can anyone help me in this question please A particle of mass 27 g and charge 32 µC is

released from rest when it is 79 cm from a
second particle of charge −22 µC.
Determine the magnitude of the initial acceleration
of the 27 g particle.
Answer in units of m/s
2
.

F = m a = k Q1 Q2/d^2

so
a = F/m

m = 27 g = .027 kg

Q1 = 32*10^-6
Q2 = -22*10^-6
so F is attractive between them

F=(1/.027)k (32)(22)(10^-12)/.79^2

Oh

k is about 9*10^9

To determine the magnitude of the initial acceleration of the 27 g particle, we can use the basic principles of electrostatics and Newton's second law of motion.

First, let's calculate the electrostatic force between the two particles using Coulomb's Law:

F = k * ((q1 * q2) / r^2)

Where:
F is the electrostatic force between the particles,
k is the electrostatic constant (9 x 10^9 N m^2/C^2),
q1 is the charge of the first particle (32 µC = 32 x 10^-6 C),
q2 is the charge of the second particle (-22 µC = -22 x 10^-6 C),
r is the distance between the particles (79 cm = 79 x 10^-2 m).

Plugging in these values:

F = (9 x 10^9 N m^2/C^2) * ((32 x 10^-6 C) * (-22 x 10^-6 C)) / (79 x 10^-2 m)^2

F = (9 x 10^9 N m^2/C^2) * (32 x -22 x 10^-12 C^2) / (79^2 x 10^-4 m^2)

F ≈ -1.407 x 10^-3 N

The negative sign indicates that the electrostatic force is attractive between the particles.

Next, we can calculate the acceleration of the 27 g particle using Newton's second law:

F = m * a

Where:
F is the force acting on the particle (electrostatic force),
m is the mass of the particle (27 g = 27 x 10^-3 kg),
a is the acceleration of the particle.

Plugging in the values:

-1.407 x 10^-3 N = (27 x 10^-3 kg) * a

a = (-1.407 x 10^-3 N) / (27 x 10^-3 kg)

a ≈ -0.052 m/s^2

The magnitude of the initial acceleration of the 27 g particle is approximately 0.052 m/s^2. (Note that the negative sign indicates that the acceleration is directed towards the second particle, as it is an attractive force.)

Sure, I can help you with this question.

To determine the magnitude of the initial acceleration of the 27 g particle, we can use the principles of Coulomb's law and Newton's second law.

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, k is the electrostatic constant (9 * 10^9 N*m^2/C^2), q1 and q2 are the charges of the two particles, and r is the distance between them.

In this case, the force between the two particles is:

F = (9 * 10^9 N*m^2/C^2) * |((32 * 10^-6 C) * (-22 * 10^-6 C))| / (0.79 m)^2

Now, according to Newton's second law, the force acting on a particle is equal to the mass of the particle multiplied by its acceleration:

F = m * a

where F is the force, m is the mass of the particle, and a is the acceleration.

Rearranging the equation, we can solve for the acceleration:

a = F / m

Now we can substitute the values into the equation:

a = [(9 * 10^9 N*m^2/C^2) * |((32 * 10^-6 C) * (-22 * 10^-6 C))| / (0.79 m)^2] / 0.027 kg

Simplifying the expression and calculating it, you will get the magnitude of the initial acceleration of the 27 g particle in units of m/s^2.