Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

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

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, $\vec{E} = \frac{q}{4 {πε}_{o} R^{2}} \hat{R}$ (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, $\vec{E} = \sum_{i = 1}^{n} \vec{E_{i}} = \sum_{i = 1}^{n} \frac{q}{4 {πε}_{o} r_{i}^{2}} \hat{r_{i}}$ (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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $z / \sqrt{r}$ ) 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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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of electric field

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

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

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

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

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

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of electric field

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

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

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

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

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

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