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← JEE Advanced 2018 · Paper 2
PhysicsNumerical+3Question 9 of 54

JEE Advanced 2018 · Paper 2, Physics question 9

A particle, of mass 10−310^{-3} kg and charge 1.0 C, is initially at rest. At time t=0t = 0, the particle comes under the influence of an electric field E⃗(t)=E0sin⁡ωt i^\vec{E}(t) = E_0 \sin\omega t\, \hat{i}, where E0=1.0 N C−1E_0 = 1.0\ \mathrm{N\,C^{-1}} and ω=103 rad s−1\omega = 10^3\ \mathrm{rad\,s^{-1}}. Consider the effect of only the electrical force on the particle. Then the maximum speed, in m s−1\mathrm{m\,s^{-1}}, attained by the particle at subsequent times is ____________.

Answer and worked solution

Official answer: 2

Acceleration.
a=qE0msin⁡ωt=1×110−3sin⁡ωt=103sin⁡ωt.a=\frac{qE_0}m\sin\omega t=\frac{1\times1}{10^{-3}}\sin\omega t=10^3\sin\omega t.

Velocity, starting from rest:
v=103ω(1−cos⁡ωt)=1−cos⁡ωt.v=\frac{10^3}\omega(1-\cos\omega t)=1-\cos\omega t.
This ranges from 0 to 2. The particle never reverses; it moves forward in surges.

Maximum speed: 2 m s−12\ \mathrm{m\,s^{-1}}.

Answer: 2