Physics - Electrostatics Question with Solution | TestHub

The rod is bent into a semicircle of radius $R=L/\pi$. Let the center of the semicircle be the origin. An element of arc length $dx$ at an angle $\varphi$ from one end has charge $dq=\lambda (x)dx=(kx)dx$. Since $x=R\varphi$, $dx=Rd\varphi$, so $dq=k(R\varphi )Rd\varphi =k{R}^2\varphi d\varphi$. The electric field $dE$ due to $dq$ at the center is $dE=\frac{1}{4\pi {ϵ}_0}\frac{dq}{R^2}=\frac{k\varphi d\varphi }{4\pi {ϵ}_0}$. The components are $d{E}_x=dE\cos \varphi$ and $d{E}_y=dE\sin \varphi$. Integrating from $\varphi =0$ to $\varphi =\pi$: $E_x={\int }_0^{\pi }\frac{k\varphi \cos \varphi d\varphi }{4\pi {ϵ}_0}=\frac{k}{4\pi {ϵ}_0}[\varphi \sin \varphi +\cos \varphi {]}_0^{\pi }=\frac{k}{4\pi {ϵ}_0}[(\pi \sin \pi +\cos \pi )-(0\sin 0+\cos 0)]=\frac{k}{4\pi {ϵ}_0}[-1-1]=-\frac{k}{2\pi {ϵ}_0}$. $E_y={\int }_0^{\pi }\frac{k\varphi \sin \varphi d\varphi }{4\pi {ϵ}_0}=\frac{k}{4\pi {ϵ}_0}[-\varphi \cos \varphi +\sin \varphi {]}_0^{\pi }=\frac{k}{4\pi {ϵ}_0}[(-\pi \cos \pi +\sin \pi )-(0\cos 0+\sin 0)]=\frac{k}{4\pi {ϵ}_0}[\pi -0]=\frac{k}{4{ϵ}_0}$. The magnitude is $E=\sqrt{E_x^2+E_y^2}=\sqrt{(-\frac{k}{2\pi {ϵ}_0}{)}^2+(\frac{k}{4{ϵ}_0}{)}^2}=\frac{k}{4\pi {ϵ}_0}\sqrt{\frac{4}{{\pi }^2}+1}=\frac{k}{4\pi {ϵ}_0}\sqrt{\frac{4+{\pi }^2}{{\pi }^2}}=\frac{k}{4\pi {ϵ}_0}\sqrt{4+{\pi }^2}$.