Q83P - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

a) The electric field on the x-axis at x = 20.0 cm is 1.55N/C

b) Proved below

c) $E_p$=1.574N/C so the field due to disk is smaller than the field due to line.

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 $V{E}_{10}\%$=4.46,$V{E}_{20}\%$=1.5% simultaneously.

Step 1:

Given:

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

Required:

             $$\begin{gathered} E at t = 20 cm \\ E at x ? a \end{gathered}$$

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

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

Step 2: Concluding equation by given data

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: 

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

Based on previous assumptions equation (2) becomes

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

Step 3: Putting values and calculation

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

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