Question

A thin disk with a circular hole at its centre, called an annulus, has inner radius R₁ and outer radius R₂ (Fig. P21.91). The disk has a uniform positive surface charge density σ on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the yz-plane, with its canter at the origin. For an arbitrary point on the x-axis (the axis of the annulus), find the magnitude and direction of the electric field $\overrightarrow{E}$. Consider points both above and below the annulus. (c) Show that at points on the x-axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the canter of the annulus and the point. How close is “sufficiently close”? (d) A point particle with mass m and negative charge -q is free to move along the x-axis (but cannot move off the axis). The particle is originally placed at rest at x = 0.01R1 and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

Short answer

1. The total electric charge on the annulus of inner radius $R_1$ and outer radius $R_2$ is $\sigma \pi ({R_2}^2-{R_1}^2)$
2. For an arbitrary point on the x-axis of the annulus the magnitude and the direction of the electric field $\overrightarrow{E}$ is $E=\frac{\sigma x}{2{\epsilon }_{\circ }}\left(-\frac{1}{{(x^2+{R_2}^2)}^{\frac{1}{2}}}+\frac{1}{{(x^2+{R_1}^2)}^{\frac{1}{2}}}\right)$
3. The magnitude of the electric field is approximately proportional to the distance between the canter of the annulus and the point, The distance is:
   $E_{app}=\frac{\sigma x}{2{\epsilon }_{\circ }}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
4. The frequency of the oscillation of the particle is: $\frac{1}{2\pi }\sqrt{\frac{\sigma q}{2m{\epsilon }_{\circ }}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}$

Step-by-step solution

Calculating the total charge:

a)

The charge in uniformly distributed, therefore the total charge is:

$\begin{array}{r}Q=\sigma A_A\\ =\sigma (A_{\circ }-A_i)\\ =\sigma \pi ({R_2}^2-{R_1}^2)\end{array}$

Therefore, the total charge is $\sigma \pi ({R_2}^2-{R_1}^2)$

Calculating Magnitude and direction of the electric field

b)

The electric field of the annuls is similar to the dick, but instead the radius is from $0\to R$instead from $R_1\to R_2$.

Therefore,

$\begin{array}{r}E=E_x\\ E_x=\frac{\sigma x}{4{\epsilon }_{\circ }}{\int }_{R_1}^{R_2}\frac{2rdr}{{(x^2+r^2)}^{\frac{3}{2}}}\\ =\frac{\sigma x}{2{\epsilon }_{\circ }}{\left(-\frac{1}{{(x^2-r^2)}^{\frac{1}{2}}}\right)}_{R_2}^{R_1}\\ =\frac{\sigma x}{2{\epsilon }_{\circ }}\left(-\frac{1}{{(x^2+{R_2}^2)}^{\frac{1}{2}}}+\frac{1}{{(x^2+{R_1}^2)}^{\frac{1}{2}}}\right)\end{array}$

Hence the required value is $E=\frac{\sigma x}{2{\epsilon }_{\circ }}\left(-\frac{1}{{(x^2+{R_2}^2)}^{\frac{1}{2}}}+\frac{1}{{(x^2+{R_1}^2)}^{\frac{1}{2}}}\right)$

c)

If $x<<R_1$and $x<<R_2$then the above equation can be reduced to

$E=\frac{\sigma x}{2{\epsilon }_{\circ }}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Here, $x$is a variable and the electric field is proportional to it at a very close distance.

Calculating the frequency

d)

The restoring force according to simple harmonic motion is given by: $F=-kd\dots \dots \dots .\left(i\right)$

It is given that the equilibrium distance$x=-0.01R_1$, that is $x<<R_1$.

Therefore, the electric force at equilibrium position is

$kd=\frac{\sigma qx}{2{\epsilon }_{}}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)..........(ii)$

But it is also given that $d=x$

Therefore $k$ is given by:

$k=\frac{\sigma q}{2{\epsilon }_{\circ }}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)..........(iii)$

Therefore, the frequency is given by:

$\begin{array}{r}f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\\ =\frac{1}{2\pi }\sqrt{\frac{\sigma q}{2m{\epsilon }_{\circ }}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}\end{array}$

Hence the frequency is $\frac{1}{2\pi }\sqrt{\frac{\sigma q}{2m{\epsilon }_{\circ }}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}$