Problem 7: Small Oscillations on a Ring
1. Derivation of Electric Field in the Plane
We need the Electric Field \(E\) at a point \(P\) a distance \(x\) from the center, lying on the diameter. Let the linear charge density be \(\lambda = \frac{Q}{2\pi R}\).
Consider a small charge element \(dq = \lambda R d\phi\) at an angle \(\phi\) on the ring.
The vector distance from \(dq\) to point \(P\) (at \(x,0\)) is \(\vec{r} = (x – R\cos\phi)\hat{i} – (R\sin\phi)\hat{j}\).
The distance cubed is \( |\vec{r}|^3 = (R^2 + x^2 – 2Rx\cos\phi)^{3/2} \).
The electric field component along the x-axis (diameter) is:
$$ dE_x = \frac{1}{4\pi\epsilon_0} \frac{dq (x – R\cos\phi)}{(R^2 + x^2 – 2Rx\cos\phi)^{3/2}} $$Small x approximation (\(x \ll R\)):
1. Neglect \(x^2\) in the denominator.
2. Use binomial expansion for the denominator:
$$ (R^2 – 2Rx\cos\phi)^{-3/2} \approx R^{-3} (1 – \frac{2x}{R}\cos\phi)^{-3/2} \approx \frac{1}{R^3}(1 + \frac{3x}{R}\cos\phi) $$Multiply this with the numerator \((x – R\cos\phi)\) and neglect higher order terms of \(x\):
$$ \text{Integrand} \propto (x – R\cos\phi)(1 + \frac{3x}{R}\cos\phi) \approx x – R\cos\phi – 3x\cos^2\phi $$Now integrate from \(0\) to \(2\pi\):
- \(\int_{0}^{2\pi} x d\phi = 2\pi x\)
- \(\int_{0}^{2\pi} \cos\phi d\phi = 0\)
- \(\int_{0}^{2\pi} \cos^2\phi d\phi = \pi\)
Summing these: \( 2\pi x – 0 – 3x(\pi) = -\pi x \).
Substituting constants back in:
$$ E_x = \frac{1}{4\pi\epsilon_0} \frac{\lambda R}{R^3} (-\pi x) = – \frac{1}{4\pi\epsilon_0} \frac{Q}{2\pi R} \frac{\pi x}{R^2} $$ $$ E_x = – \frac{Q x}{8\pi\epsilon_0 R^3} $$2. Restoring Force & Time Period
The force on the bead is \( F = qE = – \left( \frac{Qq}{8\pi\epsilon_0 R^3} \right) x \).
This is the form \( F = -k_{eff} x \) with effective spring constant:
$$ k_{eff} = \frac{Qq}{8\pi\epsilon_0 R^3} $$The time period \( T = 2\pi \sqrt{\frac{m}{k_{eff}}} \) is:
$$ T = 2\pi \sqrt{ \frac{m (8\pi\epsilon_0 R^3)}{Qq} } $$