An electron with velocity v = 2.4 × 106 m/s i encounters an electric field E = 1250 N/C i. How long will it take for the speed to be one fourth its original value?
To find the time it takes for the speed of the electron to be one-fourth its original value, we can use the equations of motion in the presence of a constant electric field.
First, let's break down the given information:
- Velocity of the electron, v = 2.4 × 106 m/s i (where 'i' represents the x-direction).
- Electric field, E = 1250 N/C i.
We know that the motion of an electron in an electric field is governed by the equation:
F = q * E
where F is the force experienced by the electron, q is its charge (in this case, the charge of an electron is -1.6 × 10^-19 C), and E is the electric field.
Since the electric field is given in the x-direction and the velocity of the electron is also in the same direction, we can say that the net force acting on the electron is F = m * (dv/dt), where m is the mass of the electron.
Since the force acting on the electron is due to the electric field:
q * E = m * (dv/dt)
Substituting the values:
(-1.6 × 10^-19 C) * (1250 N/C i) = (mass of electron) * (dv/dt)
We know that mass of electron ≈ 9.11 × 10^-31 kg, so we can solve for (dv/dt):
(2.0 × 10^−16 N) / (9.11 × 10^−31 kg) = (dv/dt)
Simplifying this equation, we find that:
dv/dt ≈ 2.2 × 10^14 m/s^2
Now, we can use the equation of motion relating acceleration, time, and change in velocity:
v = u + (dv/dt) * t
where
- v is the final velocity (one-fourth of the initial velocity),
- u is the initial velocity (given as 2.4 × 10^6 m/s),
- dv/dt is the constant acceleration of the electron,
- t is the time taken.
Rearranging the equation to solve for t:
t = (v - u) / (dv/dt)
Substituting the known values:
t = (1/4 * 2.4 × 10^6 m/s - 2.4 × 10^6 m/s) / (2.2 × 10^14 m/s^2)
Simplifying this equation, we find that:
t ≈ -5.45 × 10^-9 s
Since time cannot be negative, we need to take the absolute value, resulting in:
t ≈ 5.45 × 10^-9 s
Therefore, it will take approximately 5.45 nanoseconds for the speed of the electron to be one-fourth its original value.