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 \( E_0=2 \) N/C and \( \omega=1000 \) rad/sec. The maximum speed attained by the particle is

• A. \( 2\text{ m/s} \)
• B. \( 4\text{ m/s} \)
• C. \( 6\text{ m/s} \)
• D. \( 8\text{ m/s} \)

Hint:

For sinusoidal force: \begin{itemize} \item Integrate acceleration. \item Max velocity when cosine = -1. \end{itemize}

Answer 1

The Correct Option is B

Concept: Force due to electric field: \[ F=qE(t) \Rightarrow ma=qE_0\sin\omega t \] Step 1: {\color{red}Acceleration.} \[ a(t)=\frac{qE_0}{m}\sin\omega t \] Given: \[ m=1\text{ gm}=10^{-3}\text{ kg},\quad q=1 \] \[ a_0=\frac{2}{10^{-3}}=2000 \] Step 2: {\color{red}Velocity.} \[ v(t)=\int a(t)\,dt =\frac{a_0}{\omega}(1-\cos\omega t) \] Maximum when \( \cos\omega t=-1 \): \[ v_{\max}=\frac{2a_0}{\omega} \] Step 3: {\color{red}Substitute values.} \[ v_{\max}=\frac{2\times2000}{1000}=4\text{ m/s} \]