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Question #42910

Difficulty: 8/10 | 🟢 Free

The Question

A continuous flexible thin ring of initial radius $R$ , cross-sectional area $A$ , and Young's modulus $Y$ has a total charge $Q$ uniformly distributed over its circumference. A point charge $q$ of the same sign is placed at the exact center of the ring. Due to the electrostatic repulsion, the ring expands slightly to reach a new equilibrium. Assuming the expansion is small ( $\Delta R \ll R$ ), determine the elastic potential energy stored in the ring.

Options

A.
$\frac{q^2 Q^2}{64 \pi^3 \epsilon_0^2 Y A R^3}$
B.
$\frac{q^2 Q^2}{16 \pi^3 \epsilon_0^2 Y A R^3}$
C.
$\frac{q^2 Q^2}{64 \pi \epsilon_0^2 Y A R^3}$
D.
$\frac{q^2 Q^2}{32 \pi^3 \epsilon_0^2 Y A R^3}$

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

Continuous Charge Distribution|Work Done in Stretching|Young's Modulus Sensitivity|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly bridge continuous electrostatics with solid mechanics by evaluating hoop stress.

A common JEE trap is misapplying Newton's Laws by treating the total scalar electrostatic force ( $F_{total} = qQ/4\pi\epsilon_0 R^2$ ) as a unidirectional force and incorrectly setting $2T = F_{total}$ .

This ignores the vector integration of the radial Coulomb force over the ring's arc, leading to a missing factor of $\pi$ in the tension.

Furthermore, students often forget the geometric volume factor or the $1/2$ in the elastic strain energy formula, leading to dimensionally similar but numerically incorrect distractors.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electric Charges and Fields - Class 12
Concept	Continuous Charge Distribution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free