an electron with a speed of 5.00 x 10^8 cm/s enters an electric field of magnitude 1.00 x 10^3 N/C, traveling along field lines in the direction that s its motion.
a) How far will the electron travel in the field before stopping momentarily
b) how much time will have elapsed
c) in the region with the E field is only 8.00 mm long, what fraction of the electron's initial KE will be lost in that region?
F=ma
F=eE
ma = eE
a=eE/m= 1.6•10⁻¹⁹•10³/9.1•10⁻³¹= =1.76•10¹⁴ m/s²
s=v₀²/2a = v₀²m/2eE=
=(5•10⁶)²•9.1•10⁻³¹/2•1.6•10⁻¹⁹•10³= =0.071 m
v=v₀-at,
v=0,
t= v₀/a=5•10⁶/1.76•10¹⁴=2.84•10⁻ ⁸ s.
s₁=(v₀²-v₁²)/2a,
v₁² = v₀²-2s₁a.
KE₁/KE₀=(m v₁²/2)/ (m v₀²/2) =
= (v₁²/ v₀)² =( v₀²-2s₁a)/ v₀²=
=1-(2s₁a/ v₀²)= 1- (2•8•10⁻³•1.76•10¹⁴/25•10¹⁶)=
= 1-1.13•10⁻⁵=0.99999.
To solve these problems, we need to use the equations of motion for an object under constant acceleration, along with the given values for speed, electric field, and distance.
Given:
Speed of the electron, v = 5.00 x 10^8 cm/s = 5.00 x 10^6 m/s (since 1 m = 100 cm)
Electric field magnitude, E = 1.00 x 10^3 N/C
Length of the region with the electric field, d = 8.00 mm = 8.00 x 10^-3 m
a) To calculate the distance the electron will travel in the field before stopping momentarily, we need to find the acceleration of the electron first. The electric force experienced by the electron is given by F = qE, where q is the charge of the electron (1.60 x 10^-19 C).
Using Newton's second law, F = ma, where m is the mass of the electron (9.11 x 10^-31 kg), we can equate the electric force to the product of mass and acceleration to solve for acceleration:
qE = ma
a = qE/m
a = (1.60 x 10^-19 C)(1.00 x 10^3 N/C) / (9.11 x 10^-31 kg)
a = 17.55 x 10^11 m/s^2
Now, we can use the equation of motion to find the stopping distance of the electron:
v^2 = u^2 + 2as
Since the electron starts from rest (u = 0), and we want to find the distance traveled (s), we rearrange the equation to solve for s:
s = v^2 / (2a)
s = (5.00 x 10^6 m/s)^2 / (2 × 17.55 x 10^11 m/s^2)
s = 7.16 x 10^-6 m
Therefore, the electron will travel approximately 7.16 x 10^-6 m in the field before stopping momentarily.
b) To calculate the time elapsed, we can use the equation of motion: v = u + at.
Since the electron starts from rest (u = 0), and we know the acceleration (a = 17.55 x 10^11 m/s^2), we can solve for t:
t = v / a
t = (5.00 x 10^6 m/s) / (17.55 x 10^11 m/s^2)
t = 2.85 x 10^-6 s
Therefore, approximately 2.85 x 10^-6 seconds will elapse.
c) To find the fraction of the electron's initial kinetic energy (KE) lost in the 8.00 mm region, we can use conservation of energy. Initially, the electron has only kinetic energy, given by KE = (1/2)mv^2.
The work done on the electron by the electric field is given by W = QEd, where Q is the charge of the electron. Since the electron is negatively charged, the work done by the electric field is negative (i.e., it reduces the kinetic energy).
The fraction of initial kinetic energy lost is:
Fraction lost = |W| / KE
Fraction lost = |-QEd| / KE
Fraction lost = |-(-1.60 x 10^-19 C)(1.00 x 10^3 N/C)(8.00 x 10^-3 m)| / [(1/2)(9.11 x 10^-31 kg)(5.00 x 10^6 m/s)^2]
Fraction lost = |1.28 x 10^-25 J| / 2.28 x 10^-12 J
Therefore, the fraction of the electron's initial kinetic energy lost in the 8.00 mm region is approximately 5.614 x 10^-14.
To find the answers to these questions, we need to use kinematic equations and principles of electric fields. Let's break down each question and explain step-by-step how to find the answers:
a) How far will the electron travel in the field before stopping momentarily?
To determine how far the electron will travel before stopping, we need to find the deceleration and use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity (0 cm/s since the electron stops), u is the initial velocity (5.00 x 10^8 cm/s), a is the acceleration (opposite of the electric field magnitude), and s is the distance traveled.
Given:
Initial velocity, u = 5.00 x 10^8 cm/s
Electric field magnitude, E = 1.00 x 10^3 N/C
The electric field exerts a force on the electron given by the equation: F = q * E, where F is the force, q is the charge, and E is the electric field magnitude. The charge of an electron is -e, where e is the elementary charge (-1.6 x 10^-19 C). Therefore, the force exerted on the electron is: F = (-1.6 x 10^-19 C) * (1.00 x 10^3 N/C).
Since force is equal to mass times acceleration (F = m * a), we can rearrange the equation to find the acceleration, a = (F / m).
The mass of an electron, m = 9.11 x 10^-31 kg.
Using the found value of acceleration, we can substitute it into the equation of motion: v^2 = u^2 + 2as. Rearrange the equation to solve for s:
s = (v^2 - u^2) / (2a)
s = (0 - (5.00 x 10^8 cm/s)^2) / (2 * a)
Now substitute the calculated acceleration to find the distance traveled, s.
b) How much time will have elapsed?
To find the time elapsed, we can use the formula t = (v - u) / a, where t is the time, v is the final velocity (0 cm/s), u is the initial velocity (5.00 x 10^8 cm/s), and a is the acceleration (opposite of the electric field magnitude).
c) In the region where the electric field is only 8.00 mm long, what fraction of the electron's initial kinetic energy will be lost?
To find the fraction of the electron's initial kinetic energy lost, we need to calculate the initial and final kinetic energies. The initial kinetic energy, KE_initial = 0.5 * m * u^2, and the final kinetic energy, KE_final = 0.5 * m * v^2, where m is the mass, u is the initial velocity, and v is the final velocity.
The fraction of kinetic energy lost can be defined as (KE_initial - KE_final) / KE_initial. Substitute the known values to find the fraction.