For a disk of radius 20 cm with uniformly distributed charge $\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{C}$, calculate the magnitude of the electric field on the axis of the disk, 5 mm from the center of the disk, using each of the following equations:

(a)$\mathbf{E}\mathbf{=}\frac{\left(Q/A\right)}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\left[1-\frac{z}{{\left(R^2+z^2\right)}^{1/2}}\right]$

(b)$\mathbf{E}\mathbf{\approx }\frac{\mathbf{Q}\mathbf{/}\mathbf{A}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\left[1-\frac{z}{R}\right]$

(c)$\mathbf{E}\mathbf{\approx }\frac{\mathbf{Q}\mathbf{/}\mathbf{A}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}$

(d) How good are the approximate equations at this distance? (e) At what distance does the least accurate approximation for the electric field give a result that is closest to the most accurate: at a distance R/2, close to the disk, at a distance R, or far from the disk?

The given data can be listed below,

• The radius of disk is,$\mathrm{R}=20\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$
• The charge on the disk is,$\mathrm{Q}=7\times {10}^{-6}\mathrm{C}$

The electric field intensity E is measured in volts per metre and is a type of electric energy that occurs in vacuum.

It is most closely linked to the concepts of voltage and force.

The electric field on the axis of disk in given equation is given by,

$E=\frac{\left(Q/A\right)}{2{\epsilon }_0}\left[1-\frac{z}{{\left(R^2+z^2\right)}^{v2}}\right]$ …(i)

Substitute all the values in the above equation

$$\begin{gathered} \mathrm{E}=\frac{7\times {10}^{-6}\mathrm{C}}{2\mathrm{\pi }{\left(0.20\mathrm{m}\right)}^2{\mathrm{\epsilon }}_0}\left[1-\frac{5\times {10}^{-3}\mathrm{m}}{\sqrt{{\left(0.20\mathrm{m}\right)}^2+{\left(5\times {10}^{-3}\mathrm{m}\right)}^2}}\right] \\ =\frac{5.57\times {10}^{-5}\mathrm{C}/{\mathrm{m}}^2}{2{\mathrm{\epsilon }}_0}\left[1-.242\right] \\ =3.068\times {10}^6\mathrm{N}/\mathrm{m} \end{gathered}$$

Thus, the electric field on the axis of disk is$3.068\times {10}^6\mathrm{N}/\mathrm{m}$.

The given equation for electric field is expressed as,

$E=\frac{\left(Q/A\right)}{2{\epsilon }_0}\left[1-\frac{z}{R}\right]$ …(ii)

Here, Q is the charge on the disk, A is the area of disk,${\epsilon }_0$ is the permittivity of free space, R is the radius of the disk and z is the radial distance from centre of disk.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{E}=\frac{5.57\times {10}^{-5}}{2{\mathrm{\epsilon }}_0}\left(1-\frac{5\times {10}^{-3}\mathrm{m}}{0.20\mathrm{m}}\right) \\ =3.067\times {10}^6\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the value of electric field by given equation is$3.067\times {10}^6\mathrm{N}/\mathrm{C}$ .

The given equation for electric field is expressed as,

$E=\frac{Q/A}{2{\epsilon }_0}$ …(iii)

Substitute all the values in the equation.

localid="1656933814468" $$\begin{gathered} E=\frac{5.57\times {10}^{-5}\mathrm{C}/{\mathrm{m}}^2}{2{\mathrm{\epsilon }}_0} \\ =3.145\times {10}^6\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the electric field by given equation is$3.145\times {10}^6\mathrm{N}/\mathrm{C}$.

The second approximation in par(b) yields a very accurate result at this distance, with the exact equation and approximated equation in part (b) being the same when rounded to three significant numbers, but the approximated equation in part (c) yields a result with a 2.6% increase in the original value.

Thus, the accurate answer is given by equation (b) and in the equation (c) the value is increase by 2.6%.

$E=\frac{\left(Q/A\right)}{2{\epsilon }_0}\left[1-\frac{z}{{\left(R^2+z^2\right)}^{1/2}}\right]$

substitute z = R/2 in the above,

$$\begin{gathered} E=\frac{\left(Q/A\right)}{2{\epsilon }_0}\left[1-\frac{R/2}{{\left(R^2+{\left({\displaystyle \frac{R}{2}}\right)}^2\right)}^{1/2}}\right] \\ =\frac{\left(Q/A\right)}{2{\epsilon }_0}\left(1-\frac{1}{\sqrt{5}}\right) \\ =\left(0.553\right)\frac{\left(Q/A\right)}{2{\epsilon }_0} \end{gathered}$$

When z≫R,${\left(R^2+z^2\right)}^{1/2}$can be approximated as R so the equation will become,

$E=\frac{\left(Q/A\right)}{2{\epsilon }_0}\left(1-\frac{z}{R}\right)$

When R≫z z/R term can be neglected. So,the equation will be given by,

$E=\frac{\left(Q/A\right)}{2{\epsilon }_0}\left[1\right]$

$$\begin{gathered} E=\frac{\left(Q/A\right)}{2{\epsilon }_0}\left[1-\frac{R}{{\left(R^2+R^2\right)}^{1/2}}\right] \\ =\frac{\left(Q/A\right)}{2{\epsilon }_0}\left(1-\frac{1}{\sqrt{}}\right) \\ =\left(0.293\right)\frac{\left(Q/A\right)}{2{\epsilon }_0} \end{gathered}$$

$$\begin{gathered} E=\frac{\left(Q/A\right)}{2{\epsilon }_0}\left[1-\frac{z}{{\left(R^2+z^2\right)}^{1/2}}\right] \\ =\frac{\left(Q/A\right)}{2{\epsilon }_0}\left(1-\frac{z}{z}\right) \\ =0 \end{gathered}$$

The electric field far from disk is zero.

Thus, the value of electric field is approximately same when the observational point is closer to the disk.