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Question #42847

Difficulty: 8/10 | 🟢 Free

The Question

An isolated, perfectly conducting sphere of radius aa in vacuum is initially charged to a potential V0V_0. At t=0t=0, it is connected to ground through a series combination of a time-varying internal resistor and a variable external load resistor RLR_L. The internal resistor is a cylinder of length LL and cross-sectional area AA, filled with a material of constant permittivity ϵ\epsilon. Due to heating, the dielectric relaxation time of this material decreases as τ(t)=τ01+αt\tau(t) = \frac{\tau_0}{1 + \alpha t}. A control system continuously adjusts RLR_L such that the instantaneous power dissipated in RLR_L is always maximized. Assuming the capacitance of the cylindrical resistor is negligible, which of the following gives the potential of the sphere V(t)V(t)?

Options

  • A.

    V0exp⁡(−ϵA(2t+αt2)16πϵ0aLτ0)V_0 \exp\left( -\frac{\epsilon A (2t + \alpha t^2)}{16 \pi \epsilon_0 a L \tau_0} \right)

  • B.

    V0exp⁡(−ϵA(2t+αt2)8πϵ0aLτ0)V_0 \exp\left( -\frac{\epsilon A (2t + \alpha t^2)}{8 \pi \epsilon_0 a L \tau_0} \right)

  • C.

    V0exp⁡(−ϵAt8πϵ0aLτ0)V_0 \exp\left( -\frac{\epsilon A t}{8 \pi \epsilon_0 a L \tau_0} \right)

  • D.

    V0exp⁡(−ϵA(t+αt2)8πϵ0aLτ0)V_0 \exp\left( -\frac{\epsilon A (t + \alpha t^2)}{8 \pi \epsilon_0 a L \tau_0} \right)

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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

Charged Conducting Sphere|Capacitance Independence|Maximum Power Transfer|Relaxation Time Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Maximum Power Transfer theorem in a dynamic, time-varying system coupled with an RC discharge circuit.

The primary trap is that students often memorize RL=RintR_L = R_{int} for static DC circuits but fail to realize that when applying it instantaneously to a time-varying source, the total equivalent resistance of the circuit becomes Req=2Rint(t)R_{eq} = 2R_{int}(t).

Many students will incorrectly use Req=Rint(t)R_{eq} = R_{int}(t) (forgetting the load itself in the discharge path) leading to Option B.

Additionally, it tests the fundamental relationship between dielectric relaxation time, permittivity, and conductivity, requiring students to dynamically integrate the time-varying RC constant rather than treating it algebraically.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptCharged Conducting Sphere
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree