In Fig. 22-64a, a particle of charge$\mathbf{+}\mathit{Q}$produces an electric field of magnitude ${\mathit{E}}_{\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{t}}$at point P, at distance Rfrom the particle. In Fig. 22-64b, that same amount of charge is spread uniformly along a circular arc that has radius Rand subtends an angle$\theta$. The charge on the arc produces an electric field of magnitudeat its center of curvatureP.For what value ofdoes the electric field${\mathit{E}}_{\mathbf{a}\mathbf{r}\mathbf{c}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.500}{\mathit{E}}_{\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{t}}$? (Hint:You will probably resort to a graphical solution.)

Charge of the particle is$+Q$, produces an electric field of magnitude$E_{part}$, at distance, R.

The same amount of charge subtends an angle along a circular arc.

Charge on arc produces electric field,$E_{arc}$

The given data,$E_{arc}=0.5E_{part}$

Using the concept of the electric field of a charged rod, we can get the desired angle y comparing the electric field relation given.

Formulae:

Electric field due to charged circular rod, $E=\frac{\lambda \sin \theta }{4\pi {\epsilon }_{o}r}$ (i)

The electric field of a particle, $E=\frac{Q}{4\pi {\epsilon }_o{r}^2}$ (ii)

Electric field due to a charged circular rod which subtend and angle between – θ/2 to + θ/2 is given by using equation (i) as follows:

$\begin{array}{c}{E}_{arc}=\frac{\lambda }{4\pi {\epsilon }_{o}r}[sin(\theta /2)-sin(-\theta /2)]\\ =\frac{2\lambda sin(\theta /2)}{4\pi {\epsilon }_{o}r}\end{array}$

Now, the arc length is$L=r\theta$with$\theta$expressed in radians. Thus, using R instead of, we obtain the above equation as follows:

$\begin{array}{c}{E}_{arc}=\frac{2(Q/L)sin(\theta /2)}{4\pi {\epsilon }_{o}R}\\ =\frac{2(Q/R\theta )sin(\theta /2)}{4\pi {\epsilon }_{o}R}\\ =\frac{2Qsin(\theta /2)}{4\pi {\epsilon }_o{R}^2\theta }\mathrm{..............................}\left(a\right)\end{array}$

Thus, the problem requires, $E_{arc}=\left(\mathrm{½}\right)E_{particle}$where$E_{particle}$is given by Eq. of the electric field. Thus, it can be given by using equations (ii) and (a) as:

$\begin{array}{c}\frac{2Qsin(\theta /2)}{4\pi {\epsilon }_o{R}^2\theta }=\frac{1}{2}\frac{Q}{4\pi {\epsilon }_o{R}^2}\\ sin(\theta /2)=\theta /4\end{array}$

where we note, again, that the angle is in radians. The approximate solution to this equation is.$3.791\text{rador}{217}^o$