Physics - Electrostatics Question with Solution | TestHub

Ans (A)

$\vec{E}=-E_x\hat{x}-E_y\hat{y}$

where $E_x={\int }_{line}dE\cos \theta$, $E_y={\int }_{line}dE\sin \theta$.

$dE=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dl}{R^2}$

$E_x={\int }_{line}\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dl}{R^2}\cos \theta =\frac{\lambda }{4\pi {\epsilon }_0}{\int }_0^{\pi /2}\cos \theta \frac{Rd\theta }{R^2}$

$E_x=\frac{\lambda }{4\pi {\epsilon }_{0}R}[\sin \theta {]}_0^{\pi /2}=\frac{\lambda }{4\pi {\epsilon }_{0}R}$

similarly $E_y={\int }_{line}\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dl}{R^2}\sin \theta ={\int }_0^{\pi /2}\sin \theta \frac{Rd\theta }{R^2}$

$\Rightarrow E_y=\frac{\lambda }{4\pi {\epsilon }_{0}R}[-\cos \theta {]}_0^{\pi /2}=\frac{\lambda }{4\pi {\epsilon }_{0}R}$

Thus $\vec{E}=-E_x\hat{x}-E_y\hat{y}=\frac{\lambda }{4\pi {\epsilon }_{0}R}(-\hat{x}-\hat{y})$