Question

The charge per unit length of the four quadrant of the ring is 2$\lambda$, -2$\lambda$, $\lambda$ and -$\lambda$ respectively. The electric field at the centre is

• A. $-\frac{\lambda }{2\pi {\epsilon }_{0}R}\hat{i}$
• B. $\frac{\lambda }{2\pi {\epsilon }_{0}R}\hat{j}$
• C. $\frac{\sqrt{2}\lambda }{4\pi {\epsilon }_{0}R}\hat{i}$
• D. None

Answer 1

The correct option is A $-\frac{\lambda }{2\pi {\epsilon }_{0}R}\hat{i}$

We know for a quarter ring, taking a small element dq

$dE=\frac{Kdq}{R^2}$, where $K=\frac{4}{4\pi {\epsilon }_0}$

components at centre is obtained as:

$E_x=\int dE\cos \theta =(-)\int \frac{K\lambda }{R}\cos \theta d\theta$ (as $Ex$ is along $-x$ axis)

$E_y=\int dE\sin \theta =(-)\frac{k\lambda }{R}\int \sin \theta d\theta$

integrating over are quarter to components of $\overrightarrow{E}$:

$E_x=\frac{-K}{R}\left[\right(2\lambda ){\int }_0^{\frac{\pi }{2}}\cos \theta d\theta +(-2\lambda ){\int }_{\frac{\pi }{2}}^{\pi }\cos \theta d\theta +(\lambda ){\int }_{\pi }^{3\frac{\pi }{2}}\cos \theta d\theta +(-\lambda \left){\int }_{3\frac{\pi }{2}}^{2\pi }\cos \theta d\theta \right]$

$E_x=\frac{-K}{R}\left(2\lambda \right)$

Similarly,

$E_y=\frac{-K}{R}\left[\right(2\lambda ){\int }_0^{\frac{\pi }{2}}\sin \theta d\theta +(-2\lambda ){\int }_{\frac{\pi }{2}}^{\pi }\sin \theta d\theta +(\lambda ){\int }_{\pi }^{3\frac{\pi }{2}}\sin \theta d\theta +(-\lambda \left){\int }_{3\frac{\pi }{2}}^{2\pi }\sin \theta d\theta \right]$

which give,

$E_y=0$

So net $\overrightarrow{E}$ field at centre

$\overrightarrow{E}=Ex\hat{i}+E_y\hat{i}=\frac{-k}{R}\left(2\lambda \right)\hat{i}+0\hat{j}$

$\Rightarrow \overrightarrow{E}=\frac{-\lambda }{2n{\epsilon }_{0}R}\hat{i}$