P19R2015 – Semicircle with Non-Uniform Charge Density

We consider an infinitesimal slice of length $ds=Rd\theta $ located at an arbitrary angle $\theta $ as shown below.

The infinitesimal charge $dq$ of this slice is equal to $dq=\lambda ds={\lambda }_0R\theta d\theta $ and it creates an electric field of magnitude $dE$ at the center of semicircle given by

dE=\frac{kdq}{R^2}=\frac{{\lambda }_0k}{R}{\mathrm{cos} \left(\theta \right)\ }d\theta

The $x$-component of the electric field at the center of the semicircle is equal to

dE_x=dE{\mathrm{sin} \left(\theta \right)\ }=\frac{{\lambda }_0k}{R}{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }d\theta

and integrating over the semicircle from $\theta =-\pi $ to $\theta =+\pi $ then yield the $x$-component of the electric field $E_x$ as follows

E_x=\int^{+\pi /2}_{-\pi /2}{\frac{{\lambda }_0k}{R}{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }d\theta }=\frac{{\lambda }_0k}{R}\int^{+1}_{-1}{udu}=0

where we used the substitution $\left\{ \begin{array}{c}
u={\mathrm{sin} \left(\theta \right)\ } \\
du={\mathrm{cos} \left(\theta \right)\ }d\theta \end{array}
\right.$. Thus, we conclude that there is no $x$-component of the electric field at the center of the semicircle and we write

\boxed{E_x=0}

The $y$-component of the electric field at the center of the semicircle is equal to

dE_y=dE{\mathrm{cos} \left(\theta \right)\ }=\frac{{\lambda }_0k}{R}{{\mathrm{cos}}^{\mathrm{2}} \left(\theta \right)\ }d\theta

and integrating over the semicircle from $\theta =-\pi /2$ to $\theta =+\pi /2$ then yield the $x$-component of the electric field $E_y$ as follows

\begin{aligned}
E_y&=\int^{+\pi /2}_{-\pi /2}{\frac{{\lambda }_0k}{R}{{\mathrm{cos}}^{\mathrm{2}} \left(\theta \right)\ }d\theta } \\
\\ 
&=\frac{{\lambda }_0k}{R}\int^{+\pi /2}_{-\pi /2}{\left(\frac{1+{\mathrm{cos} \left(2\theta \right)\ }}{2}\right)}d\theta \\
\\
&=\frac{{\lambda }0k}{2R}{\left[\theta +\frac{{\mathrm{sin} \left(2\theta \right)\ }}{2}\right]}^{\pi /2}_{-\pi /2} \\
\\
&=\frac{{\lambda }_0k}{2R}\left[\frac{\pi }{2}+\frac{{\mathrm{sin} \left(\pi \right)\ }}{2}+\frac{\pi }{2}-\frac{{\mathrm{sin} \left(-\pi \right)\ }}{2}\right] \\
\\
&=\frac{{\lambda }_0k\pi }{2R}
\end{aligned}

where we used the trig identity $\displaystyle{{{\mathrm{cos}}^2 \left(\theta \right)\ }=\frac{1+{\mathrm{cos} \left(2\theta \right)\ }}{2}}$. Thus, we conclude that the $y$-component of the electric field at the center of the semicircle is equal to

\boxed{E_y=\frac{{\lambda }_0k\pi }{2R}}

Finally, the electric field at the center of the semicircle is given by

\boxed{\overrightarrow{E}=\frac{{\lambda }_0k\pi }{2R}\ \hat{y}}