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A particle charge +15.2μC and mass 1.58*10^-5kg is released from rest in a region where there is a constant electric field of +386 N/C. What is the displacement of the particle after a time of 2.87*10^-2s?

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Answer:

Step-by-step explanation:

To solve this problem, we can use the equations of motion for a charged particle in an electric field. The equation we'll use is:

y = y₀ + v₀yt + 0.5at²

Where:

- y is the displacement of the particle after time t.

- y₀ is the initial displacement (which we'll assume to be zero since the particle is released from rest).

- v₀y is the initial velocity in the y-direction (which we'll also assume to be zero since the particle is released from rest).

- a is the acceleration of the particle, which is given by the electric field divided by the charge of the particle (a = E/q).

- t is the time.

Given:

- Particle charge (q) = +15.2 μC = +15.2 × 10⁻⁶ C

- Particle mass (m) = 1.58 × 10⁻⁵ kg

- Electric field (E) = +386 N/C

- Time (t) = 2.87 × 10⁻² s

First, let's calculate the acceleration (a):

a = E/q

a = 386 N/C / 15.2 × 10⁻⁶ C

a = 2.55 × 10⁴ m/s²

Now, we can calculate the displacement (y):

y = 0 + 0 + 0.5at²

y = 0.5 × (2.55 × 10⁴ m/s²) × (2.87 × 10⁻² s)²

y ≈ 10.5 m

Therefore, the displacement of the particle after a time of 2.87 × 10⁻² s is approximately 10.5 meters.

User PrzemekTom
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