An electron with initial speed = 2.54×10^7 m/s is traveling parallel to an electric field of magnitude = 1.66×10^4 N/C .
How far will the electron travel before it stops?
How much time will elapse before it returns to its starting point?
find its original KE.
then, work done=force*distance
where force=Eq
set them equal, solve for distance
To find the distance the electron will travel before it stops, we can use the formula for the deceleration due to electric force:
a = qE/m
where:
- a is the acceleration
- q is the charge of the electron
- E is the magnitude of the electric field
- m is the mass of the electron
The initial velocity, v, is given as 2.54×10^7 m/s, and the magnitude of the electric field, E, is given as 1.66×10^4 N/C. The charge of the electron, q, is 1.6×10^-19 C, and the mass of the electron, m, is 9.1×10^-31 kg.
First, let's calculate the acceleration:
a = (1.6×10^-19 C)(1.66×10^4 N/C) / (9.1×10^-31 kg)
a = 2.92×10^14 m/s^2
To find the distance the electron will travel until it stops, we can use the equation of motion:
v^2 = u^2 + 2as
where:
- u is the initial velocity
- v is the final velocity (which is 0 in this case since the electron stops)
- a is the acceleration
- s is the distance traveled
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (0 - (2.54×10^7 m/s)^2) / (2(2.92×10^14 m/s^2))
s = -6.50×10^-13 m
Since the distance cannot be negative, the electron will travel approximately 6.50×10^-13 m before it stops.
Now, let's calculate the time required for the electron to return to its starting point.
The time required for the electron to return to its starting point can be calculated using the equation of motion:
s = ut + (1/2)at^2
where:
- s is the distance traveled
- u is the initial velocity
- a is the acceleration
- t is the time
In this case, the distance traveled, s, is equal to 0 since the electron returns to its starting point. The initial velocity, u, is 2.54×10^7 m/s, and the acceleration, a, is -2.92×10^14 m/s^2 (negative because it acts in the opposite direction of the initial velocity).
0 = (2.54×10^7 m/s)t + (1/2)(-2.92×10^14 m/s^2)t^2
Rearranging and factoring:
(1/2)(-2.92×10^14 m/s^2)t^2 + (2.54×10^7 m/s)t = 0
Simplifying:
-1.46×10^14 m/s^2t^2 + 2.54×10^7 m/s t = 0
t(-1.46×10^14 m/s^2 t + 2.54×10^7 m/s) = 0
From this equation, we can see that there are two possible solutions for t:
1) t = 0 (which is the starting point)
2) -1.46×10^14 m/s^2 t + 2.54×10^7 m/s = 0
Solving for t in the second equation:
-1.46×10^14 m/s^2 t + 2.54×10^7 m/s = 0
t = (2.54×10^7 m/s) / (1.46×10^14 m/s^2)
t ≈ 1.739×10^-7 s
Therefore, it will take approximately 1.739×10^-7 seconds for the electron to return to its starting point.
To find the distance the electron will travel before it stops, we need to determine the acceleration of the electron. The force experienced by a charged particle in an electric field is given by the equation:
F = q * E
where F is the force, q is the charge of the particle, and E is the electric field strength. In this case, the charge of an electron is -1.6 × 10^-19 C, and the electric field strength is 1.66 × 10^4 N/C.
Therefore, the force experienced by the electron is:
F = (-1.6 × 10^-19 C) * (1.66 × 10^4 N/C)
= -2.656 × 10^-15 N
Since the force acts in the opposite direction to the motion of the electron, it will cause deceleration. The acceleration (a) is given by the equation:
F = m * a
where m is the mass of the electron, which is approximately 9.11 × 10^-31 kg. Rearranging the equation, we can solve for the acceleration:
a = F / m
= (-2.656 × 10^-15 N) / (9.11 × 10^-31 kg)
= -2.91 × 10^15 m/s^2
Since the acceleration is negative, the electron will decelerate. The final velocity (v) of the electron can be calculated using the equation of motion:
v^2 = u^2 + 2a * s
where u is the initial velocity, v is the final velocity, a is the acceleration, and s is the distance traveled. We want to find the distance traveled, so we rearrange the equation:
s = (v^2 - u^2) / (2a)
= (0 - (2.54 × 10^7 m/s)^2) / (2 * -2.91 × 10^15 m/s^2)
= 2.775 × 10^-3 m
Therefore, the electron will travel a distance of approximately 2.775 × 10^-3 meters before it stops.
To calculate the time it takes for the electron to return to its starting point, we need to find the time it takes for the electron to decelerate and come to a stop. We can use the equation of motion:
v = u + at
where u is the initial velocity, v is the final velocity (in this case, zero), a is the acceleration, and t is the time taken. Rearranging the equation, we can solve for t:
t = (v - u) / a
= (0 - 2.54 × 10^7 m/s) / (-2.91 × 10^15 m/s^2)
= 8.72 × 10^-9 s
Therefore, the time taken for the electron to return to its starting point is approximately 8.72 × 10^-9 seconds.