A particle (m = 50 g, q = 5.0 μC) is released from rest when it is 50 cm from a second particle (Q = −20 μC). Determine the magnitude of the initial acceleration of the 50-g particle. A) 54 m/s² B) 90 m/s² C) 72 m/s² D) 65 m/s² E) 36 m/s²

Question

Quizplus · Accepted Answer

The force of attraction between the two particles can be found using Coulomb's law:
$F = \dfrac{kq_1q_2}{r^2}$
where k is Coulomb's constant, $q_1$ and $q_2$ are the charges of the particles, and r is the distance between them.

Plugging in the values given in the problem, we get:
$F = \dfrac{(9	imes 10^9 Nm^2/C^2)(5.0	imes 10^{-6} C)(-20	imes 10^{-6} C)}{(0.5m)^2}$
$F = -36 N$

The force is negative because it is an attractive force. This means that the 50-g particle will accelerate towards the -20 μC particle. Using Newton's second law:
$F = ma$
where m is the mass of the particle and a is its acceleration.

Solving for a, we get:
$a = \dfrac{F}{m} = \dfrac{-36 N}{0.050 kg}$
$a = -720 m/s^2$

The negative sign means that the acceleration is in the opposite direction to the initial displacement. Taking the magnitude of the acceleration, we get:
$|a| = 720 m/s^2 = 72	imes 10^1 m/s^2 \approx \boxed{72\ m/s^2}$, which is choice C.

A Particle (M = 50 G, Q = 5