Question

A uniformly charged and infinitely long line having a linear charge density ' $\lambda$ ' 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)

Electric flux $\oint_{S} \vec{E} \cdot \vec{dS}=\frac{q_{\text {in }}}{\varepsilon_{0}}, q_{\text {in }}$ 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 $\Delta OO'\,\, A\,\,\, R ^{2}= y ^{2}+\left( O ' A \right)^{2}$
$\therefore O'\,\, A=\sqrt{R^{2}-y^{2}}$
and and $A B=2 \sqrt{R^{2}-y^{2}}$
Charge on length $AB =2 \sqrt{ R ^{2}- y ^{2}} \times \lambda$
$\therefore $ electric flux $=\oint_{S} \vec{E} \cdot \vec{d S}=\frac{2 \lambda \sqrt{R^{2}-y^{2}}}{\varepsilon_{0}}$