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Question #42764

Difficulty: 8/10 | 🟢 Free

The Question

A large tank contains a liquid of uniform density ρ\rho. The tank accelerates horizontally to the right (+i^+\hat{i} direction) with a constant acceleration a=ga = g. A rigid, thin-walled, uncharged conducting spherical shell of outer radius RR is completely submerged in the liquid and held fixed relative to the tank. The cavity inside the shell contains a gas of negligible mass. A point charge +Q+Q is held fixed at the exact center of the cavity. A uniform external electric field E⃗0=E0i^\vec{E}_0 = E_0 \hat{i} is applied horizontally to the right. Let F⃗e\vec{F}_e be the net electrostatic force exerted on the point charge +Q+Q. Let ΔPout=Pout,L−Pout,R\Delta P_{out} = P_{out, L} - P_{out, R} be the difference in mechanical pressure between the leftmost point (Pout,LP_{out, L}) and the rightmost point (Pout,RP_{out, R}) on the outer surface of the shell. Let ΔPin=Pin,L−Pin,R\Delta P_{in} = P_{in, L} - P_{in, R} be the difference in gas pressure between the leftmost point (Pin,LP_{in, L}) and the rightmost point (Pin,RP_{in, R}) on the inner surface of the shell. Which of the following sets of values is correct?

Options

  • A.

    F⃗e=0⃗,ΔPout=2ρgR,ΔPin=0\vec{F}_e = \vec{0}, \Delta P_{out} = 2\rho g R, \Delta P_{in} = 0

  • B.

    F⃗e=QE0i^,ΔPout=2ρgR,ΔPin=2ρgR\vec{F}_e = Q E_0 \hat{i}, \Delta P_{out} = 2\rho g R, \Delta P_{in} = 2\rho g R

  • C.

    F⃗e=0⃗,ΔPout=ρgR,ΔPin=ρgR\vec{F}_e = \vec{0}, \Delta P_{out} = \rho g R, \Delta P_{in} = \rho g R

  • D.

    F⃗e=QE0i^,ΔPout=2ρgR,ΔPin=0\vec{F}_e = Q E_0 \hat{i}, \Delta P_{out} = 2\rho g R, \Delta P_{in} = 0

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Cavity Field Shielding|Conservative Field Path Independence|Accelerated Frame Hydrostatics

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to decouple mechanical and electrostatic equilibrium in an accelerated frame.

A common trap is assuming that the external tilted effective potential (due to non-inertial acceleration) penetrates the cavity, leading students to incorrectly assign a pressure gradient to the internal massless gas (ΔPin=2ρgR\Delta P_{in} = 2\rho g R).

Furthermore, students often misapply the superposition principle, thinking the external electric field E⃗0\vec{E}_0 acts on the internal charge +Q+Q, failing to recognize that the conducting shell provides perfect electrostatic shielding (Faraday cage effect) while simultaneously providing mechanical shielding against the external pressure gradient.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectrostatic Potential and Capacitance - Class 12
ConceptCavity Field Shielding
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree