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 = 7.25×104 N/C. What is the resulting speed of the particle?

To find the resulting speed of the particle, we can use the equations of motion for a particle under constant acceleration. In this case, the electric field provides the constant acceleration.

The equation that relates the final velocity (vf) to the initial velocity (vi), acceleration (a), and displacement (d) is:

vf^2 = vi^2 + 2ad

Here, the initial velocity (vi) is zero since the particle starts at rest. The acceleration (a) can be calculated using the equation:

a = qE / m

where q is the charge of the particle and E is the electric field strength. Finally, the displacement (d) is given as 0.102 m.

Let's substitute the given values into the equations:

a = (6.40×10^(-19) C * 7.25×10^4 N/C) / (3.65×10^(-24) kg)
a = (4.64×10^(-14) C.N) / (3.65×10^(-24) kg)
a ≈ 1.27×10^10 m/s^2

Substituting these values into the equation for final velocity:

vf^2 = (0 m/s)^2 + 2 * (1.27×10^10 m/s^2) * (0.102 m)
vf^2 = 2.6×10^9 m^2/s^2

Taking the square root of both sides:

vf ≈ √(2.6×10^9 m^2/s^2)
vf ≈ 5.1×10^4 m/s

Therefore, the resulting speed of the particle is approximately 5.1×10^4 m/s.