Question

Consider a particle of mass 1 gm and charge 1.0 Coulomb at rest. Now, the particle is subjected to an electric field \( E(t) = E_0 \sin(\omega t) \) in the x-direction, where \( E_0 = 2 \, \text{N/C} \) and \( \omega = 1000 \, \text{rad/sec} \). The maximum speed attained by the particle is:

• A. 2 m/s
• B. 4 m/s
• C. 6 m/s
• D. 8 m/s

Hint:

The maximum speed of a particle under the influence of a time-varying electric field can be found by integrating the acceleration, considering the maximum value of the sine function.

Answer 1

The Correct Option is B

Step 1: Particle Force The force on the particle from the electric field is: \[ F = q E(t) \] Where: - \( q = 1.0 \, \text{C} \) (particle charge) - \( E(t) = E_0 \sin(\omega t) \) (electric field) Therefore: \[ F = 1.0 \cdot 2 \sin(1000 t) = 2 \sin(1000 t) \, \text{N} \] Step 2: Particle Acceleration The particle's acceleration is given by Newton's second law: \[ F = m a \] Where: - \( m = 1 \, \text{gm} = 1 \times 10^{-3} \, \text{kg} \) (particle mass) - \( a \) is the particle's acceleration Thus: \[ a = \frac{F}{m} = \frac{2 \sin(1000 t)}{1 \times 10^{-3}} = 2000 \sin(1000 t) \, \text{m/s}^2 \] Step 3: Maximum Speed Maximum speed occurs when acceleration is maximal. The maximum value of \( \sin(1000 t) \) is 1, thus: \[ a_{\text{max}} = 2000 \, \text{m/s}^2 \] Since the particle starts at rest, maximum speed occurs after maximum acceleration time. Velocity is the integral of acceleration: \[ v(t) = \int a(t) \, dt = \int 2000 \sin(1000 t) \, dt \] The integral of \( \sin(1000 t) \) is: \[ v(t) = -\frac{2000}{1000} \cos(1000 t) + C \] At \( t = 0 \), the particle is at rest, so \( C = 2 \). Thus: \[ v(t) = 2 - 2 \cos(1000 t) \] Maximum speed occurs when \( \cos(1000 t) = -1 \), resulting in: \[ v_{\text{max}} = 2 - 2(-1) = 4 \, \text{m/s} \] Step 4: Conclusion The particle's maximum speed is \( 4 \, \text{m/s} \). Thus, the answer is: \[ \boxed{(B)} \, 4 \, \text{m/s} \]