A particle with a mass of 3.65×10^−24 kg and a charge of 6.40×10^−19 C starts at rest and is accelerated for a distance of 0.102 m through a uniform electric field having strength E = 7.25×10^4 N/C. What is the resulting speed of the particle(m/s)?
To find the resulting speed of the particle, we can use the equations of motion in combination with the principles of mechanics and electromagnetism.
1. First, let's determine the net force acting on the particle. We can use the equation F = qE, where F is the force, q is the charge, and E is the electric field strength. Plugging in the values given, we have F = (6.40×10^−19 C) × (7.25×10^4 N/C).
2. Next, we can calculate the acceleration of the particle using Newton's second law of motion, F = ma. Rearranging the equation, we have a = F/m, where m is the mass of the particle. Substituting the known values, the acceleration is a = [(6.40×10^−19 C) × (7.25×10^4 N/C)] / (3.65×10^−24 kg).
3. Now, we can use the kinematic equation v^2 = u^2 + 2ad, where v is the final velocity, u is the initial velocity (which is 0 since the particle starts at rest), a is the acceleration, and d is the distance traveled. Rearranging the equation, we have v^2 = 2ad, since u^2 is zero.
4. Plugging in the known values, we get v^2 = 2 * [(6.40×10^−19 C) × (7.25×10^4 N/C)] / (3.65×10^−24 kg) * 0.102 m.
5. Solving the equation for v^2, we find v^2 ≈ 2.56 × 10^6 m^2/s^2.
6. Finally, taking the square root of both sides, we find v ≈ 1.60 × 10^3 m/s. Therefore, the resulting speed of the particle is approximately 1.60 × 10^3 m/s.
To find the resulting speed of the particle, we can use the equation for the work done on a charged particle by an electric field.
1. First, we need to find the work done on the particle. The work done (W) is given by the equation:
W = q * E * d
where
W is the work done (in joules),
q is the charge of the particle (in coulombs),
E is the electric field strength (in N/C), and
d is the distance through which the particle is accelerated (in meters).
Plugging in the values given in the problem, we have:
W = (6.40×10^−19 C) * (7.25×10^4 N/C) * (0.102 m)
2. Calculate the work done:
W = (6.40×10^−19 C) * (7.25×10^4 N/C) * (0.102 m)
≈ 4.75×10^−17 J
3. The work done on the particle by the electric field is equal to the change in kinetic energy of the particle. Since the particle starts at rest, the initial kinetic energy is zero. Therefore, the work done is equal to the final kinetic energy of the particle:
W = ΔKE = (1/2) * m * v^2
where
ΔKE is the change in kinetic energy (in joules),
m is the mass of the particle (in kilograms), and
v is the final speed of the particle (in m/s).
Plugging in the values given in the problem, we have:
4.75×10^−17 J = (1/2) * (3.65×10^−24 kg) * v^2
4. Rearrange the equation to solve for the final speed (v):
v^2 = (2 * 4.75×10^−17 J) / (3.65×10^−24 kg)
v^2 ≈ 2.60 × 10^7 m^2/s^2
5. Take the square root of both sides to find v:
v ≈ √(2.60 × 10^7 m^2/s^2)
v ≈ 5.10 × 10^3 m/s
Therefore, the resulting speed of the particle is approximately 5.10 × 10^3 m/s.