need help with this problem
A charge of -2.73 uC is fixed in place. From a horizontal distance of 0.0465 m, a particle of mass 9.50 x 10-3 kg and charge -9.56 uC is fired with an initial speed of 97.2 m/s directly toward the fixed charge. What is the distance of closest approach?
RKOOOOO
To solve this problem, we can use the principles of electrostatics and projectile motion.
First, let's understand the forces acting on the particle due to the fixed charge. The electrostatic force between two charges is given by Coulomb's law:
F = k * (|q1| * |q2|) / r^2
Where F is the force, k is the electrostatic constant (9 x 10^9 N*m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
In this case, the force experienced by the particle is attractive, as the charges have opposite signs. So we can rewrite Coulomb's law as:
F = -k * (|q1| * |q2|) / r^2
Now, let's consider the projectile motion of the particle. The force acting on the particle in the horizontal direction is zero since there are no external forces. In the vertical direction, the only force acting is the gravitational force:
F_gravity = m * g
Where F_gravity is the gravitational force, m is the mass of the particle, and g is the acceleration due to gravity.
The motion of the particle can be separated into horizontal and vertical components. In the horizontal direction, the particle will undergo uniform motion due to the absence of any forces. In the vertical direction, the particle will experience constant acceleration due to gravity.
Now, let's find the distance of closest approach, which is the minimum distance between the particle and the fixed charge.
To find this distance, we need to find the point where the electrostatic force is equal in magnitude to the gravitational force. At this point, the particle will stop moving toward the fixed charge and start moving away.
Now, let's set up an equation for the electrostatic force and the gravitational force at the distance of closest approach:
-k * (|q1| * |q2|) / r^2 = m * g
Substituting the given values:
- (9 x 10^9 N*m^2/C^2) * (2.73 x 10^(-6) C) * (9.56 x 10^(-6) C) / r^2 = (9.50 x 10^(-3) kg) * (9.8 m/s^2)
Now, solve the equation for r:
r^2 = [(9 x 10^9 N*m^2/C^2) * (2.73 x 10^(-6) C) * (9.56 x 10^(-6) C)] / [(9.50 x 10^(-3) kg) * (9.8 m/s^2)]
r^2 = (219.3 N*m^2/kg) / (92.6 m^2/s^2)
r^2 = 2.37 m^2
r ≈ 1.54 m
Therefore, the distance of closest approach is approximately 1.54 meters.