question_answer

A uniformly charged and infinitely long line having a liner charge density \[\lambda \] is placed at a normal distance \[y\] from a point O. Consider a sphere of radius \[R\] with O as the center and \[R\] > \[y\]. Electric flux through the surface of the sphere is

A) Zero                 

B) \[\frac{2\lambda R}{{{\varepsilon }_{0}}}\]

C) \[\frac{2\lambda \sqrt{{{R}^{2}}-{{y}^{2}}}}{{{\varepsilon }_{0}}}\]

D) \[\frac{\lambda \sqrt{{{R}^{2}}+{{y}^{2}}}}{{{\varepsilon }_{0}}}\]

Correct Answer: C

Solution :

[c] Electric flux \[\oint\limits_{s}{{\vec{E}}}.\vec{d}s=\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}\] \[{{q}_{in}}\] is the charge enclosed by the Gaussian Surface, which, in the present case, is the surface of the given sphere, As shown, length AB of line lies inside the sphere, in \[\Delta OO'A\], \[{{R}^{2}}={{Y}^{2}}+{{(O'A)}^{2}}\] \[\therefore O'A=\sqrt{{{R}^{2}}-{{y}^{2}}}\] and \[AB=2\sqrt{{{R}^{2}}-{{y}^{2}}}\] Charge on length AB is \[2\sqrt{{{R}^{2}}-{{y}^{2}}}\times \lambda \] Therefore, electric flux is \[\oint\limits_{S}{{\vec{E}}}.\vec{d}s=\frac{2\lambda \sqrt{{{R}^{2}}-{{y}^{2}}}}{{{\varepsilon }_{0}}}\]