In a cathode ray tube, electrons initially at rest are accelerated by a uniform electric field of magnitude 4.71 x 105 N/C during the first 5.49 cm of the tube's length; then they move at essentially constant velocity another 47.9 cm before hitting the screen. Find the speed of the electrons when they hit the screen.
How long does it take them to travel the length of the tube?
Use the same method that is explained by BobPursley here:
http://www.jiskha.com/display.cgi?id=1222244515
(The numbers are different)
kjlk
To find the speed of the electrons when they hit the screen, we can use the concept of acceleration in a uniform electric field.
Initially, the electrons are at rest, so their initial velocity (u) is 0 m/s. The electric field (E) is given as 4.71 x 10^5 N/C. The distance traveled (d) in the first 5.49 cm is 0.0549 m. We can use the equation for motion with constant acceleration:
v^2 = u^2 + 2ad
Where v is the final velocity, u is the initial velocity, a is the acceleration, and d is the distance traveled.
In this case, the acceleration is due to the electric field. The force on the electrons is given by F = qE, where q is the charge of the electron (e) and E is the electric field strength.
From F = ma, we have qE = ma
Since the mass (m) of the electrons cancel out on both sides of the equation, we have a = qE
The acceleration a = qE can be substituted into the equation for motion:
v^2 = u^2 + 2(qE)d
Substituting the values, we have:
v^2 = 0 + 2(1.6 x 10^-19 C)(4.71 x 10^5 N/C)(0.0549 m)
Simplifying, we find:
v^2 = 0.09
Taking the square root of both sides, we get:
v = 0.3 m/s
So, the speed of the electrons when they hit the screen is 0.3 m/s.
To find the time it takes for the electrons to travel the length of the tube, we need to calculate the time taken for both the accelerated and constant velocity motion.
For the accelerated motion, we use the equation:
v = u + at
Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity u is 0 m/s, the final velocity v is the answer we found earlier (0.3 m/s), the acceleration a is given by qE, the electric field E is given as 4.71 x 10^5 N/C, and the distance traveled d is 0.0549 m.
Substituting the values, we have:
0.3 = 0 + (1.6 x 10^-19 C)(4.71 x 10^5 N/C)t
Simplifying, we find:
t = 3.19 x 10^-15 s
For the motion at constant velocity, we know the distance traveled is 47.9 cm = 0.479 m.
We can use the equation:
v = d/t
Where v is the velocity, d is the distance traveled, and t is the time.
Substituting the values, we have:
0.3 = 0.479/t
Solving for t, we find:
t = 1.597 s
Therefore, the total time taken for the electrons to travel the length of the tube is the sum of the times for the accelerated motion and constant velocity motion:
Total time = 3.19 x 10^-15 s + 1.597 s
Total time = 1.597 s (approximately)
So, it takes approximately 1.597 seconds for the electrons to travel the length of the tube.