Answer:
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
