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?

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 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

Electrostatics PYQ — JEE Advanced 2018 · Shift 2 · Q39 with Solution

View written solutionFree

Correct answer: A, B

Use Gauss's law for the flux through the spherical shell

The electric flux through any closed surface is

\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}.

So we only need the charge of the wire segment lying inside the spherical shell.

Find the length of the wire inside the sphere

Let the perpendicular distance of the wire from the center $O$ be $d$ .

Since the wire cuts the sphere at points $P$ and $Q$ , the chord $PQ$ of the sphere subtends angle $120^\circ$ at the center.

For a chord subtending angle $\theta$ at the center, its length is

PQ = 2R\sin\frac{\theta}{2}.

Here $\theta=120^\circ$ , so

PQ = 2R\sin 60^\circ = 2R\cdot \frac{\sqrt{3}}{2} = \sqrt{3}R.

This chord lies along the straight line of the wire, so the length of wire inside the sphere is

\ell = \sqrt{3}R.

Hence enclosed charge is

q_{\text{enclosed}} = \lambda \ell = \lambda \sqrt{3}R.

Therefore,

\Phi = \frac{\lambda \sqrt{3}R}{\varepsilon_0}.

So Option A is correct and Option C is incorrect.

Check the $z$ -component of electric field on the spherical surface

The wire is parallel to the $z$ -axis. The electric field due to an infinite line charge is always perpendicular to the wire.

Since the wire is along the $z$ -direction, the field has no component along $z$ anywhere in space. Thus

E_z = 0

at every point on the spherical surface as well.

So Option B is correct.

Check whether the electric field is normal to the spherical surface everywhere

The electric field due to a line charge is directed radially outward from the wire, not from the center of the sphere.

But the normal to a spherical surface at a point is along the radius through the center of the sphere.

These two directions are generally different, except possibly at a few special points. Therefore the field is not normal to the spherical surface at all points.

So Option D is incorrect.

Final evaluation of options

A: True
B: True
C: False
D: False

Thus the correct answer is:

\boxed{A,\ B}

Comparison with stored correct answer

Stored correct answer: A, B

Our derived answer matches the stored answer exactly.

More from Electrostatics

View written solutionFree

Correct answer: A, B

Use Gauss's law for the flux through the spherical shell

The electric flux through any closed surface is

\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}.

So we only need the charge of the wire segment lying inside the spherical shell.

Find the length of the wire inside the sphere

Let the perpendicular distance of the wire from the center $O$ be $d$ .

Since the wire cuts the sphere at points $P$ and $Q$ , the chord $PQ$ of the sphere subtends angle $120^\circ$ at the center.

For a chord subtending angle $\theta$ at the center, its length is

PQ = 2R\sin\frac{\theta}{2}.

Here $\theta=120^\circ$ , so

PQ = 2R\sin 60^\circ = 2R\cdot \frac{\sqrt{3}}{2} = \sqrt{3}R.

This chord lies along the straight line of the wire, so the length of wire inside the sphere is

\ell = \sqrt{3}R.

Hence enclosed charge is

q_{\text{enclosed}} = \lambda \ell = \lambda \sqrt{3}R.

Therefore,

\Phi = \frac{\lambda \sqrt{3}R}{\varepsilon_0}.

So Option A is correct and Option C is incorrect.

Check the $z$ -component of electric field on the spherical surface

The wire is parallel to the $z$ -axis. The electric field due to an infinite line charge is always perpendicular to the wire.

Since the wire is along the $z$ -direction, the field has no component along $z$ anywhere in space. Thus

E_z = 0

at every point on the spherical surface as well.

So Option B is correct.

Check whether the electric field is normal to the spherical surface everywhere

The electric field due to a line charge is directed radially outward from the wire, not from the center of the sphere.

But the normal to a spherical surface at a point is along the radius through the center of the sphere.

These two directions are generally different, except possibly at a few special points. Therefore the field is not normal to the spherical surface at all points.

So Option D is incorrect.

Final evaluation of options

A: True
B: True
C: False
D: False

Thus the correct answer is:

\boxed{A,\ B}

Comparison with stored correct answer

Stored correct answer: A, B

Our derived answer matches the stored answer exactly.

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