Physics Solution – Question 11

Solution to Question 11

1. System Setup and Coordinate Definition

Consider a ring of radius $R$ uniformly charged with $+Q$ fixed in the $xy$-plane, centered at the origin. A thin rod of length $2R$, mass $m$, and uniform charge $-q$ lies along the $z$-axis (the axis of the ring).

In the equilibrium position, the center of the rod coincides with the center of the ring (origin). The rod extends from $z = -R$ to $z = +R$. The linear charge density of the rod is:

$$\lambda = \frac{-q}{2R}$$

Figure 1: Configuration showing the negatively charged rod displaced by a small distance $x$ along the axis of the positively charged ring.

2. Electric Field on the Axis

The electric field $E$ at a distance $z$ from the center of a uniformly charged ring of radius $R$ and charge $Q$ along its axis is given by:

$$E(z) = \frac{1}{4\pi\epsilon_0} \frac{Q z}{(R^2 + z^2)^{3/2}}$$

3. Calculating the Net Force

Let the rod be displaced by a small distance $x$ along the axis from the center. The center of the rod is now at coordinate $x$. Since the rod has length $2R$, its ends are located at coordinates:

Lower end: $z_1 = x – R$
Upper end: $z_2 = x + R$

Consider a small element of the rod of length $dz$ at position $z$. The charge on this element is $dQ_{rod} = \lambda dz$. The force $dF$ on this element due to the ring’s electric field is:

$$dF = (dQ_{rod}) E(z) = (\lambda dz) \frac{kQ z}{(R^2 + z^2)^{3/2}}$$

The total force $F$ is obtained by integrating over the length of the rod:

$$F = \int_{x-R}^{x+R} \frac{kQ\lambda z}{(R^2 + z^2)^{3/2}} \, dz$$

Substituting constants out and performing the integration $\int \frac{z}{(R^2+z^2)^{3/2}} dz = -\frac{1}{\sqrt{R^2+z^2}}$:

$$F = kQ\lambda \left[ -\frac{1}{\sqrt{R^2 + z^2}} \right]_{x-R}^{x+R}$$ $$F = -kQ\lambda \left[ \frac{1}{\sqrt{R^2 + (x+R)^2}} – \frac{1}{\sqrt{R^2 + (x-R)^2}} \right]$$

4. Small Oscillation Approximation

For small oscillations, $x \ll R$. We expand the terms inside the bracket using binomial approximation $(1+\epsilon)^n \approx 1+n\epsilon$.

First Term:

$$\frac{1}{\sqrt{R^2 + (R+x)^2}} = \frac{1}{\sqrt{R^2 + R^2 + 2Rx + x^2}} \approx \frac{1}{\sqrt{2R^2 + 2Rx}}$$ $$= \frac{1}{\sqrt{2}R} \left(1 + \frac{x}{R}\right)^{-1/2} \approx \frac{1}{\sqrt{2}R} \left(1 – \frac{x}{2R}\right)$$

Second Term:

$$\frac{1}{\sqrt{R^2 + (R-x)^2}} \approx \frac{1}{\sqrt{2R^2 – 2Rx}}$$ $$= \frac{1}{\sqrt{2}R} \left(1 – \frac{x}{R}\right)^{-1/2} \approx \frac{1}{\sqrt{2}R} \left(1 + \frac{x}{2R}\right)$$

Now, substitute these back into the force equation:

$$F = -kQ\lambda \left[ \frac{1}{\sqrt{2}R}\left(1 – \frac{x}{2R}\right) – \frac{1}{\sqrt{2}R}\left(1 + \frac{x}{2R}\right) \right]$$ $$F = -kQ\lambda \frac{1}{\sqrt{2}R} \left[ – \frac{x}{2R} – \frac{x}{2R} \right] = -kQ\lambda \frac{1}{\sqrt{2}R} \left( -\frac{x}{R} \right)$$ $$F = \frac{kQ\lambda x}{\sqrt{2}R^2}$$

5. Calculation of Time Period

Substitute $\lambda = -\frac{q}{2R}$ and $k = \frac{1}{4\pi\epsilon_0}$:

$$F = \frac{Q x}{\sqrt{2}R^2 4\pi\epsilon_0} \left( -\frac{q}{2R} \right)$$ $$F = – \left( \frac{Qq}{8\sqrt{2}\pi\epsilon_0 R^3} \right) x$$

This is in the form of Simple Harmonic Motion (SHM) equation $F = -K_{eq} x$, where the effective force constant is:

$$K_{eq} = \frac{Qq}{8\sqrt{2}\pi\epsilon_0 R^3}$$

The angular frequency $\omega$ is $\sqrt{\frac{K_{eq}}{m}}$. The time period $T$ is:

$$T = 2\pi \sqrt{\frac{m}{K_{eq}}} = 2\pi \sqrt{\frac{m \cdot 8\sqrt{2}\pi\epsilon_0 R^3}{Qq}}$$ $$T = \sqrt{ \frac{4\pi^2 \cdot 8\sqrt{2}\pi\epsilon_0 m R^3}{Qq} }$$

The period of small amplitude oscillations is: $$T = 2\pi \sqrt{\frac{8\sqrt{2}\pi\epsilon_0 m R^3}{Qq}}$$