Consider a thin plastic rod bent into a semicircular arc of radius $\mathbf{R}$with center at the origin (Figure 15.57). The rod carries a uniformly distributed negative charge $\mathbf{-}\mathbf{Q}$.

(a) Determine the electric field $\overrightarrow{\mathbf{E}}$at the origin contributed by the rod. Include carefully labeled diagrams, and be sure to check your result. (b) An ion with charge $\mathbf{-}\mathbf{2}\mathbf{e}$and mass is placed at rest at the origin. After a very short time $\mathbf{∆}\mathbf{t}$the ion has moved only a very short distance but has acquired some momentum .$\overrightarrow{\mathbf{P}}$Calculate $\overrightarrow{\mathbf{P}}$.

The given data can be listed below,

• The charge on the rod is,$-Q$.
• The charge on an ion is,$-2e=3.2\times {10}^{-19}C$
• The mass of an ion is,$M$

One of physics' two most basic ideas is momentum. Unless acted on by force, momentum is a characteristic that remains constant. A force is created by a change in motion. Thus, motion and force are inextricably linked.

The arc is divided into infinitesimal arc with angle θ with y axis.

$dQ=\frac{Q}{\pi }d\theta$

Here,Q is the total charge and π is the angle in radian.

Taking all infinitesimal arcs as point charges electric field is given by,

$dE=\frac{KdQ}{r^2}\hat{r}$

Here, K is the coulomb constant,$\hat{r}$is the unit vector and r is the position of point charge.

The position vector is given by,

The magnitude of the position vector is given by,

$$\begin{gathered} \left|\overrightarrow{r}\right|=\sqrt{{\left(-R\sin \theta \right)}^2+{\left(-R\cos \theta \right)}^2} \\ =R \end{gathered}$$

The unit vector is given by,

Substitute all the values in the electric field equation.

The components of the electric field in x and y-direction is given by,

$$\begin{gathered} d{E}_x=\frac{KQ}{\pi R^2}\left(-\sin \theta \right)d\theta \\ d{E}_y=\frac{KQ}{\pi R^2}\left(-\cos \theta \right)d\theta \end{gathered}$$

The values of the above-mentioned components are calculated by integrating both sides of the equation so, for the x component of electric field is given by,

$$\begin{gathered} E_x=\frac{KQ}{\pi R^2}{\int }_0^{\pi }\left(-\sin \theta \right)d\theta \\ =\frac{KQ}{\pi R^2}({\overline{)\cos }}_0^{\pi } \\ =\frac{KQ}{\pi R^2}\left(-2\right) \\ =-\frac{KQ}{\pi R^2} \end{gathered}$$

Similarly,

role="math" localid="1656930167161" $$\begin{gathered} E_y=\frac{KQ}{\pi R^2}{\int }_0^{\pi }\left(-\cos \theta \right)d\theta \\ =\frac{KQ}{\pi R^2}({\sin }_0^{\pi } \\ =0 \end{gathered}$$

The left upper part of the arc has a component of electric field in y-direction which have same magnitude as the lower part of the arc but opposite in direction, they will cancel each other thus, the net electric field at origin with only x-direction.

Thus, the electric field is $\frac{2KQ}{\pi R^2}$pointing in left direction.

The force exerted on the charged ion due to the electric field of the charged semicircle is given by,

$F=qE$ …(i)

The force is mass times acceleration; thus equation (i) can be written as,

$Ma=qE$ …(ii)

Here,ais the acceleration, m is the mass of ion, q is the charge on ion and E is the magnitude of electric field

The acceleration is the rate of change of the velocity which can be given by,

$a=\frac{dv}{dt}$.

Substitute all the values in equation (ii).

$$\begin{gathered} M\frac{dv}{dt}=q\frac{2KQ}{\pi R^2} \\ Mdv=\frac{2KQ}{\pi R^2}dt \\ p=\frac{2KQ}{\pi R^2}dt \end{gathered}$$

Thus, the momentum of the ion is$\frac{2KQ}{\pi R^2}dt$ .