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

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step-by-step explanation: