Question

Half ring of radius $a$ is uniformly charged with linear charge density $\lambda$. Find out electric field intensity at point $O$.

• A. $\frac{\lambda }{2\pi {ϵ}_{0}a}(-\hat{i})$
• B. $\frac{\lambda }{4\pi {ϵ}_{0}a}(+\hat{i})$
• C. $\frac{\lambda }{2\pi {ϵ}_{0}a}(+\hat{i})$
• D. $\frac{\lambda }{4\pi {\epsilon }_{0}a}(-\hat{i})$

Answer 1

The correct option is A $\frac{\lambda }{2\pi {ϵ}_{0}a}(-\hat{i})$


Taking an infinitely small element $dl$ on the upper side of ring which have charge $dq$ as shown in figure. Since charge is uniformly distributed over the ring.

$\therefore dq=\lambda dl=\lambda \left(ad\theta \right)$

Now electric field at centre of ring due to $dq$ will be

$dE=\frac{kdq}{a^2}=\frac{k\lambda ad\theta }{a^2}=\frac{k\lambda }{a}d\theta$

From symmetry, $d{E}_y$ will get cancelled on integration.

$\therefore E_y=\int d{E}_y=0$

So, net electric field at the centre will be

$E_{net}=\int d{E}_x=\int dE\cos \theta (-\hat{i})$

$\Rightarrow E_{net}={\int }_{-\pi /2}^{\pi /2}\frac{k\lambda }{a}\cos \theta d\theta (-\hat{i})=\frac{k\lambda }{a}{\left(\sin \theta \right)}_{-\pi /2}^{+\pi /2}(-\hat{i})$

$\Rightarrow E_{net}=\frac{\lambda }{4\pi {ϵ}_{0}a}.\left(2\right)(-\hat{i})=\frac{\lambda }{2\pi {ϵ}_{0}a}(-\hat{i})$

Alternative:

Net electric firld due to a arc subtending angle $\alpha$ at the centre, $E_{net}=\frac{2k\lambda }{R}\sin \frac{\alpha }{2}$

Here $\alpha ={180}^{\circ }$

$E_{net}=\frac{2k\lambda }{a}\sin {90}^{\circ }$

$E_{net}=\frac{2k\lambda }{a}=\frac{\lambda }{4\pi {ϵ}_{0}a}.\left(2\right)=\frac{\lambda }{2\pi {ϵ}_{0}a}$

Since ring is symmetric to $x-$ axis, $E_y=0$ and $E_x$ will be along negative $x-$ axis

Hence, option $\left(a\right)$ is correct.