An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.6 x 10-7 C/m2, and the plates are separated by a distance of 1.4 x 10-2 m. How fast is the electron moving just before it reaches the positive plate?
To calculate the speed of the electron just before it reaches the positive plate, we can use the principles of electrostatics and kinetic energy.
The electric field between the parallel plates can be calculated using the formula:
E = V / d
Where:
E is the electric field (in N/C)
V is the potential difference between the plates (in volts)
d is the distance between the plates (in meters)
In this case, the potential difference can be determined using the formula:
V = Ed
Where:
V is the potential difference (in volts)
E is the electric field (in N/C)
d is the distance between the plates (in meters)
The electric field between the plates can be calculated using the formula:
E = σ / ε0
Where:
E is the electric field (in N/C)
σ is the charge per unit area on each plate (in C/m^2)
ε0 is the permittivity of free space (approximated as 8.85 x 10^-12 C^2/Nm^2)
Substituting the values given:
E = (2.6 x 10^-7 C/m^2) / (8.85 x 10^-12 C^2/Nm^2)
E ≈ 2.93 x 10^4 N/C
Now, substituting the value of E into the formula for potential difference:
V = (2.93 x 10^4 N/C) * (1.4 x 10^-2 m)
V ≈ 4.10 x 10^2 V
Next, we need to find the kinetic energy of the electron just before reaching the positive plate. The kinetic energy can be calculated using the formula:
KE = (1/2) * m * v^2
Where:
KE is the kinetic energy (in joules)
m is the mass of the electron (in kilograms)
v is the velocity of the electron (in meters per second)
The mass of an electron is approximately 9.11 x 10^-31 kg.
Since the electron is released from rest, its initial velocity is 0 m/s. Therefore, the kinetic energy just before reaching the positive plate is equal to the potential energy:
KE = V
Finally, we can solve for the velocity of the electron using the calculated potential energy:
V = (1/2) * m * v^2
(4.10 x 10^2 V) = (1/2) * (9.11 x 10^-31 kg) * v^2
Solving for v^2:
v^2 = (2 * 4.10 x 10^2 V) / (9.11 x 10^-31 kg)
v ≈ 8.38 x 10^6 m/s
Therefore, the electron is moving approximately 8.38 x 10^6 meters per second just before it reaches the positive plate.