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Question #42761
The Question
A solid, uncharged conducting sphere of radius is placed inside a rocket that is accelerating in deep space with a constant acceleration . Let the center of the sphere be at the origin . Let and be the mass and charge of an electron, respectively. Assume that the positive ions of the conductor's lattice are fixed relative to the sphere. Once electrostatic equilibrium is reached in the frame of the rocket, what is the potential difference between the point on the surface of the sphere and the center , and what is the induced surface charge density at point ?
Options
- A.
- B.
- C.
- D.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Variable_Constraint
Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.
Concepts Behind This Question
Conductor Interior Potential|Frame of Reference Inertial vs Non-Inertial|Pseudo Force Directionality
Why This Question Is Tricky
Common Mistakes
This question tests the limits of the 'equipotential volume' mnemonic for conductors.
Students are heavily conditioned to assume that the electric field inside a conductor in electrostatic equilibrium is always zero (), which would lead them to incorrectly choose Option C.
However, this rule only applies in an inertial frame without non-electrical body forces.
In an accelerating frame, the pseudo-force acts as an effective uniform gravitational field on the electron gas.
To maintain equilibrium, the electrons must redistribute to create a non-zero internal electric field that exactly cancels the pseudo-force.
A secondary trap (Option B) tests whether students correctly apply the boundary conditions for a polarized sphere (where ) rather than incorrectly applying the infinite slab formula ().
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Electrostatic Potential and Capacitance - Class 12 |
| Concept | Conductor Interior Potential |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |