To understand why the electron emitted from the negative plate of a parallel plate capacitor fails to strike the upper plate under certain conditions, we need to analyze the forces acting on the electron and the motion it undergoes in the presence of both electric and magnetic fields. Let's break this down step by step.
Understanding the Forces on the Electron
When the electron is emitted from the negative plate, it experiences an electric force due to the electric field created by the potential difference \( V \) between the plates. The electric field \( E \) can be calculated using the formula:
E = V/d
Here, \( V \) is the potential difference and \( d \) is the separation between the plates. The force \( F_e \) acting on the electron due to this electric field is given by:
F_e = eE = e(V/d)
where \( e \) is the charge of the electron.
Magnetic Force and Motion
In addition to the electric force, the electron also experiences a magnetic force when it moves through the magnetic field \( B \). The magnetic force \( F_m \) acting on a charged particle moving with velocity \( v \) in a magnetic field is given by:
F_m = evB
Since the electron is emitted with negligible speed, we can assume its initial velocity \( v \) is very small. However, as it moves through the electric field, it will accelerate due to the electric force.
Analyzing the Motion of the Electron
The net force acting on the electron in the vertical direction (towards the upper plate) is the electric force, while the magnetic force acts perpendicular to its motion. The electron will follow a curved path due to the magnetic force, which can be analyzed using the equations of motion.
To determine whether the electron will strike the upper plate, we need to consider the time it takes for the electron to travel the distance \( d \) between the plates. The acceleration \( a \) of the electron due to the electric field is:
a = F_e/m = (eV)/(md)
where \( m \) is the mass of the electron. The time \( t \) it takes to travel the distance \( d \) can be found using the equation of motion:
d = (1/2)at^2
Substituting for \( a \), we get:
d = (1/2)(eV/md)t^2
Condition for Missing the Upper Plate
Now, we need to consider the horizontal motion of the electron due to the magnetic field. The radius of curvature \( r \) of the electron's path in the magnetic field can be derived from the balance of forces:
F_m = F_e
From this, we can derive that the radius of curvature is:
r = mv/(eB)
For the electron to miss the upper plate, the distance it travels horizontally while moving vertically through the plates must be less than the separation \( d \). This gives us the condition:
d > 2r
Substituting for \( r \) and rearranging gives:
d > 2(mv)/(eB)
Now, we can relate the velocity \( v \) to the potential energy gained by the electron as it moves through the electric field:
KE = eV = (1/2)mv^2
From this, we can express \( v \) as:
v = sqrt(2eV/m)
Substituting this back into our condition for missing the upper plate, we find:
d > 2(m * sqrt(2eV/m))/(eB)
After simplifying, we arrive at:
d > [2m * eV/(eB^2)]^(1/2)
Final Thoughts
This inequality shows that if the separation \( d \) between the plates exceeds the derived expression, the electron will not strike the upper plate. The interplay between the electric and magnetic forces, along with the initial conditions of the electron's emission, determines its trajectory. This analysis highlights the fascinating dynamics of charged particles in electric and magnetic fields, which is a fundamental concept in electromagnetism.
