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Question #42651

Difficulty: 8/10 | 🟢 Free

The Question

A solid non-conducting sphere of radius $R$ has a total charge $Q$ uniformly distributed throughout its volume. A thin, uncharged conducting spherical shell of radius $R/2$ is embedded concentrically within the non-conducting sphere (assume the shell is infinitesimally thin and does not alter the uniform volume charge distribution). This embedded shell is connected via a long, thin, insulated conducting wire to a distant, initially uncharged solid conducting sphere of radius $R$ . The wire contains a switch which is initially open. When the switch is closed, charge flows until electrostatic equilibrium is reached. Calculate the total heat dissipated in this process. (Assume the permittivity of free space is $\epsilon_0$ and the distant sphere is far enough that proximity effects are negligible).

Options

A.
$\frac{121 Q^2}{1536 \pi \epsilon_0 R}$
B.
$\frac{Q^2}{24 \pi \epsilon_0 R}$
C.
$\frac{3 Q^2}{32 \pi \epsilon_0 R}$
D.
$\frac{121 Q^2}{512 \pi \epsilon_0 R}$

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

Non-conducting Charged Sphere|Common Potential Redistribution|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the precise application of the superposition principle in a non-uniform potential field and the Work-Energy Theorem for continuous charge distributions.

Students often fall into the trap of treating the non-conducting charged sphere as an equipotential volume, incorrectly assigning the surface potential $V = kQ/R$ (Option B) or the center potential $V = 3kQ/2R$ (Option C) to the embedded shell.

Furthermore, students may forget to account for the self-potential generated by the newly acquired charge $-q$ on the embedded shell (Option D).

Correct modeling requires integrating the potential contribution of the volume charge density up to the specific boundary and recognizing that the heat dissipated is the work done by the conservative electrostatic field.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electric Charges and Fields - Class 12
Concept	Non-conducting Charged Sphere
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free