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Found in: Page 653

### Fundamentals Of Physics

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 ec at the disk center and electrons es 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 $\stackrel{\mathbf{\to }}{{\mathbf{E}}_{\mathbf{c}}}$ due to electron ec and (b) the net electric field ${\stackrel{\mathbf{\to }}{\mathbf{E}}}_{\mathbf{s}\mathbf{,}\mathbf{net}}$ due to electrons es? The proton is then moved to z=R/10.0. What then are the magnitudes of (c) ${\stackrel{\mathbf{\to }}{\mathbf{E}}}_{\mathbf{c}}$ and ${\stackrel{\mathbf{\to }}{\mathbf{E}}}_{\mathbf{s}\mathbf{,}\mathbf{net}}$ (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 ${\stackrel{\mathbf{\to }}{\mathbf{E}}}_{{\mathbf{c}}}$ increases, as expected. Why does the magnitude of ${\stackrel{\mathbf{\to }}{\mathbf{E}}}_{\mathbf{s}\mathbf{,}\mathbf{net}}$ from the two side electrons decrease, as we see from (b) and (d)?

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

## Step 1: The given data

• Charge of electron, ${\mathrm{q}}_{\mathrm{e}}=-1.6×{10}^{-19}\mathrm{C}$
• 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, $\stackrel{\to }{\mathrm{E}}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^{2}}\stackrel{^}{\mathrm{R}}$ (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, $\stackrel{\to }{\mathrm{E}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\stackrel{\to }{{\mathrm{E}}_{\mathrm{i}}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}_{\mathrm{i}}^{2}}\stackrel{^}{{\mathrm{r}}_{\mathrm{i}}}$ (ii)

Distance of the field point from the charge =r

## Step 3: a) Calculation of the electric field due to electron ec

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:

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

## Step 4: b) Calculation of the electric field due to electrons es→

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

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

## 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:

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

## 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:

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

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

Since, Ec is inversely proportional to z2, thus, the value of the electric field gets smaller than that of part (a). For the net electric field Es,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.