An electron is projected into a uniform electric field E = (1000 N/C)i with an initial velocity (2.00* 10^6m/s)i in the direction of the field. How far the electron travel before it is brought momentarily to rest?
retarding force = E e = m a where m is electron mass and e is electron charge so
a = Ee/m
constant Force --- > constant de acceleration, just like gravity
if you throw something up with speed Vi
F = -m g = - m a
v = Vi - g t, when v = 0, t = Vi/g
x = Vi t - (1/2) g t^2
x = Vi^2/g - (1/2) g Vi^2/g^2 = (1/2) Vi^2/g
here instead of g we have for acceleration E e/m
so
x = (1/2) Vi^2 [ m / Ee ]
the field is horizontal
To find the distance the electron travels before it is brought momentarily to rest, we need to calculate the time it takes for the electron to reach zero velocity.
We can use the equation of motion: v = u + at, where:
- v is the final velocity (zero in this case)
- u is the initial velocity (2.00 * 10^6 m/s)
- a is the acceleration
- t is the time
In this case, the electron experiences a force due to the electric field, causing it to decelerate until it comes to rest.
The force experienced by the electron is given by the equation: F = qE, where:
- F is the force experienced by the electron
- q is the charge of the electron (1.6 * 10^(-19) C)
- E is the electric field (1000 N/C)
Since the force is equal to ma (where m is the mass of the electron), we can rewrite the equation as: ma = qE.
The acceleration (a) can be rearranged from the equation as: a = qE/m.
Substituting the values, we get: a = (1.6 * 10^(-19) C)(1000 N/C) / (9.11 * 10^(-31) kg).
Now, we can substitute the values of a and u into the equation v = u + at and solve for t:
0 = (2.00 * 10^6 m/s) + [(1.6 * 10^(-19) C)(1000 N/C) / (9.11 * 10^(-31) kg)] * t
Solving for t, we get: t = - (2.00 * 10^6 m/s) / [(1.6 * 10^(-19) C)(1000 N/C) / (9.11 * 10^(-31) kg)]
Calculating this expression, t = - (2.00 * 10^6 m/s) / [(1.6 * 10^(-19) C)(1000 N/C) / (9.11 * 10^(-31) kg)] = -11.19 x 10^(-12) s.
Since we are interested in the distance traveled, we can use the equation: s = ut + (1/2)at^2.
Substituting the values, we get: s = (2.00 * 10^6 m/s) * (-11.19 x 10^(-12) s) + (1/2) * [(1.6 * 10^(-19) C)(1000 N/C) / (9.11 * 10^(-31) kg)] * (-11.19 x 10^(-12) s)^2
Calculating this expression, s = -1.24 x 10^(-4) m.
Therefore, the electron travels approximately -1.24 x 10^(-4) meters before it is brought momentarily to rest. Note that the negative sign indicates the direction of the electron's motion is opposite to the direction of the electric field.
To find the distance traveled by the electron before it is brought momentarily to rest, we need to determine the deceleration of the electron and then use the kinematic equation for distance.
The electric force experienced by the electron can be calculated using the equation F = qE, where F is the force, q is the charge of the electron, and E is the electric field strength. In this case, the charge of an electron is -1.6 × 10^-19 C (coulombs), and the electric field strength is given as E = (1000 N/C)i.
So, substituting these values into the equation, we have:
F = qE
F = (-1.6 × 10^-19 C)(1000 N/C)i
F = (-1.6 × 10^-16 N)i
Since force is related to acceleration through Newton's second law (F = ma), and in this case, the acceleration is in the opposite direction of the initial velocity, we can write:
ma = (-1.6 × 10^-16 N)i
Dividing both sides of the equation by the electron mass (m = 9.1 × 10^-31 kg), we get:
a = (-1.6 × 10^-16 N)i / (9.1 × 10^-31 kg)
a = (-1.76 × 10^14 m/s^2)i
Now, since the electron is initially moving in the direction of the electric field, the acceleration will act against the direction of motion. So, the acceleration vector can be represented as a = (-1.76 × 10^14 m/s^2)i.
To find the distance traveled, we can use the kinematic equation for distance:
d = (v^2 - u^2) / (2a)
where d is the distance, v is the final velocity, u is the initial velocity, and a is the acceleration.
Since the final velocity is zero (because the electron is brought momentarily to rest), the equation becomes:
d = -u^2 / (2a)
d = -[(2.00 × 10^6 m/s)^2] / [2(-1.76 × 10^14 m/s^2)]
Evaluating the equation, we find:
d ≈ 2.27 × 10^-10 meters
Therefore, the electron travels approximately 2.27 x 10^-10 meters before it is brought momentarily to rest.