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Question #42731

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor has rectangular plates of length LL, width ww, and separation dd. It is connected across an ideal DC voltage source of EMF ε\varepsilon. A dielectric slab of the same dimensions (L×w×dL \times w \times d), with dielectric constant KK and finite resistivity ρ\rho, initially fills the entire space between the plates. At time t=0t=0, the slab starts being pulled out along its length at a constant velocity vv. Edge effects are negligible, and the space vacated by the slab becomes vacuum (permittivity ϵ0\epsilon_0). What is the total current II drawn from the voltage source at the instant when a length xx of the slab remains inside the capacitor?

Options

  • A.

    εwd[xρ−ϵ0v(K−1)]\frac{\varepsilon w}{d} \left[ \frac{x}{\rho} - \epsilon_0 v (K-1) \right]

  • B.

    εwd[xρ+ϵ0v(K−1)]\frac{\varepsilon w}{d} \left[ \frac{x}{\rho} + \epsilon_0 v (K-1) \right]

  • C.

    εwxρd\frac{\varepsilon w x}{\rho d}

  • D.

    εwd[Lρ−ϵ0v(K−1)]\frac{\varepsilon w}{d} \left[ \frac{L}{\rho} - \epsilon_0 v (K-1) \right]

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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

Resistivity vs Resistance|Kirchhoff’s Loop Rule|Dielectric Slab Displacement

Why This Question Is Tricky

Common Mistakes

Students incorrectly assume that because the voltage across the capacitor is constant, the circuit is in a quasi-static steady state where the current is simply determined by Ohm's law I=ε/R(x)I = \varepsilon/R(x).

They fail to apply Kirchhoff's Junction Rule correctly, which requires accounting for the displacement current IC=d(CV)/dtI_C = d(CV)/dt induced by the time-varying geometry of the dielectric.

Furthermore, a secondary trap involves a sign error in the kinematic relation; students often use dx/dt=+vdx/dt = +v instead of −v-v, failing to realize that as the slab is removed, the decreasing capacitance forces charge back into the battery, thereby reducing the net current drawn.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptResistivity vs Resistance
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree