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Question #42774

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor has rectangular plates of length $L$ and width $w$ , separated by a distance $d$ . A solid dielectric slab of mass $m$ , dielectric constant $K$ , and identical dimensions ( $L \times w \times d$ ) is connected to a fixed vertical wall by a massless, non-conducting spring of spring constant $k$ . Initially, the spring is in its relaxed state, and the dielectric slab is just at the edge of the capacitor, inserted by a negligible distance. A battery of constant electromotive force $V$ is connected across the capacitor plates and remains connected. The system is released from rest. Assuming the slab does not fully enter the capacitor during its motion and neglecting friction and edge effects, what is the total work done by the battery from the moment of release until the spring reaches its maximum extension?

Options

A.
$\frac{\epsilon_0^2 w^2 (K-1)^2 V^4}{k d^2}$
B.
$\frac{\epsilon_0^2 w^2 (K-1)^2 V^4}{2 k d^2}$
C.
$\frac{\epsilon_0^2 w^2 (K-1)^2 V^4}{4 k d^2}$
D.
$\frac{\epsilon_0^2 w^2 (K-1)^2 V^4}{8 k d^2}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Dielectric Polarization Logic|Work Done in Rotating Dipole|Conservative vs Non-conservative

Why This Question Is Tricky

Common Mistakes

The core trap lies in the distinction between the equilibrium position and the maximum displacement of an oscillating system, compounded by the energy contributions in a closed circuit.

Students often incorrectly equate the electrostatic force directly to the spring force ( $F_e = kx$ ) to find the maximum extension, which actually yields the equilibrium position ( $x_{eq}$ ).

Furthermore, they frequently misapply energy conservation by neglecting the work done by the battery ( $\Delta W_{batt} = V \Delta Q = 2\Delta U_{cap}$ ).

By failing to account for the battery's continuous energy input, or by confusing $x_{eq}$ with $x_{max}$ , students will calculate a battery work that is off by a factor of 2 or 4.

The correct approach requires applying the Work-Energy Theorem ( $W_{net} = 0$ at $x_{max}$ ) to find the true maximum extension ( $x_{max} = 2x_{eq}$ ), and then calculating the battery work using $\Delta W_{batt} = V^2 \Delta C$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Dielectric Polarization Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free