WBJEE · Physics · Electrostatics
Consider a particle of mass 1 gm and charge 1.0 Coulomb is at rest. Now the particle is subjected to an electric field \(E(t)=E_0 \sin \omega t\) in the \(x\)-direction, where \(E_0=2\) Newton/Coulomb and \(\omega=1000 \mathrm{rad} / \mathrm{sec}\). The maximum speed attained by the particle is
- A 2 m/sec
- B 4 m/sec
- C 6 m/sec
- D 8 m/sec
Answer & Solution
Correct Answer
(B) 4 m/sec
Step-by-step Solution
Detailed explanation
\(V_{\max }=\frac{1}{m} \int_0^{\frac{T}{2}} F d t,=\frac{1}{m} \int_0^{\frac{T}{2}} q E_0 \sin (\omega t) d t\) \(=\frac{2 \mathrm{qE}_0}{\mathrm{~m} \omega}=4 \mathrm{~m} / \mathrm{s}\)
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