Animated Solution for Physics - Electrostatics: A charged shell of radius R carries a total charge Q. Given Φ as the flux of electric field through a closed cylindrical surface of height h, radius r and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct ? [ϵ0 is the permittivity of free space]
Select Answer:
* Multiple Correct
Visualized Solution
Φ=ϵ0qenclosed
Φ=ϵ0qenclosed
Option (A)
h>2R,r>R
Cylinder completely encloses the shell.
qenclosed=Q⟹Φ=ϵ0Q
Option (B)
h<58R,r=53R
d=r2+(2h)2<(53R)2+(54R)2=R
Cylinder is completely inside the shell.
Flux for Option (B)
Charge resides only on the surface of the shell.
qenclosed=0⟹Φ=0
Options (C) & (D)
h>2R,r<R
Cylinder cuts the shell, enclosing two spherical caps.
Area of one cap=2πR2(1−cosθ)
where sinθ=Rr
Total Enclosed Charge
Total area of two caps=4πR2(1−cosθ)
Fraction of total area=4πR24πR2(1−cosθ)=1−cosθ
qenclosed=Q(1−cosθ)
Option (C)
r=54R⟹sinθ=54⟹cosθ=53
qenclosed=Q(1−53)=52Q
Φ=5ϵ02Q=5ϵ0Q
Option (D)
r=53R⟹sinθ=53⟹cosθ=54
qenclosed=Q(1−54)=5Q
Φ=5ϵ0Q
00:00 / 00:00
The Sigma Insight: Electric Field Lines, Flux and Gauss's Law
Solution Diagram
The Beauty of Gauss's Law
Imagine you have a hollow spherical shell, and you've sprinkled a total charge Q uniformly over its surface. Now, you are handed a magical, invisible cylinder and asked to calculate the electric flux passing through it when it's placed right at the center of the sphere.
This might sound like a nightmare of complex surface integrals, but thanks to the sheer elegance of Gauss's Law, it becomes a beautiful geometric puzzle. Gauss's Law tells us that the total electric flux Φ through any closed surface is simply the total charge enclosed by that surface divided by the permittivity of free space, ϵ0.
Φ=ϵ0qenclosed
Our entire mission boils down to one simple question: How much of the shell's charge is trapped inside the cylinder?
Option A
The Giant Cylinder
Let's evaluate the first scenario. We are given a cylinder with a height h>2R and a radius r>R.
Visualize this: the cylinder is taller than the sphere's diameter and wider than the sphere itself. If you place this giant cylinder perfectly centered over the sphere, it will completely swallow the spherical shell.
Because the entire shell is sitting comfortably inside the cylinder, the total charge enclosed is simply the total charge of the shell, Q.
qenclosed=Q
Plugging this into Gauss's Law, the flux is:
Φ=ϵ0Q
This perfectly matches Option A. So, Option A is correct!
Option B
The Tiny Cylinder
Now, let's shrink our cylinder. In Option B, the radius is r=53R and the height is h<58R.
We need to know if this cylinder pokes out of the sphere or if it's completely hidden inside the hollow void. To check this, we calculate the distance from the center to the furthest point of the cylinder—its corners.
Using the Pythagorean theorem, the distance d to a corner is:
d=r2+(2h)2
Let's plug in the maximum possible height to see the worst-case scenario:
d<(53R)2+(54R)2=259R2+2516R2=2525R2=R
Since the distance to the corners is strictly less than R, the entire cylinder fits perfectly inside the hollow space of the shell.
Here is the catch: all the charge of a conducting or insulating shell resides only on its surface. The inside is completely empty of charge. Therefore, our tiny cylinder encloses absolutely zero charge.
qenclosed=0⟹Φ=0
This matches Option B perfectly. Option B is correct!
The Intersecting Cylinders
Options C and D
Things get incredibly interesting in Options C and D. Here, the cylinder is taller than the sphere (h>2R), but its radius is smaller than the sphere's radius (r<R).
Imagine pushing a tall apple corer through an apple. The cylinder pierces right through the top and bottom of the spherical shell. The only parts of the charged shell that end up inside the cylinder are the two circular "caps" at the top and bottom.
To find the enclosed charge, we need to find the surface area of these two spherical caps. The surface area of a single spherical cap subtending a half-angle θ at the center is given by:
Area of one cap=2πR2(1−cosθ)
Since the cylinder cuts out a top cap and a bottom cap, the total enclosed area is double that:
Total Enclosed Area=4πR2(1−cosθ)
To find the fraction of the total charge enclosed, we divide this enclosed area by the total surface area of the sphere (4πR2):
Fraction=4πR24πR2(1−cosθ)=1−cosθ
So, the master formula for the enclosed charge when the cylinder pierces the sphere is:
qenclosed=Q(1−cosθ)
We can find θ using the radius of the cylinder. Looking at the right triangle formed by the radius of the cylinder r and the radius of the sphere R, we see that:
sinθ=Rr
Testing Option C
In Option C, the radius is r=54R.
This means sinθ=54. Using the classic 3-4-5 right triangle, if the sine is 4/5, the cosine must be 3/5.
cosθ=53
Let's plug this into our master formula for the enclosed charge:
qenclosed=Q(1−53)=Q(52)=52Q
By Gauss's Law, the flux should be:
Φ=5ϵ02Q
However, Option C claims the flux is 5ϵ0Q. Therefore, Option C is incorrect.
Testing Option D
Finally, let's test Option D, where the radius is r=53R.
This means sinθ=53. Again, using our 3-4-5 triangle, the cosine must be 4/5.
cosθ=54
Plugging this into the master formula:
qenclosed=Q(1−54)=Q(51)=5Q
By Gauss's Law, the flux is:
Φ=5ϵ0Q
This matches Option D perfectly! So, Option D is correct.
The Final Verdict
By systematically applying Gauss's Law and visualizing the geometric intersections, we've cracked the code. The correct statements are A, B, and D. This problem is a fantastic reminder of how powerful and elegant Gauss's Law can be when combined with a little bit of spatial reasoning!
