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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Figure 22-39 shows an uneven arrangement of electrons (e) and protons (p) on a circular arc of radius r=2.00 cm, with angles , , , and . What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the net electric field produced at the center of the arc?

1. The magnitude of the net electric field produced at the center of the arc is $3.92×{10}^{-6}\mathrm{N}/\mathrm{C}$.
2. The direction of the net electric field produced at the center of the arc is data-custom-editor="chemistry" $-76.{4}^{\circ }$.
See the step by step solution

## Step 1: The given data

1. Radius of the circular disc, $\mathrm{r}=2.00\mathrm{cm}$
2. Arrangement of electrons and protons at angles ${\mathrm{\theta }}_{1}=30.{0}^{\circ }$, ${\mathrm{\theta }}_{2}=50.{0}^{\circ }$, ${\mathrm{\theta }}_{3}=30.{0}^{\circ }$, and ${\mathrm{\theta }}_{4}=20.{0}^{\circ }$.

## Step 2: Understanding the concept of electric field

Using the concept of the electric field at a given point, we can get the value of an individual electric field by a charge. Again using the superposition law, we can get the value of the electric field in its direction and this determines the net electric field at that point.

The magnitude of the electric field is given as-

$\left|\stackrel{\to }{\mathrm{E}}\right|=\frac{\left|\mathrm{q}\right|}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{R}}$ (i)

where, R = The distance of the field point from the charge

q = charge of the particle,

data-custom-editor="chemistry" ${\mathrm{\epsilon }}_{\mathrm{o}}$ is the permittivity of vacuum.

The magnitude of the resultant vector,

$\left|\mathrm{V}\right|=\sqrt{{\mathrm{V}}_{\mathrm{x}}^{2}+{\mathrm{V}}_{\mathrm{y}}^{2}}$ (ii)

where, Vx is the horizontal component of the vector and Vy is the vertical component of the vector.

The angle or the direction of the electric field made with the x-axis,

data-custom-editor="chemistry" $\mathrm{\theta }={\mathrm{tan}}^{-1}\left(\frac{{\mathrm{V}}_{\mathrm{y}}}{{\mathrm{V}}_{\mathrm{x}}}\right)$ (iii)

where, Vx is the horizontal component of the vector and Vy is the vertical component of the vector.

## Step 3: a) Calculation of the magnitude of the net electric field.

As the magnitude of the charge is the same on all the given particles and all the charges are at the same distance from the center, the magnitude of electric fields at the center will also be the same. The field of each charge can be given using equation (i) and the given values, as:

The directions of the electric fields due to the charge particles are indicated in the phasor diagram below.

Now, the components of electric field in both the x- and y-directions due to the produced fields can be given as: (for clockwise directions)

 x-component y-component ${\mathrm{E}}_{\mathrm{A}}$ $\mathrm{E}\mathrm{cos}{0}^{\circ }$ $\mathrm{E}\mathrm{sin}{0}^{\circ }$ ${\mathrm{E}}_{\mathrm{B}}$ $\mathrm{Ecos}\left(-{150}^{\mathrm{o}}\right)$ $\mathrm{Esin}\left(-{150}^{\mathrm{o}}\right)$ ${\mathrm{E}}_{\mathrm{C}}$ $\mathrm{Ecos}\left(-{100}^{\mathrm{o}}\right)$ $\mathrm{Esin}\left(-{100}^{\mathrm{o}}\right)$ ${\mathrm{E}}_{\mathrm{D}}$ $\mathrm{Ecos}\left({130}^{\mathrm{o}}\right)$ $\mathrm{Esin}\left({130}^{\mathrm{o}}\right)$ ${\mathrm{E}}_{\mathrm{E}}$ $\mathrm{Ecos}\left(-{20}^{\mathrm{o}}\right)$ $\mathrm{Esin}\left(-{20}^{\mathrm{o}}\right)$

Thus, the net electric field along x-axis can be given as the sum of all the x-directional electric fields as follows:

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{x},\mathrm{net}}& =& \left\{\mathrm{Ecos}{0}^{\mathrm{o}}+\mathrm{Ecos}\left(-{150}^{\mathrm{o}}\right)+\mathrm{Ecos}\left(-{100}^{\mathrm{o}}\right)+\mathrm{Ecos}\left({130}^{\mathrm{o}}\right)+\mathrm{Ecos}\left(-{20}^{\mathrm{o}}\right)\right\}\\ & =& \left(3.6×{10}^{-9}\mathrm{N}/\mathrm{C}\right)\left\{\mathrm{cos}{0}^{\mathrm{o}}+\mathrm{cos}\left(-{150}^{\mathrm{o}}\right)+\mathrm{cos}\left(-{100}^{\mathrm{o}}\right)+\mathrm{cos}\left({130}^{\mathrm{o}}\right)+\mathrm{cos}\left(-{20}^{\mathrm{o}}\right)\right\}\\ & =& 9.26×{10}^{-7}\mathrm{N}/\mathrm{C}\end{array}$

Now, the net electric field along y--axis can be given as the sum of all the x-directional electric fields as follows:

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{y},\mathrm{net}}& =& \left\{\mathrm{Esin}{0}^{\mathrm{o}}+\mathrm{Esin}\left(-{150}^{\mathrm{o}}\right)+\mathrm{Esin}\left(-{100}^{\mathrm{o}}\right)+\mathrm{Esin}\left({130}^{\mathrm{o}}\right)+\mathrm{Esin}\left(-{20}^{\mathrm{o}}\right)\right\}\\ & =& \left(3.6×{10}^{-9}\mathrm{N}/\mathrm{C}\right)\left\{\mathrm{sin}{0}^{\mathrm{o}}+\mathrm{sin}\left(-{150}^{\mathrm{o}}\right)+\mathrm{sin}\left(-{100}^{\mathrm{o}}\right)+\mathrm{sin}\left({130}^{\mathrm{o}}\right)+\mathrm{sin}\left(-{20}^{\mathrm{o}}\right)\right\}\\ & =& -3.81×{10}^{-6}\mathrm{N}/\mathrm{C}\end{array}$

Thus, the magnitude of the resultant electric field is given using equation as follows:

localid="1662519095157" $\begin{array}{rcl}\left|{\mathrm{E}}_{\mathrm{net}}\right|& =& \sqrt{{\left({\mathrm{E}}_{\mathrm{x},\mathrm{net}}\right)}^{2}+{\left({\mathrm{E}}_{\mathrm{y},\mathrm{net}}\right)}^{2}}\\ & =& \sqrt{{\left(9.26×{10}^{-7}\mathrm{N}/\mathrm{C}\right)}^{2}+{\left(-3.81×{10}^{-6}\mathrm{N}/\mathrm{C}\right)}^{2}}\\ & =& 3.92×{10}^{-6}\mathrm{N}/\mathrm{C}\end{array}$

Hence, the magnitude of the net electric field is $\begin{array}{rcl}& & 3.92×{10}^{-6}\mathrm{N}/\mathrm{C}\end{array}$

## Step 4: b) Calculation of the direction of the net electric field

Now, the direction of the net electric field using the data obtained in part (a) calculations can be given using equation (iii) as:

$\begin{array}{rcl}\mathrm{\theta }& =& {\mathrm{tan}}^{-1}\left(\frac{-3.81×{10}^{-6}\mathrm{N}/\mathrm{C}}{9.26×{10}^{-7}\mathrm{N}/\mathrm{C}}\right)\\ & =& -76.{4}^{\mathrm{o}}\end{array}$

Hence, the direction of the field is $-76.{4}^{\mathrm{o}}$ from the positive of x-axis.