Question

Negative charge -Q is distributed uniformly around a quarter-circle of radius a that lies in the first quadrant, with the center of curvature at the origin. Find the x- and y-components of the net electric field at the origin.

Short answer

The x- and y- component of the net electric field at the origin is${\in }_{\mathrm{x}}={\in }_{\mathrm{y}}=\frac{\mathrm{Q}}{2{\mathrm{\pi }}^2{\in }_0{\mathrm{a}}^2}$

Step-by-step solution

Formula

The formula for the quadrant, with the center of curvature at the origin

Therefore, for x- and y- components;

$\overrightarrow{\mathrm{dE}}=\frac{\mathrm{K\lambda }\left(\mathrm{d\theta }\right)}{\mathrm{a}}\cos \theta \widehat{\left(\mathrm{i}\right)}+\frac{\mathrm{K\lambda }\left(\mathrm{d\theta }\right)}{\mathrm{a}}\sin \theta \widehat{\left(j\right)}$

Solution

$$\begin{gathered} \Rightarrow \mathrm{d}{\overrightarrow{\in }}_{\mathrm{x}}=\frac{\mathrm{kkd\theta }}{0}\mathrm{cos\theta }\left(+\hat{\mathrm{x}}\right)\left\{\begin{array}{l}\mathrm{d}{\overrightarrow{\in }}_{\mathrm{y}}=\frac{\mathrm{khd\theta }}{0}\mathrm{sen\theta }\left(+\overrightarrow{\mathrm{J}}\right)\\ {\mathrm{p}}^{\mathrm{\pi }/2}\mathrm{kd\theta }\end{array}\right. \\ ={\int }_0^{\mathrm{\pi }/2}\frac{\mathrm{kd\theta }}{\mathrm{a}}\mathrm{cos\theta }\left(+\hat{\mathrm{x}}\right)+\frac{\mathrm{kh}}{\mathrm{a}}{\int }_0^{\mathrm{\pi }/2}\mathrm{cos\theta d\theta }\left(\hat{\mathrm{x}}\right){\int }_0^{\mathrm{\pi }/2}\frac{\mathrm{kd\theta }}{\mathrm{a}}\mathrm{sen\theta }\left(+\mathrm{j}\right)=\frac{\mathrm{kh}}{\mathrm{a}}{\int }_0^{\mathrm{\pi }/2}\mathrm{sen\theta t\theta }\left(\mathrm{j}\right) \\ =\frac{\mathrm{kh}}{\mathrm{a}}\left({\mathrm{sen\theta I}}_0^{\mathrm{\pi }/2}\right)\left(+\mathrm{\lambda }\right)\Rightarrow \frac{\mathrm{kh}}{\mathrm{a}}\left(+\mathrm{\lambda }\right)\frac{\mathrm{kh}}{\mathrm{a}}\left(-{\mathrm{cos\theta I}}_0^{\mathrm{\pi }/2}\right)\left(-\mathrm{j}\right)\Rightarrow \frac{\mathrm{kh}}{\mathrm{a}}\left(+\mathrm{j}\right) \\ \mathrm{k}=\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}{\in }_{\mathrm{x}}={\in }_{\mathrm{y}}=\frac{2\mathrm{Q}}{4{\mathrm{\pi }}^2{\in }_{\mathrm{o}}{\mathrm{a}}^2}=\frac{2\mathrm{Q}}{2{\mathrm{\pi }}^2{\in }_{\mathrm{o}}{\mathrm{a}}^2} \end{gathered}$$

Hence, the x- and y- component of the net electric field at the origin is

${\in }_x={\in }_{y=}\frac{2\mathrm{Q}}{2{\mathrm{\pi }}^2{\in }_{\mathrm{o}}{\mathrm{a}}^2}$