Figure 22-28 shows two disks and a flat ring, each with the same uniform charge Q. Rank the objects according to the magnitude of the electric field they create at points P(which are at the same vertical heights), greatest first.

The transverse speed of the wave is the displacement of the wave in the given period of its oscillation and the period is given by the reciprocal of the frequency of the wave. Thus, using the given formulae with the given data, the required values can be calculated.

The electric field at a point on the central axis through a uniformly charged disk,

$E=\frac{\sigma }{2{\epsilon }_0}(1-\frac{z}{\sqrt{z^2+R^2}})$………..(i)

The electric field at a point due to uniformly charged ring,

$E=\frac{qz}{4\pi {\epsilon }_0{(z^2+R^2)}^{3/2}}$ …………. (ii)

The surface charge density of a material,

$\sigma =\frac{q}{A}$………..(iii)

Let, the vertical height of Point P form the center of each given disk and ring be$h$.

For object (a),

Using the given data and equation (ii) in equation (i), the magnitude of the electric field at the point P for disk 1 will be given as:

$\begin{array}{c}{E}_1=\frac{Q}{4\pi {\epsilon }_{0}R}(1-\frac{h}{\sqrt{h^2+R^2}})(\because \text{Fromequation(iii),}\sigma =Q/2\pi R)\\ =\frac{Q}{4\pi {\epsilon }_{0}R}(1-\frac{h}{R})(let,h<<R)\end{array}$

For object (b),

Using the given data and equation (ii) in equation (i), the magnitude of the electric field at the point P for disk 2 will be given as:

$\begin{array}{c}{E}_2=\frac{Q}{8\pi {\epsilon }_{0}R}(1-\frac{h}{\sqrt{h^2+4R^2}})\left(\because \text{Fromequation(iii),}\sigma =Q/2\pi \left(2R\right)\right)\\ =\frac{Q}{8\pi {\epsilon }_{0}R}(1-\frac{h}{2R})(let,h<<R)\end{array}$

For object (c),

Using the given data and equation (ii) in equation (i), the magnitude of the electric field at the point P for flat ring (can be treated as bigger disk with a deduction of small disk) will be given as:

$\begin{array}{c}{E}_3=\frac{Q}{8\pi {\epsilon }_{0}R}(1-\frac{h}{\sqrt{h^2+4R^2}})-\frac{Q}{4\pi {\epsilon }_{0}R}(1-\frac{h}{\sqrt{h^2+R^2}})\\ \left(\because \text{Fromequation(iii),}{\sigma }_{outer}=Q/2\pi \left(2R\right),{\sigma }_{inner}=Q/2\pi \left(2R\right)\right)\\ =\frac{Q}{8\pi {\epsilon }_{0}R}(1-\frac{h}{2R})-\frac{Q}{4\pi {\epsilon }_{0}R}(1-\frac{h}{R})\text{(let,h<<R)}\\ =-\frac{Q}{4\pi {\epsilon }_{0}R}-\frac{Q}{16\pi {\epsilon }_{0}R}\left(\frac{h}{R}\right)+\frac{Q}{4\pi {\epsilon }_{0}R}\left(\frac{h}{R}\right)\\ =\frac{Q}{4\pi {\epsilon }_{0}R}(\frac{3h}{4R}-1)\end{array}$

If the vertical height is considered zero, the rank of the electric fields will be$E_1>E_2>E_3$.

Hence, the rank of the magnitude of electric fields will be$E_1>E_2>E_3$ .