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Question #42659

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass $m$ and charge $+q$ is placed in a one-dimensional region where the electric potential is given by $V(x,t) = \frac{1}{2} k (x - ut)^2$ , with $k$ and $u$ being positive constants. The reference potential $V=0$ is defined at the moving boundary $x = ut$ . At $t=0$ , the particle is released from rest at $x = x_0$ (where $x_0 > 0$ ). Defining $\omega = \sqrt{\frac{qk}{m}}$ , what is the kinetic energy of the particle in the laboratory frame at time $t = \frac{\pi}{2\omega}$ ?

Options

A.
$\frac{1}{2} m (u - x_0 \omega)^2$
B.
$\frac{1}{2} m (u^2 + x_0^2 \omega^2)$
C.
$\frac{1}{2} m (u + x_0 \omega)^2$
D.
$\frac{1}{2} m \omega^2 x_0^2$

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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

Reference Potential Convention|Trajectory of Charged Particle|Phasor Rotation Convention|Calculus

Why This Question Is Tricky

Common Mistakes

This question targets the misapplication of the Work-Energy Theorem and Koenig's theorem in time-varying fields.

A common trap is assuming mechanical energy ( $K + qV$ ) is conserved in the laboratory frame, or incorrectly applying Koenig's theorem by omitting the dot product term ( $m \vec{u} \cdot \vec{v}_{rel}$ ) to get $K_{lab} = K_{rel} + \frac{1}{2}mu^2$ (Option B).

Because the potential explicitly depends on time ( $\partial V/\partial t \neq 0$ ), the electric field does non-conservative work.

The elegant solution requires shifting to a moving inertial reference frame where the potential becomes static, solving the resulting simple harmonic motion to find the relative velocity, and applying the Galilean velocity addition to find the true kinetic energy in the lab frame.

Quick Facts About This Question

Question Information

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