A horizontal beam of electrons initially moving at 4.3 107 m/s is deflected vertically by the vertical electric field between oppositely charged parallel plates as shown in the diagram. The magnitude of the field is 1.90 104 N/C.
What is the vertical deflection d of the electrons as they leave the plates?
To find the vertical deflection (d) of the electrons as they leave the plates, we can use the following equation:
d = (1/2) * a * t^2
Where:
a is the acceleration of the electrons perpendicular to their initial velocity,
t is the time taken by the electrons to pass through the plates.
First, let's find the acceleration (a) of the electrons:
The force acting on the electrons can be calculated using the equation:
F = q * E
Where:
F is the force,
q is the charge of the electron (1.6 * 10^-19 C),
E is the magnitude of the electric field (1.90 * 10^4 N/C).
Substituting the values:
F = (1.6 * 10^-19 C) * (1.90 * 10^4 N/C)
F = 3.04 * 10^-15 N
Next, we can use Newton's second law to find the acceleration:
F = m * a
Where:
m is the mass of the electron (9.11 * 10^-31 kg),
a is the acceleration.
Substituting the values:
3.04 * 10^-15 N = (9.11 * 10^-31 kg) * a
Solving for a:
a = (3.04 * 10^-15 N) / (9.11 * 10^-31 kg)
a = 3.34 * 10^15 m/s^2
Now, we need to find the time (t) taken by the electrons to pass through the plates.
The initial horizontal velocity of the electrons is given as 4.3 * 10^7 m/s.
We can use the equation:
v = u + a * t
Where:
v is the final velocity (0 m/s),
u is the initial velocity (4.3 * 10^7 m/s),
a is the acceleration (3.34 * 10^15 m/s^2),
t is the time.
Substituting the values:
0 = (4.3 * 10^7 m/s) + (3.34 * 10^15 m/s^2) * t
Solving for t:
t = - (4.3 * 10^7 m/s) / (3.34 * 10^15 m/s^2)
t ≈ - 1.29 * 10^-8 s
Since time cannot be negative in this context, we take the absolute value:
t ≈ 1.29 * 10^-8 s
Finally, we can calculate the vertical deflection (d) using the first equation:
d = (1/2) * a * t^2
Substituting the values:
d = (1/2) * (3.34 * 10^15 m/s^2) * (1.29 * 10^-8 s)^2
d ≈ 2.01 * 10^-5 m
Therefore, the vertical deflection of the electrons as they leave the plates is approximately 2.01 * 10^-5 meters.
To find the vertical deflection of the electrons as they leave the plates, we need to use the basic principles of electromagnetism and the equation for the vertical deflection of charged particles in an electric field.
The equation for the vertical deflection of charged particles in an electric field can be derived from Newton's second law. It states that the vertical force exerted on the electrons (Fy) is equal to the charge of the particle (q) multiplied by the vertical electric field (E), and can be represented as Fy = qE.
Given:
Initial velocity of the electrons (v0) = 4.3 × 10^7 m/s
Magnitude of the vertical electric field (E) = 1.90 × 10^4 N/C
Next, we need to calculate the time the electrons spend between the plates. Since the vertical deflection is caused by the vertical acceleration, which is constant, we can use the equation for the horizontal displacement of an object under constant acceleration:
d = v0 * t + (1/2) * a * t^2
Where:
d = vertical deflection (what we want to find)
v0 = initial velocity of the electrons
t = time spent between the plates (what we want to find, assuming the electrons are uniformly accelerated)
a = vertical acceleration
Since the vertical acceleration of the electrons is caused by the vertical electric field, we can rewrite the equation as:
d = v0 * t + (1/2) * (q/m) * E * t^2
Where:
m = mass of the electrons
However, we can also rewrite the equation as:
d = v0 * t + (1/2) * (q/m) * (E/m) * m * t^2
Where:
(E/m) = electric field per unit mass (a constant)
To find the value of (E/m), we can use the fact that the acceleration of the electrons in the electric field is given by:
a = (q/m) * E
Since (E/m) = a / (q/m), we can substitute this in the equation:
d = v0 * t + (1/2) * (q/m) * (E/m) * m * t^2
d = v0 * t + (1/2) * a * (E/m) * m * t^2
d = v0 * t + (1/2) * a * (E/m) * m * t^2
The term (E/m) * m is just the electric field strength E, so we can simplify the equation to:
d = v0 * t + (1/2) * a * E * t^2
Now, we have an equation that relates the vertical deflection (d) to the initial velocity (v0), the acceleration (a), the electric field (E), and the time spent between the plates (t).
Given that the initial velocity of the electrons is 4.3 × 10^7 m/s, the magnitude of the vertical electric field is 1.90 × 10^4 N/C, and assuming the electrons are uniformly accelerated between the plates, we can substitute these values into the equation and solve for the vertical deflection (d) of the electrons as they leave the plates.