Electrostatics Question 662
Question: 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
Options:
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}}} $
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Answer:
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}}} $
BETA
