A particle of mass 57 g and charge 70 µC is
released from rest when it is 34 cm from a
second particle of charge −12 µC.
Determine the magnitude of the initial acceleration
of the 57 g particle.
Answer in units of m/s2.
To determine the magnitude of the initial acceleration of the 57 g particle, we can use Coulomb's Law and Newton's second law of motion.
Coulomb's Law states that the force between two charged particles is given by:
F = (k * |q1 * q2|) / r^2
where F is the force, q1 and q2 are the charges of the particles, r is the distance between them, and k is the electrostatic constant (9 × 10^9 N m^2/C^2).
In this case, the force between the two particles is:
F = (k * |q1 * q2|) / r^2
= (9 × 10^9 N m^2/C^2 * |70 × 10^-6 C * (-12 × 10^-6 C)|) / (0.34 m)^2
Now, using Newton's second law of motion, we know that the force experienced by an object is equal to its mass multiplied by its acceleration:
F = m * a
where F is the force, m is the mass, and a is the acceleration.
Rearranging the equation to solve for acceleration:
a = F / m
Plugging in the values:
a = ((9 × 10^9 N m^2/C^2 * |70 × 10^-6 C * (-12 × 10^-6 C)|) / (0.34 m)^2) / 0.057 kg
Now we can calculate the acceleration:
a = ((9 × 10^9 * 70 * 12) / (0.34)^2) / 0.057
Simplifying further:
a ≈ 1336.68 m/s^2
Therefore, the magnitude of the initial acceleration of the 57 g particle is approximately 1336.68 m/s^2.