Question 794: Electrostatics & Capacitance

Question

Two beads, each with charge $q$ and mass $m$, are on a horizontal, frictionless, non-conducting, circular hoop of radius $a$. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by
[ $\varepsilon_0$ is the permittivity of free space.]

Accepted Answer

Consider a circular hoop of radius $a$ with centre $O$.
Bead $A$ is fixed to the hoop, while bead $B$ can slide freely along its circumference.
Forces on bead $B$:
Electrostatic force
due to bead $A$, acting along the line $AB$.
Normal reaction
from the hoop, directed radially inward toward $O$.
Because these two forces are the only interactions, bead $B$ reaches equilibrium at the point $C$Ã¢ÂÂthe position diametrically opposite $A$Ã¢ÂÂwhere the Coulomb repulsion is exactly balanced by the hoopÃ¢ÂÂs normal reaction.
Displace bead $B$ slightly from its equilibrium point $C$ by a small angle $	heta$ measured at the centre $O$.
1. Geometry of the configuration
In the isosceles triangle $OAB$ (see figure)
$\angle A=\dfrac{	heta}{2}$
$\angle O = 180^{\circ}-	heta$
The side opposite $\angle O$ is $AB$.  Using the cosine rule,
$$ AB^{2}=a^{2}+a^{2}-2a^{2}\cos(180^{\circ}-	heta)=4a^{2}\cos^{2}\!\frac{	heta}{2}. $$
Hence the separation of the charges is
$$ r=AB=2a\cos\frac{	heta}{2}. $$
2. Coulomb force on $B$
$$ F=\frac{q^{2}}{4\pi\varepsilon_{0}r^{2}}   =\frac{q^{2}}{16\pi\varepsilon_{0}a^{2}\cos^{2}\dfrac{	heta}{2}}. $$
3. Components along the hoop
The force makes an angle $\dfrac{	heta}{2}$ with the radial line $OB$.
Normal component: $F\cos\dfrac{	heta}{2}$  (balanced by the normal reaction of the hoop).
Tangential component: $F\sin\dfrac{	heta}{2}$  (provides the restoring acceleration).
Applying NewtonÃ¢ÂÂs second law tangentially,
$$ ma\frac{d^{2}	heta}{dt^{2}}       =-F\sin\frac{	heta}{2}       =-\frac{q^{2}}{16\pi\varepsilon_{0}a^{2}\cos^{2}\dfrac{	heta}{2}}\,         \sin\frac{	heta}{2}. $$
(The minus sign shows the couple is restoring.)
4. Small-angle approximation
$ \sin\frac{	heta}{2}\simeq\frac{	heta}{2},\qquad \cos\frac{	heta}{2}\simeq1. $
Substituting,
$ \frac{d^{2}	heta}{dt^{2}}       =-\frac{q^{2}}{32\pi\varepsilon_{0}ma^{3}}\;	heta       =-\omega^{2}	heta, $
where
$ \boxed{\displaystyle \omega^{2}         =\frac{q^{2}}{32\pi\varepsilon_{0}ma^{3}} }. $
The motion is simple harmonic, with angular frequency $\omega$ given above.

Answer

$${{{q^2}} \over {4\pi {\varepsilon _0}{a^3}m}}$$

Answer

$${{{q^2}} \over {32\pi {\varepsilon _0}{a^3}m}}$$

Answer

$${{{q^2}} \over {8\pi {\varepsilon _0}{a^3}m}}$$

Answer

$${{{q^2}} \over {16\pi {\varepsilon _0}{a^3}m}}$$

Electrostatics & Capacitance

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