Question:
An infinitely long thin non-conducting wire is parallel to the $z$-axis and carries a uniform line charge density $\lambda$. It pierces a thin non-conducting spherical shell of radius $R$ in such a way that the arc $PQ$ subtends an angle $120^{\circ}$ at the centre $O$ of the spherical shell, as shown in the figure. The permittivity of free space is $\in_0$. Which of the following statements is (are) true?
An infinitely long thin non-conducting wire is parallel to the $z$-axis and carries a uniform line charge density $\lambda$. It pierces a thin non-conducting spherical shell of radius $R$ in such a way that the arc $PQ$ subtends an angle $120^{\circ}$ at the centre $O$ of the spherical shell, as shown in the figure. The permittivity of free space is $\in_0$. Which of the following statements is (are) true?
Updated On: Jun 14, 2022
- The electric flux through the shell is $\sqrt{3} R \lambda / \in_0$
- The $z$-component of the electric field is zero at all the points on the surface of the shell
- The electric flux through the shell is $\sqrt{2} R \lambda / \in_0$
- The electric field is normal to the surface of the shell at all points
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The Correct Option is B
Solution and Explanation
Field due to straight wire is perpendicular to the wire & radially outward. Hence $E_z = 0$
Length, $PQ = 2R \, \sin \, 60 = 3R$ According to Gauss's law
total flux = $\oint \vec{E} . \overrightarrow{ds} = \frac{q_{in}}{\in_0} = \frac{ \lambda \sqrt{3} R}{\in_0}$
Length, $PQ = 2R \, \sin \, 60 = 3R$ According to Gauss's law
total flux = $\oint \vec{E} . \overrightarrow{ds} = \frac{q_{in}}{\in_0} = \frac{ \lambda \sqrt{3} R}{\in_0}$
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).