Q25P - Fundamentals Of Physics

Question

Figure 22-49 shows three circular arcs centered on the origin of a coordinate system. On each arc, the uniformly distributed charge is given in terms of$\mathit{Q}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mu \mathit{C}$. The radii are given in terms of$\mathit{R}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$. What are the (a) magnitude and (b) direction (relative to the positive x direction) of the net electric field at the origin due to the arcs?

Step-by-Step Solution

Answer

1. The magnitude of the net electric field at the origin due to the arcs is$1.62\times {10}^8 \mathrm{N}/\mathrm{C}$.
2. The direction (relative to positive x-direction) of the net electric field at the origin due to the arcs is– 45º, measured counter-clockwise from the +x-axis.

Step 1: The given data

Three arcs of each having uniformly distributed charge $Q=2.00\mu C$ and separation of radii, $R=10.0 \mathrm{cm}$.

Step 2: Understanding the concept of electric field

Using the concept of the electric field, we can get the net electric field by the three circular arcs using linear charge density and the lengths of the circular arcs.

 

Formulae:

The magnitude of the electric field, $E=\frac{\lambda \mathit{sin}\theta }{4{\mathrm{\pi \epsilon }}_{0}r}$                                                     (i)  

Where $R$  is the distance of field point from the charge, $q$ is the charge of the particle.

The length of an arc of a circle, $L=r\theta (\theta \mathrm{in} \mathrm{radians})$                                       (ii)

The linear density of a distribution, $\lambda =q/L$                                                 (iii) 

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

Using equation (ii), the length of the smallest arc is given as:

$L=\mathrm{\pi }R/2$

Similarly, the length of the middle-sized arc is given as:

$$\begin{gathered} L_2=\mathrm{\pi }\left(2R\right)/2 \\ =\mathrm{\pi }R \end{gathered}$$

 Similarly, the length of the largest arc is given as:

$L_3=\mathrm{\pi }\left(3R\right)/2$

Substituting the above values and equation (ii) in equation (i), we can get the net electric field as follows:

$$\begin{gathered} E_{net}=\frac{{\lambda }_1(2sin45°)}{4{\mathrm{\pi \epsilon }}_0{r}_1}+\frac{{\lambda }_2(2\sin 45°)}{4{\mathrm{\pi \epsilon }}_0{r}_2}+\frac{{\lambda }_3\left(2sin45°\right)}{4{\mathrm{\pi \epsilon }}_0{r}_{\mathit{3}}} \\ =\frac{Q}{\sqrt{2}{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0{R}^2} \\ =\frac{2\times {10}^{-6 }C}{\sqrt{2}\times {\mathrm{\pi }}^2\times {\mathrm{\epsilon }}_0{(0.1 \mathrm{m})}^2} \\ =1.62\times {10}^{6 }\mathrm{N}/\mathrm{C} \end{gathered}$$ 

Hence, the value of the net electric field is $1.62\times {10}^6 \mathrm{N}/\mathrm{C}$.

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

From the calculations of part (a), the direction of the electric field is found to be – 45º, measured counter-clockwise from the +x-axis.