
$$ E=\sqrt{(E_0)^2+(2E_0)^2+(2E_0)^2} $$
$$ E=E_0\sqrt{1+4+4}=3E_0 $$
Therefore new frequency is
$$ f_2 \propto \sqrt{3E_0}=\sqrt{3}\,f_1 $$
So percentage change in frequency is
$$ \frac{f_2-f_1}{f_1}	imes 100 = (\sqrt{3}-1)	imes 100 $$
Using $\sqrt{3}\approx 1.732$,
$$ (\sqrt{3}-1)	imes 100 \approx 0.732	imes 100=73.2\% $$
Approximate percentage change:
$$ \boxed{73\%} $$
So, the correct option is:
$$ \boxed{	ext{Option A}} $$

Answer

73%

Answer

63%

Answer

83%

Answer

53%

Electrostatics & Capacitance

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