Q65P - Fundamentals Of Physics

Question

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 R from the particle. In Fig. 22-64b, that same amount of charge is spread uniformly along a circular arc that has radius R and subtends an angle$\theta$. The charge on the arc produces an electric field of magnitude  at its center of curvatureP. For what value of does 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.)

Step-by-Step Solution

Answer

The value of for which, $E_{arc}=0.5E$parts is.$3.791\text{ rad or }{217}^o$

Step 1: The given data

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,$E_{arc}=0.5E_{part}$

 

Step 2: Understanding the concept of the electric field

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)

Step 3: Calculation of the angle value

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.$3.791\text{ rad or }{217}^o$