Q13P - Fundamentals Of Physics

Question

Figure 22-40 shows a proton (p) on the central axis through a disk with a uniform charge density due to excess electrons. The disk is seen from an edge-on view. Three of those electrons are shown: electron e_c at the disk center and electrons e_s at opposite sides of the disk, at radius R from the center. The proton is initially at distance z=R=2.00 cm from the disk. At that location, what are the magnitudes of (a) the electric field $\overrightarrow{{\mathbf{E}}_{\mathbf{c}}}$ due to electron e_c and (b) the net electric field ${\overrightarrow{\mathbf{E}}}_{\mathbf{s}\mathbf{,}\mathbf{net}}$ due to electrons e_s? The proton is then moved to z=R/10.0. What then are the magnitudes of (c) ${\overrightarrow{\mathbf{E}}}_{\mathbf{c}}$ and ${\overrightarrow{\mathbf{E}}}_{\mathbf{s}\mathbf{,}\mathbf{net}}$ (d)  at the proton’s location? (e) From (a) and (c) we see that as the proton gets nearer to the disk, the magnitude of ${\overrightarrow{\mathbf{E}}}_{\mathbf{c}}$ increases, as expected. Why does the magnitude of ${\overrightarrow{\mathbf{E}}}_{\mathbf{s}\mathbf{,}\mathbf{net}}$ from the two side electrons decrease, as we see from (b) and (d)?

Step-by-Step Solution

Answer

1. The magnitude of the electric field due to electron e_c is ${\mathrm{E}}_{\mathrm{c}}=3.6\times {10}^{-6} \mathrm{N}/\mathrm{C}$
2. The net electric field due to electrons e_s is ${\mathrm{E}}_{\mathrm{s}}=2.55\times {10}^{-6} \mathrm{N}/\mathrm{C}$
3. The magnitude of the electric field when the proton gets nearer to the disk is $3.60\times {10}^{-4} \mathrm{N}/\mathrm{C}$
4. The magnitude of the electric field as the net electric field increases as expected is $7.09\times {10}^{-7} \mathrm{N}/\mathrm{C}$
5. From the relation of distance and electric filed being proportional, the field decreases for the side two electrons.

Step 1: The given data

• Charge of electron, ${\mathrm{q}}_{\mathrm{e}}=-1.6\times {10}^{-19} \mathrm{C}$
• Distance of the filed point from the charge, $\mathrm{R}=0.020 \mathrm{m}$

Step 2: Understanding the concept of electric field

Electric field is a vector field that represents the direction of force on the charge placed at some point in the field.

Formulae:

The magnitude of the electric field, $\overrightarrow{\mathrm{E}}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2}\hat{\mathrm{R}}$                                                   (i)  

where, R is the distance of field point from the charge and q is the charge of the particle

According the superposition principle, the electric field at a point due to more than one charges, $\overrightarrow{\mathrm{E}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\overrightarrow{{\mathrm{E}}_{\mathrm{i}}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}_{\mathrm{i}}^2}\widehat{{\mathrm{r}}_{\mathrm{i}}}$                                                                     (ii)

Distance of the field point from the charge =r  

Step 3: a) Calculation of the electric field due to electron e c

The electron is a distance R=0.020 m away from the centre of the disk. Thus, the electric field at that point can be given using equation (i) as follows: 

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{c}}& =& \frac{\left(8.99\times {10}^9 \mathrm{N}.{\displaystyle \frac{\mathrm{m}}{{\mathrm{C}}^2}}\right)}{\left(0.020 {\mathrm{m}}^2\right)}\left[1.60\times {10}^{-19} \mathrm{C}\right]\\ & =& 3.6\times {10}^{-6} \mathrm{N}/\mathrm{C}\end{array}$

Hence, the magnitude of the electric field is $3.6\times {10}^{-6} \mathrm{N}/\mathrm{C}$ and it is directed towards e_c. 

Step 4: b) Calculation of the electric field due to electrons e s →

The horizontal components of the individual fields (due to the two $\overrightarrow{{\mathrm{e}}_{\mathrm{s}}}$ charges) cancel, and the vertical components add to give the net electric field using equations (i) and (ii) as given:

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{s},\mathrm{net}}& =& \frac{2\mathrm{ez}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\left({\mathrm{R}}^2+{\mathrm{z}}^2\right)}^{3/2}}\\ & =& \frac{2\left(8.99\times {10}^9.\mathrm{N}{\displaystyle \frac{{\mathrm{m}}^2}{{\mathrm{C}}^2}}\right)\left(1.60\times {10}^{-19} \mathrm{C}\right)\left(0.020 \mathrm{m}\right)}{{\left[{\left(0.020 \mathrm{m}\right)}^2+{\left(0.020 \mathrm{m}\right)}^2\right]}^{3/2}}\\ & =& 2.55\times {10}^{-6} \mathrm{N}/\mathrm{C}\end{array}$          

Hence, the value of the electric field is $2.55\times {10}^{-6} \mathrm{N}/\mathrm{C}$

Step 5: c) Calculation of the electric field when proton gets nearer to the disk

Now, the electron is at a distance $\frac{\mathrm{R}}{10}=\frac{0.020 \mathrm{m}}{10}=0.002 \mathrm{m}$ away from the centre of the disk. Thus, the electric field at that point can be given using equation (i) as follows: 

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{c}}& =& \frac{\left(8.99\times {10}^9. \mathrm{N}{\displaystyle \frac{\mathrm{m}}{{\mathrm{C}}^2}}\right)}{{\left(0.0020 \mathrm{m}\right)}^2}\left[1.60\times {10}^{-19} \mathrm{C}\right]\\ & =& 3.6\times {10}^{-4} \mathrm{N}/\mathrm{C}\end{array}$

Hence, the value of the electric field is $3.6\times {10}^{-4} \mathrm{N}/\mathrm{C}$

Step 6: d) Calculation of the electric field due to the side charges

For $\mathrm{z}=\frac{\mathrm{R}}{10}=\frac{0.020 \mathrm{m}}{10}=0.002 \mathrm{m}$, the electric field at that point can be given using equation (i) as follows: 

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{s},\mathrm{net}}& =& \frac{2\mathrm{ez}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\left({\mathrm{R}}^2+{\mathrm{z}}^2\right)}^{3/2}}\\ & =& \frac{2\left(8.99\times {10}^9 \mathrm{N}.{\displaystyle \frac{{\mathrm{m}}^2}{{\mathrm{C}}^2}}\right)\left(1.6\times {10}^{-19} \mathrm{C}\right)\left(0.020 \mathrm{m}\right)}{{\left[{\left(0.020 \mathrm{m}\right)}^2+{\left(0.020 \mathrm{m}\right)}^2\right]}^{3/2}}\\ & =& 7.09\times {10}^{-7} \mathrm{N}/\mathrm{C}\\ {\mathrm{E}}_{\mathrm{s},\mathrm{net}}& =& 7.09\times {10}^{-7} \mathrm{N}/\mathrm{C}\end{array}$

 

Hence, the required value of the electric field is $7.09\times {10}^{-7} \mathrm{N}/\mathrm{C}$

Step 7: e) Comparing the field strengths in part (b) and part (d)

Since, E_c is inversely proportional to z², thus, the value of the electric field gets smaller than that of part (a). For the net electric field E_s,net, due to the side charges, the trigonometric factor for the y-component shrinks to almost a linear value (that is $\mathrm{z}/\sqrt{\mathrm{r}}$) for very small z and thus, it leads to the decrease in the net electric field for x-components cancelling each other.