a) An electron starts at rest and accelerates through an electric field
established by a set of parallel plates with a potential difference of 35 V.
What is the speed of the electron the instant before it hits the negative plate?
(e = 1.6 × 10^-19 C, melectron = 9.1 × 10^-31 kg)
b) Instead of hitting the negative plate, the electron, travelling East, escapes the
parallel plates through a small hole and enters a magnetic field of 0.75 T
directed downward. What will be the magnetic force (magnitude and
direction) on the charge?
c) Once the electron has entered the magnetic field, it is in circular motion.
What is the radius of the electron’s circular path?
energy in Joules = charge in Coulombs * Potential difference in Volts
= 1.6 * 10^-19 * 35 = 56 * 10^-19 Joules
so
(1/2) m v^2 = 56 * 10^-19
(1/2)(9.1*10^-31) v^2 = 56 * 10^-19
v^2 = 12.3 * 10^12
v = 3.5 * 10^6 m/s (luckily well below speed of light)
F = q V cross B = 1.6*10^-19 *3.5*10^6 * 0.75
V is East . B is down so V cross B is North
m v^2 / R = F
Whoops. q is NEGATIVE so q V cross B is SOUTH
Okay, makes more sense now. Thanks!
a) Well, let's calculate the speed of the electron before it hits the negative plate. We can use the equation v = √(2qV / m), where v is the speed, q is the charge, V is the potential difference, and m is the mass. Plugging in the values, we get v = √(2 * 1.6 × 10^-19 C * 35 V / 9.1 × 10^-31 kg). Math, do your thing!
b) Now let's talk about the magnetic force on the electron. The magnetic force can be calculated using the equation F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field strength. Since the electron is traveling East, and the magnetic field is directed downward, I'm sorry to tell you, but it seems like these two just can't get along. They're perpendicular to each other, so the force will be 0. No magnetic force for this electron, it's just floating around unaffected.
c) Since the electron is now in circular motion, we can use the equation F = mv^2/r to find the radius (r). However, we just established that there is no magnetic force on the electron, so the net force is 0. A force of 0 means the electron is not experiencing any centripetal force, which means that it won't be moving in a circular path. So, in this case, the radius of the electron's circular path would be zero. It's just chilling, no circles for this electron!
a) To find the speed of the electron just before it hits the negative plate, we can use the equation for the kinetic energy of a charged particle in an electric field. The kinetic energy is equal to the work done on the particle by the electric field.
The work done on the electron is equal to the change in its electric potential energy. The potential difference between the plates is given as 35 V, so the change in electric potential energy is:
ΔPE = qΔV
where q is the charge of the electron and ΔV is the potential difference. Plugging in the values, we have:
ΔPE = (1.6 × 10^-19 C)(35 V)
Next, we equate the change in potential energy to the change in kinetic energy:
ΔKE = ΔPE
The change in kinetic energy is equal to the final kinetic energy (K) minus the initial kinetic energy (K0).
ΔKE = K - K0
Since the electron starts at rest, its initial kinetic energy is zero (K0 = 0).
Therefore, we can rewrite our equation as:
K = ΔPE
Now, we can use the equation for the kinetic energy of a particle:
K = (1/2)mv^2
where m is the mass of the electron and v is its velocity (we will assume the electron's mass to be constant).
Setting these two equations equal to each other:
(1/2)mv^2 = ΔPE
Simplifying and solving for v, we get:
v = √(2ΔPE / m)
Substituting the values for ΔPE and m, we can calculate the velocity.
b) To find the magnetic force on the charge, we can use the equation for the magnetic force on a charged particle moving in a magnetic field.
The equation for the magnetic force is given by:
F = qvBsinθ
where F is the magnetic force, q is the charge of the electron, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
In this case, the electron is traveling East, so the angle θ is 90 degrees because the velocity vector is perpendicular to the magnetic field vector.
Therefore, we have:
F = qvBsin90
Since sin90 is equal to 1, we can simplify the equation to:
F = qvB
Plugging in the values for q, v, and B, we can calculate the magnitude of the magnetic force.
c) Once the electron enters the magnetic field, it undergoes circular motion. The force on the electron due to the magnetic field acts as the centripetal force, keeping it in a circular path.
The centripetal force is given by the equation:
F = mω^2r
where F is the centripetal force, m is the mass of the electron, ω is the angular velocity, and r is the radius of the circular path.
Since the magnetic force is acting as the centripetal force, we can equate the two:
qvB = mω^2r
Simplifying and solving for r, we get:
r = (mv)/(qB)
By substituting the values for m, v, q, and B, we can calculate the radius of the electron's circular path.