91Ó°ÊÓ

Given:

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

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

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

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

Where: $\mathrm{δ}$is the charge per unit area and is given by

$\mathrm{σ}=\frac{Q}{A}=\frac{7×{10}^{−12}}{\mathrm{π}{\left(2.5×{10}^{−2}\right)}^2}=3.56×{10}^{−9}\mathrm{C}/{\mathrm{m}}^2$

Substitution in (1) results

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

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

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

Now when x >> a the following right:

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

Based on previous assumptions equation (2) becomes

$E_P=E_x=∫\frac{k\mathrm{σ}}{x^2}dA={∫}_0^R \frac{k\mathrm{σ}\left(2\mathrm{Ï€}R\right)dR}{x^2}=\frac{k\mathrm{σ}\mathrm{Ï€}{R}^2}{x^2}=\frac{kQ}{x^2}=\frac{Q}{4\mathrm{Ï€}\mathrm{Î\mu }{x}^2}$

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

$E_P=\frac{7×{10}^{−12}}{4\mathrm{π}×8.85×{10}^{−12}(.02{)}^2}=1.574\mathrm{N}/\mathrm{C}$

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

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

for the disk the electric field is

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

the percentage difference is the percentage error and is given by

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

Similarly at x =20m

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