Animated Solution for Physics - Electrostatics: A small electric dipole p0 , having a moment of inertia I about its center, is kept at a distance r from the center of a spherical shell of radius R . The surface charge density σ is uniformly distributed on the spherical shell. The dipole is initially oriented at a small angle θ as shown in the figure. While staying at a distance r , the dipole is free to rotate about its center.
If released from rest, then which of the following statement(s) is(are) correct?
[ε0 is the permittivity of free space.]
Select Answer:
* Multiple Correct
Visualized Solution
Analyzing the Setup
A spherical shell of radius R
Uniform surface charge density σ
Electric Field of a Spherical Shell
E={04πε01r2Qfor r<Rfor r>R
Electric Field at Distance r>R
Q=σ(4πR2)
E=4πε01r2σ(4πR2)=ε0r2σR2
Torque on the Dipole
τ=p0×E
∣τ∣=p0Esinθ
Restoring Torque for Small θ
For small θ,sinθ≈θ
τ≈−p0Eθ
Angular SHM Equation
τ=Iα
Iα=−p0(ε0r2σR2)θ
α=−(ε0Ir2σR2p0)θ
Angular Frequency ω
ω2=ε0Ir2σR2p0
ω=ε0Ir2σR2p0
Evaluating at Specific Distances
At r=2R:ω=4ε0Iσp0
At r=10R:ω=100ε0Iσp0
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The Sigma Insight: Electric Dipole
Solution Diagram
The Setup and the Field
Imagine a perfectly spherical shell of radius R, uniformly coated with a surface charge density σ. This charged shell creates an electric field around it.
According to Gauss's Law, the electric field inside a uniformly charged spherical shell is exactly zero. However, outside the shell, it behaves as if all its charge is concentrated at the center.
The total charge on the shell is the product of its surface charge density and its surface area:
Q=σ(4πR2)
At a distance r>R, the electric field E is given by:
E=4πε01r2Q=ε0r2σR2
This field points radially outward. For a dipole placed on the vertical axis at distance r, the electric field points vertically upward.
The Dance of the Dipole
Now, let's introduce a small electric dipole with dipole moment p0 at this distance r. The dipole is initially tilted at a small angle θ relative to the electric field.
When a dipole is placed in an electric field, it experiences a torque that tries to align it with the field. The torque τ is given by the cross product:
τ=p0×E
The magnitude of this torque is:
∣τ∣=p0Esinθ
Because the torque acts to decrease the angle θ and bring the dipole back to the vertical alignment, it is a restoring torque. We can add a negative sign to indicate this restoring nature:
τ=−p0Esinθ
The Math of the Oscillation
Since the dipole is released from a very small angle, we can use the small angle approximation, where sinθ≈θ. This simplifies our torque equation to:
τ≈−p0Eθ
According to Newton's Second Law for rotation, the torque is also equal to the moment of inertia I multiplied by the angular acceleration α:
Iα=−p0Eθ
Substituting our expression for the electric field E, we get:
Iα=−p0(ε0r2σR2)θ
Rearranging this to solve for α, we find the classic signature of Simple Harmonic Motion (SHM):
α=−(ε0Ir2σR2p0)θ
Comparing this with the standard SHM equation α=−ω2θ, we can extract the angular frequency ω:
ω=ε0Ir2σR2p0
Checking the Options
This formula tells us that the dipole will undergo small oscillations for any finite value of r>R. Inside the shell (r<R), the electric field is zero, so there is no torque and no oscillation. This makes Option (B) correct.
Let's evaluate the angular frequency at specific distances to check the remaining options.
At r=2R:
ω=ε0I(2R)2σR2p0=4ε0Iσp0
This does not match Option (C).
At r=10R:
ω=ε0I(10R)2σR2p0=100ε0Iσp0
This perfectly matches Option (D).
Therefore, the correct statements are (B) and (D).
