Question

A rod of length $L$ as a total charge $Q$ distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.

Answer 1

$\lambda =\frac{Q}{L}$
$(L=\pi \times r)$
Now,
$dq=\lambda dx$
$dx=rdQ$
$dq=\lambda rQ$
$dE=\int \frac{kdq}{r^2}$
We need
$2dE\sin Q={\int }_0^{n/2}\frac{2kdq\sin Q}{r^2}$
$={\int }_0^{n/2}\frac{2k}{n^2}rdQ.\sin Q$
$=\frac{2k}{r}\cdot d\times [-\cos Q{]}_0^2$
$=\frac{2k\times Q}{r\times L}$ (as $L=nr)$
$r=\frac{L}{n}$
$=\frac{2kQ}{L^2}n$
$=\frac{2}{4n{ϵ}_0}\frac{Q}{L^2}\times n$
$=\frac{2Q}{4{ϵ}_0{L}^2}=\frac{Q}{2{ϵ}_0{L}^2}$.