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_cat the disk center and electrons e_sat opposite sides of the disk, at radius Rfrom the center. The proton is initially at distance z=R=2.00 cmfrom the disk. At that location, what are the magnitudes of (a) the electric field $\overrightarrow{{\mathbf{E}}_{\mathbf{c}}}$ due to electron e_cand (b) the netelectric 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)?

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

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

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.

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$2.55\times {10}^{-6}\mathrm{N}/\mathrm{C}$

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$3.6\times {10}^{-4}\mathrm{N}/\mathrm{C}$

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$7.09\times {10}^{-7}\mathrm{N}/\mathrm{C}$

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.