Question

A uniformly charged and infinitely long line having a linear charge density 'l' is placed at a normal distance y from a point O. Consider a sphere of radius R with O as centre 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. $\frac{\lambda\sqrt{R^{2} + y^{2}}}{\varepsilon_{0}}$

Answer 1

Answer: (C)

$\frac{2\lambda\sqrt{R^{2} - y^{2}}}{\varepsilon_{0}}$

Answer 2

Electric flux $\oint_{S}^{}{\overset{\rightarrow}{E}.\overset{\rightarrow}{dS} = \frac{q_{in}}{\varepsilon_{0}}}$ q_in is the charge enclosed by the Gaussian-surface which, in the present case, is the surface of given sphere. As shown, length AB of the line lies inside the sphere.

In DOO'A R² = y² + (O'A)²

\ O'A = $\sqrt{R^{2} - y^{2}}$

and AB = 2$\sqrt{R^{2} - y^{2}}$

Charge on length AB = 2$\sqrt{R^{2} - y^{2}}$× l

\ electric flux = $\oint_{S}^{}{\overset{\rightarrow}{E}.\overset{\rightarrow}{dS} = \frac{2\lambda\sqrt{R^{2} - y^{2}}}{\varepsilon_{0}}}$