A uniformly charged disk like the disk in Fig. 21.25 has a radius of 2.50 cm and carries a total charge of 7.0 * 10-12 C.

(a) Find the electric field (magnitude and direction) on the x-axis at x = 20.0 cm

.

(b) Show that for x W R, becomes E = Q>4pP0x2, where Q is the total charge on the disk.

(c) Is the magnitude of the electric field you calculated in part (a) larger or smaller than the electric field 20.0 cm from a point charge that has the same total charge as this disk? In terms of the approximation used in part (b) to derive E = Q>4pP0x2 for a point charge from Eq. explain why this is so.

(d) What is the percent difference between the electric fields produced by the finite disk and by a point charge with the same charge at x = 20.0 cm and x = 10.0 cm?

Given:

$\mathrm{R}=2.5\mathrm{cm}=2.5\times {10}^{-2}\mathrm{m},\mathrm{Q}=7\times {10}^{-12}$

$$\begin{gathered} Eatt=20cm \\ Eatx?a \end{gathered}$$

Is the electric field in part (a) larger or smaller than (b)?

$VE\mathrm{\%}\text{at}x=20\mathrm{cm}\text{and}x=10\mathrm{cm}\text{.}$

a) Recall concluded expression for disk

$E=\frac{\sigma }{2e}\left[1-\frac{1}{\sqrt{{\left(\frac{R}{x}\right)}^2}+1}\right]$ ⇒ (1)

Where: $\delta$is the charge per unit area and is given by

$\sigma =\frac{Q}{A}=\frac{7\times {10}^{-12}}{\pi {\left(2.5\times {10}^{-2}\right)}^2}=3.56\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2$

Substitution in (1) results

$E=\frac{3.56\times {10}^{-9}}{2\times 8.85\times {10}^{-12}}\left[1-\frac{1}{\sqrt{(2.5/20{)}^2+1}}\right]=1.55\mathrm{N}/\mathrm{C}$

b) To get an approximation for the field wield we know that

$d{E}_x=\frac{kdQ}{r^2}\cos \left(\theta \right)$ → (2)

Now when x >> a the following right:

role="math" localid="1668257867382" $\begin{array}{c}\cos \left(\theta \right)=\frac{x}{\sqrt{x^2+R^2}}\approx \frac{x}{x}=1\text{because}x?R\\ r^2={\left(\sqrt{a^2+x^2}\right)}^2=x^2+R^2\approx x^2\\ \\ dQ=\sigma dA\end{array}$

Based on previous assumptions equation (2) becomes

$E_P=E_x=\int \frac{k\sigma }{x^2}dA={\int }_0^R \frac{k\sigma \left(2\pi R\right)dR}{x^2}=\frac{k\sigma \pi R^2}{x^2}=\frac{kQ}{x^2}=\frac{Q}{4\pi \epsilon x^2}$

The electric field $E_p$, at $x$=0.2 m is given by

$E_P=\frac{7\times {10}^{-12}}{4\pi \times 8.85\times {10}^{-12}(.02{)}^2}=1.574\mathrm{N}/\mathrm{C}$

So the electric field of the line is smaller than the approximation field at x =0.2 m

d) The approximate electric field at x = 0.1 m is given by

$E_P=\frac{9\times {10}^9\times 7\times {10}^{-12}}{(0.1{)}^2}=6.3\mathrm{N}/\mathrm{C}$

for the disk the electric field is

$E_P=\frac{3.56\times {20}^{-9}}{2\times 8.85\times {10}^{-12}}\left[1-\frac{1}{\sqrt{(2.5/10{)}^2+1}}\right]=6.00\mathrm{N}/\mathrm{C}$

the percentage difference is the percentage error and is given by

${\mathrm{VE}}_{10}\mathrm{\%}=\frac{E_P-E}{E_P}\times 100=\frac{0.3}{6.3}\times 100=4.76\mathrm{\%}$

Similarly at x =20m

${\mathrm{V}}_{20}\mathrm{\%}=\frac{E_P-E}{E_P}\times 100=\frac{0.024}{1.574}\times 100=1.5\mathrm{\%}$