Question

A thread carrying a uniform charge $\lambda$ per unit length has the configurations shown in Fig. $a$ and $b$. Assuming a curvature radius $R$ to be considerably less than the length of the thread, find the magnitude of the electric field strength at the point $O$.

Answer 1

The problem is reduced to finding $E_x$ and $E_y$ viz. the projections of $\overrightarrow{E}$ in Fig, where it is assumed that $\lambda >0$.
Let us start with $E_x$. The contribution to $E_x$ from the charge element of the segment $dx$ is
$d{E}_x=\frac{1}{4\pi {ϵ}_0}\frac{\lambda dx}{r^2}\sin \alpha ....\left(1\right)$
Let us reduce this expression to the form convenient for integration. In our case, $dx=rd\alpha /\cos \alpha ,r=y/\cos \alpha$. Then
$d{E}_x=\frac{\lambda }{4\pi {ϵ}_{0}y}\sin \alpha d\alpha$.
Integrating this expression over $\alpha$ between $0$ and $\pi /2$, we find
$E_x=\lambda /4\pi {ϵ}_{0}y$.
In order to find the projection $E_y$ it is sufficient to recall that $d{E}_y$ differs from $d{E}_x$ in that $\sin \alpha$ in (1) is simply replaced by $\cos \alpha$.
This gives
$d{E}_y=\left(\lambda \cos \alpha d\alpha \right)/4\pi {ϵ}_{0}y$ and $E_y=\lambda /4\pi {ϵ}_{0}y$.
We have obtained an interesting result:
$E_x=E_y$ independently of $y$,
i.e. $\overrightarrow{E}$ is oriented at the angle of ${45}^{\circ }$ to the rod. The modulus of $\overrightarrow{E}$ is
$E=\sqrt{E_x^2+E_y^2}=\lambda \sqrt{2}/4\pi {ϵ}_{0}y$
the net electric field strength at the point $O$ due to straight parts of the thread equals zero. For the curved part (arc) let us derive a general expression i.e. let us calculate the field strength at the centre of arc of radius $R$ and linear charge density $\lambda$ and which subtends angle ${\theta }_0$ at the centre.
From the symmetry the sought field strength will be directed along the bisector of the angle ${\theta }_0$ and is given by
$E={\int }_{-{\theta }_0/2}^{+{\theta }_0/2}\frac{\lambda R\left(d\theta \right)}{4\pi {ϵ}_0{R}^2}\cos \theta =\frac{\lambda }{2\pi {ϵ}_{0}R}\sin \frac{{\theta }_0}{2}$
In our problem ${\theta }_0=\pi /2$, thus the field strength due to the turned part at the point $E_0=\frac{\sqrt{2}\lambda }{4\pi {ϵ}_{0}R}$ which is also the sought result.
(b) Net field strength at $O$ due to straight parts equal $\sqrt{2}\left(\frac{\sqrt{2}\lambda }{4\pi {ϵ}_{0}R}\right)=\frac{\lambda }{2\pi {ϵ}_{0}R}$ and is directed vertically down, field strength due to the given curved part (semi-circle) at the point $O$ becomes $\frac{\lambda }{2\pi {ϵ}_{0}R}$ and is directed vertically upward. Hence the sought net field strength becomes zero.