Animated Solution for Physics - Electrostatics: An infinitely long thin non-conducting wire is parallel to the Z-axis and carries a uniform line charge density λ. It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 120∘ at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is ε0. Which of the following statements is (are) true?
Select Answer:
* Multiple Correct
Visualized Solution
Visualizing the Setup
Spherical shell of radius R
Wire parallel to Z-axis with charge density λ
Geometry of Intersection
Wire intersects sphere at P and Q
∠POQ=120∘
Finding the Enclosed Length
Length of wire inside sphere=PQ
OM⊥PQ
Trigonometry for the Chord
∠POM=2120∘=60∘
PM=Rsin60∘
Calculating Total Length PQ
PQ=2×PM
PQ=2Rsin60∘=2R(23)=3R
Charge Enclosed
qenclosed=λ×PQ
qenclosed=3λR
Gauss’s Law
Gauss’s Law: ϕ=ε0qenclosed
Flux Calculation
ϕ=ε03λR
Option (a) is correct
Electric Field Direction
Electric field of infinite wire is radial
E⊥Wire
Z-component of Electric Field
Wire ∥ Z-axis⟹E is in X-Y plane
Ez=0 everywhere
Option (b) is correct
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The Sigma Insight: Electric Field Lines, Flux and Gauss's Law
Solution Diagram
The journey to solving this problem begins with a clear visualization of the physical setup. Imagine a delicate, non-conducting spherical shell of radius R floating in space. Now, picture an infinitely long, charged wire piercing straight through it, perfectly parallel to the Z-axis.
This isn't just any random intersection; the problem gives us a beautiful geometric constraint. The segment of the wire trapped inside the sphere, which we'll call the chord PQ, subtends an angle of exactly 120∘ at the center O of the sphere.
Our mission is to uncover the secrets of the electric flux passing through this shell and the nature of the electric field it experiences. Let's dive in!
Analyzing the Geometry
To find the electric flux using Gauss's Law, our primary target is the total charge enclosed within the spherical shell. Since the wire has a uniform linear charge density λ, the enclosed charge is simply λ multiplied by the length of the wire inside the sphere.
So, how long is this trapped segment PQ?
Let's look at the triangle formed by the center O and the points P and Q. The sides OP and OQ are just the radii of the sphere, so OP=OQ=R. The angle between them is given as ∠POQ=120∘.
If we drop a perpendicular from the center O to the chord PQ at point M, it splits the triangle into two identical right-angled triangles. It also bisects the angle, making ∠POM=60∘.
Using basic trigonometry in △POM, we can find the half-length of the chord:
PM=Rsin60∘
Since sin60∘=23, we have:
PM=R(23)
The total length of the chord PQ is twice this value:
PQ=2×PM=3R
The Master Equation
Gauss's Law
Now that we have the length of the wire inside the sphere, finding the enclosed charge is a breeze.
qenclosed=λ×PQ=3λR
Gauss's Law is one of the most elegant principles in electromagnetism. It states that the total electric flux ϕ through a closed surface is equal to the enclosed charge divided by the permittivity of free space ε0.
ϕ=ε0qenclosed
Substituting our enclosed charge into the master equation, we get:
ϕ=ε03λR
This perfectly matches our first option! The electric flux through the shell is indeed 3Rλ/ε0.
Decoding the Electric Field
Now, let's shift our focus to the electric field itself. The wire is infinitely long and uniformly charged. By symmetry, the electric field lines must point radially outward (or inward, depending on the sign of λ) from the wire.
Crucially, the electric field is always perpendicular to the wire. Since the wire is parallel to the Z-axis, the electric field vectors must lie entirely in the X-Y plane.
What does this mean for the Z-component of the electric field?
It means it doesn't exist! The electric field has absolutely no component along the Z-axis.
Ez=0
This holds true everywhere in space, including all points on the surface of the spherical shell. Thus, the second option is also correct.
As for the fourth option, the electric field is normal to the wire, not the spherical shell. The only places where the electric field is normal to the shell are the points where the wire's radial field aligns with the sphere's radial normal, which is not true for all points.
Final Conclusion
By combining geometric intuition with the power of Gauss's Law and symmetry arguments, we've successfully navigated this problem. The correct statements are that the flux is 3Rλ/ε0 and the Z-component of the electric field is zero everywhere.
