An infinitely long thin non-conducting wire i | vedclass

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Question

(A) According to Gauss's law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{	ext{enclosed}}}{\epsilon_0}$.The length

Vedclass · Accepted Answer

$A, B$

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(A) According to Gauss's law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{	ext{enclosed}}}{\epsilon_0}$.The length of the wire segment $PQ$ inside the spherical shell is $L = 2R \sin(120^{\circ}/2) = 2R \sin(60^{\circ}) = 2R 	imes \frac{\sqrt{3}}{2} = R\sqrt{3}$.The charge enclosed by the shell is $Q_{	ext{enclosed}} = \lambda L = \lambda R \sqrt{3}$.Therefore,the electric flux through the shell is $\phi = \frac{\lambda R \sqrt{3}}{\epsilon_0}$. Thus,statement $(A)$ is true and $(C)$ is false.Since the wire is parallel to the $z$-axis,the electric field lines are directed radially outward from the wire in the $xy$-plane. Therefore,the electric field has no component along the $z$-axis. Thus,statement $(B)$ is true.The electric field is only normal to the surface of the shell if the charge distribution is spherically symmetric,which is not the case here. Thus,statement $(D)$ is false.Therefore,the correct statements are $(A)$ and $(B)$.

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