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Q13P

Expert-verifiedFound in: Page 653

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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} **

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

- 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, $\overrightarrow{\mathrm{E}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\overrightarrow{{\mathrm{E}}_{\mathrm{i}}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon}}_{\mathrm{o}}{\mathrm{r}}_{\mathrm{i}}^{2}}\hat{{\mathrm{r}}_{\mathrm{i}}}$ (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:** **

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

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:

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

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:** **

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

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:** **

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

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 z^{2}, 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.

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