91Ó°ÊÓ

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

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,$E=\frac{\mathrm{λ}\mathit{sin}\mathrm{θ}}{4{\mathrm{Ï€Î\mu }}_{0}r}$ (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, $L=r\mathrm{θ}(\mathrm{θ}\mathrm{in}\mathrm{radians})$ (ii)

The linear density of a distribution, $\mathrm{λ}=q/L$ (iii)

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

$L=\mathrm{Ï€}R/2$

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

role="math" localid="1657283259438" $$\begin{gathered} L_2=\mathrm{Ï€}\left(2R\right)/2 \\ =\mathrm{Ï€}R \end{gathered}$$

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

$L_3=\mathrm{Ï€}\left(3R\right)/2$

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

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

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

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