A particle of charge and mass kg is released from rest in a region where there is a constant electric field of +668 N/C. What is the displacement of the particle after a time of?

The displacement of the particle after a time of 3.64 × 10⁻² s = 0.366 m

Explanation:

Given q = 19.6 μC = 19.6 × 10^-6 C

m = mass of the charge = 2.37 × 10^-5 kg

E = 668 N/C

We first calculate the force on the charge

F = Electric field × Charge = Eq = 668 × 19.6 × 10⁻⁶ = 0.0131 N

Then, using Newton's law of motion's F = ma

We can obtain the acceleration of the particle

0.0131 = 2.37 × 10⁻⁵ × a

a = 0.0131/(2.37 × 10⁻⁵) = 552.4 m/s²

Then, using the equations of motion, we can obtain the displacement of the charge after t = 3.64 × 10⁻² = 0.0364 s

u = initial velocity = 0 m/s (since the particle was initially at rest)

t = 0.0364 s

a = 552.4 m/s²

d = displacement of the particle = ?

d = ut + at²/2

d = (0)(0.0364) + (552.4)(0.0364²)/2 = 0 + 0.366 = 0.366 m

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

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