An electron has an initial velocity of 2.00 ✕ 10 6 m/s in a uniform 6.00 ✕ 10 5 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity.
(a) What is the direction of the electric field?
(b) How far does the electron travel before coming to rest?
m
(c) How long does it take the electron to come to rest?
s
(d) What is the electron's speed when it returns to its starting point?
m/s
To answer these questions, we can use the equations of motion for a uniformly accelerated particle under the influence of an electric field. Here's how we can do it step by step:
(a) To determine the direction of the electric field, we need to consider that the electric field accelerates the electron in the opposite direction to its initial velocity. Since the initial velocity of the electron is in the positive direction, the electric field must be in the negative direction.
(b) To find the distance traveled by the electron before coming to rest, we can use the equation:
\[v^2 = u^2 + 2as\]
where:
v = final velocity (0 m/s, since the electron comes to rest)
u = initial velocity (2.00 ✕ 10^6 m/s)
a = acceleration (due to the electric field, which is -6.00 ✕ 10^5 N/C)
s = distance traveled
Substituting the given values into the equation, we can solve for s:
\[0^2 = (2.00 ✕ 10^6)^2 + 2(-6.00 ✕ 10^5)s\]
Simplifying the equation and solving for s, we get:
\[s = \frac{(2.00 ✕ 10^6)^2}{2(6.00 ✕ 10^5)}\]
(c) To calculate the time it takes for the electron to come to rest, we can use the equation:
\[v = u + at\]
where t is the time taken to come to rest. Since the final velocity is 0 m/s and the initial velocity is 2.00 ✕ 10^6 m/s, and the acceleration is -6.00 ✕ 10^5 N/C (opposite direction to the initial velocity), we can solve for t:
\[0 = (2.00 ✕ 10^6) + (-6.00 ✕ 10^5)t\]
Simplifying the equation and solving for t, we get:
\[t = \frac{(2.00 ✕ 10^6)}{(6.00 ✕ 10^5)}\]
(d) To find the speed of the electron when it returns to its starting point, we use the equation:
\[v^2 = u^2 + 2as\]
where:
v = final velocity (unknown)
u = initial velocity (2.00 ✕ 10^6 m/s)
a = acceleration (opposite to the initial velocity, -6.00 ✕ 10^5 N/C)
s = displacement (in this case, the distance traveled in part (b), which we calculated as s)
Substituting the values into the equation, we can solve for v:
\[v^2 = (2.00 ✕ 10^6)^2 + 2(-6.00 ✕ 10^5)s\]
Solving the equation for v, we get:
\[v = \sqrt{(2.00 ✕ 10^6)^2 + 2(-6.00 ✕ 10^5)s}\]
This equation will give the speed of the electron when it returns to its starting point.