Question 833: Electrostatics & Capacitance

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 center $$O$$ of the spherical shell, as shown in the figure. The permittivity of free space is $${ \in _0}.$$ Which of the following statement is (are) true?

Accepted Answer

Let Gaussian surface be same as the non-conducting spherical shell. The charge enclosed by this Gaussian surface is the charge on the line segment PQ i.e.,
$${q_{enc}} = PQ\lambda  = \sqrt 3 R\lambda $$,
where we used law of cosines in triangle OPQ to get $$PQ = \sqrt {{R^2} + {R^2} - 2(R)(R)\cos 120^\circ }  = \sqrt 3 R$$. The electric flux through the shell (Gaussian surface) is given by Gauss's law
$$\phi  = {q_{enc}}/{ \in _0} = \sqrt 3 R\lambda /{ \in _0}$$.
The electric field at a point due to the infinitely long line charge is radial i.e., perpendicular to the line PQ. The non-conducting spherical shell does not affect or alter the electric field. Thus, electric field at a point lying on the surface of the shell is perpendicular to the z-axis i.e., its z component is zero. The figure shows electric field on the surface of the shell for a positive line charge. Note that the magnitude of electric field is inversely proportional to the distance of the point from the line charge.

Answer

The electric flux through the shell is $$\sqrt 3 R\lambda /{ \in _0}$$

Answer

The $$z$$-component of the electric field is zero at all the points on the surface of the shell

Answer

The electric flux through the shell is $$\sqrt 2 R\lambda /{ \in _0}$$

Answer

The electric field is normal to the surface of the shell at all points

Electrostatics & Capacitance

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