In Fig. 22-55, positive charge$\mathbf{q}\mathbf{}\mathbf{=}\mathbf{7}\mathbf{.81}\mathbf{}\mathbf{pC}$is spread uniformly along a thin non-conducting rod of length$\mathbf{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{14.5}\mathbf{}\mathbf{cm}$. What are the (a) magnitude and (b) direction (relative to the positive direction of the xaxis) of the electric field produced at point P, at distance$\mathbf{R}\mathbf{}\mathbf{=}\mathbf{6.00}\mathbf{}\mathbf{cm}$from the rod along its perpendicular bisector?

• Positive charge$\mathrm{q}=7.81\text{\hspace{0.17em}pC}$ 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,

$\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}^2}\widehat{\mathrm{r}}\text{(}\because \widehat{\mathrm{r}}=\mathrm{cos\theta }\widehat{\mathrm{i}}+\mathrm{sin\theta }\widehat{\mathrm{j}}\text{)}$ (i)

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

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

The linear density of the distribution, $\mathrm{\lambda }=\mathrm{q}/\mathrm{L}$ (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$\mathrm{\lambda dx}$. By symmetry, we conclude that all horizontal field components (due to the charges) cancel and we need only “sum” (integrate) of the vertical components. Symmetry also allows us to integrate these contributions over only half the rod $(0\le \mathrm{x}\le \mathrm{L}/2)$.

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

$\mathrm{sin\theta }=\mathrm{R}/\mathrm{r}$

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:

$\begin{array}{c}\left|\overrightarrow{\mathrm{E}}\right|=2\underset{0}{\overset{\mathrm{L}/2}{}}\left(\frac{\mathrm{dq}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}^2}\right)\text{sin}\mathrm{\theta }\\ =\frac{2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\underset{0}{\overset{\mathrm{L}/2}{}}\left(\frac{\mathrm{\lambda dx}}{{\mathrm{x}}^2+{\mathrm{R}}^2}\right)\frac{\mathrm{y}}{\sqrt{{\mathrm{x}}^2+{\mathrm{R}}^2}}\\ =\frac{(\mathrm{q}/\mathrm{L})\mathrm{R}}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}}{\left[\frac{\mathrm{x}}{{\mathrm{R}}^2\sqrt{{\mathrm{x}}^2+{\mathrm{R}}^2}}\right]}_0^{\mathrm{L}/2}\\ =\frac{\mathrm{q}}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{R}}\frac{1}{\sqrt{{\mathrm{L}}^2+4{\mathrm{R}}^2}}\\ =\frac{7.81\mathrm{x}{10}^{-12}\text{C}}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}\times (0.06\text{\hspace{0.17em}m})\times \sqrt{{(0.145\text{m})}^2+{(0.06\text{m})}^2}}\\ =12.4\text{\hspace{0.17em}N/C}\end{array}$

$\mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{is}12.4\text{\hspace{0.17em}N/C}.$

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.