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 C $\frac{\sqrt{2}\lambda }{4\pi {\epsilon }_{0}R}\hat{i}$

${dE}_x=\frac{K\times 2\lambda rd\theta }{r^2}\cos \theta (-\hat{i})$

${dE}_y=\frac{K\times 2\lambda .rd\theta }{r^2}\sin \theta (-\hat{j})$

$\to E_x={\int }_0^{K/2}\frac{K\times 2\lambda rd\theta \cos \theta }{r^2}(-\hat{i})$

$=\frac{2\times \lambda }{r}(-\hat{j})$

$E_y={\int }_0^{K/2}\frac{K.2\lambda .rd\theta }{r^2}\sin \theta (-\hat{j})$

$=\frac{2K\lambda }{r}(-j)$

Similarly, for $-2\lambda$, $E_x=\frac{-2K\lambda (+\hat{i})}{r}$

$E_y=\frac{-2K\lambda }{r}(-\hat{j})$

For $\lambda :E_x=\frac{K\lambda \left(\hat{i}\right)}{r}$

$E_y=\frac{K\lambda \hat{j}}{r}$

For $-\lambda$, $E_x=\frac{-K\lambda (-\hat{i})}{r}$

$E_y=-\left(\frac{K\lambda }{r}\right)\left(\hat{j}\right)$

$\Rightarrow {\left(E_x\right)}_{net}=-\frac{2K\lambda }{r}\hat{i}{\left(E_x\right)}_{net}=0$

where $K=\frac{1}{4\pi {ϵ}_0}$

$r=R$

then ${\left(E_x\right)}_{net}=\frac{1}{4\pi {ϵ}_{0}R}\sqrt{2}\lambda \hat{i}$