91影视

• Positive charge$\mathrm{q}=7.81\text{鈥塸颁}$ is spread uniformly along a thin non-conducting rod of length,$\mathrm{L}=14.5\text{cm}$.
• The distance of the point from the rod,$\mathrm{R}=6.00\text{cm}$ .

Using the concept of the electric field for a small charge distributed over a line, the magnitude and direction of the electric field at the given point from the rod can be calculated.

Formula:

The magnitude of the electric field at a point,

$\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{蟺蔚}}_{\mathrm{o}}{\mathrm{r}}^2}\widehat{\mathrm{r}}\text{(}鈭�\widehat{\mathrm{r}}=\mathrm{肠辞蝉胃}\widehat{\mathrm{i}}+\mathrm{蝉颈苍胃}\widehat{\mathrm{j}}\text{)}$ (i)

Where, $\mathrm{r}$= The distance of field point from the charge

$\mathrm{q}$= charge of the particle

The linear density of the distribution, $\mathrm{位}=\mathrm{q}/\mathrm{L}$ (ii)

We assume q > 0. Let, the (infinitesimal) charge on an element$\mathrm{dx}$of the rod contains charge using equation (i),$\mathrm{dq}$is$\mathrm{位诲虫}$. By symmetry, we conclude that all horizontal field components (due to the charges) cancel and we need only 鈥渟um鈥� (integrate) of the vertical components. Symmetry also allows us to integrate these contributions over only half the rod $(0鈮�\mathrm{x}鈮�\mathrm{L}/2)$.

Thus, the value of sine from the figure is given as:

$\mathrm{蝉颈苍胃}=\mathrm{R}/\mathrm{r}$

where,$\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{R}}^2}$.

Using the concept of electric field due a line charge the magnitude of the electric from equation (i) can be given as:

$\begin{array}{c}\left|\stackrel{鈫�}{\mathrm{E}}\right|=2\underset{0}{\overset{\mathrm{L}/2}{}}\left(\frac{\mathrm{dq}}{4{\mathrm{蟺蔚}}_{\mathrm{o}}{\mathrm{r}}^2}\right)\text{sin}\mathrm{胃}\\ =\frac{2}{4{\mathrm{蟺蔚}}_{\mathrm{o}}}\underset{0}{\overset{\mathrm{L}/2}{}}\left(\frac{\mathrm{位诲虫}}{{\mathrm{x}}^2+{\mathrm{R}}^2}\right)\frac{\mathrm{y}}{\sqrt{{\mathrm{x}}^2+{\mathrm{R}}^2}}\\ =\frac{(\mathrm{q}/\mathrm{L})\mathrm{R}}{2{\mathrm{蟺蔚}}_{\mathrm{o}}}{\left[\frac{\mathrm{x}}{{\mathrm{R}}^2\sqrt{{\mathrm{x}}^2+{\mathrm{R}}^2}}\right]}_0^{\mathrm{L}/2}\\ =\frac{\mathrm{q}}{2{\mathrm{蟺蔚}}_{\mathrm{o}}\mathrm{R}}\frac{1}{\sqrt{{\mathrm{L}}^2+4{\mathrm{R}}^2}}\\ =\frac{7.81\mathrm{x}{10}^{鈭�12}\text{C}}{2{\mathrm{蟺蔚}}_{\mathrm{o}}脳(0.06\text{鈥尘})脳\sqrt{{(0.145\text{m})}^2+{(0.06\text{m})}^2}}\\ =12.4\text{鈥塏/颁}\end{array}$

$\mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{is}12.4\text{鈥塏/颁}.$

As noted above in the calculations of part (a), the electric field E points in the +y-direction, or $+{90}^{\mathrm{o}}$counter-clockwise from the +x-axis.