An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is σ = 1.67 10-7 C/m2, and the plate separation is 1.35 10-2 m. How fast is the electron moving just before it reaches the positive plate?
Answer in m/s
To find the speed of the electron just before it reaches the positive plate, we can use the principles of electric potential energy and conservation of energy.
The electric potential energy (PE) of a charged particle in an electric field is given by the formula:
PE = qV
where q is the charge of the particle and V is the electric potential difference (voltage) between the two plates of the capacitor.
Initially, the electron is at rest at the negative plate, so its initial kinetic energy (KE) is zero. As the electron moves towards the positive plate, the increasing electric potential energy is converted into kinetic energy.
According to the conservation of energy principle, the total mechanical energy of the electron is conserved. Therefore, the electric potential energy gained by the electron must be equal to the kinetic energy it possesses just before reaching the positive plate.
PE = KE
Now, let's calculate the electric potential energy gained by the electron:
PE = qV = eDV
where e is the charge of the electron (1.6 x 10^-19 C) and DV is the change in voltage between the plates.
The voltage (V) between the plates can be calculated using the formula:
V = Ed
where E is the electric field strength and d is the plate separation.
The electric field strength is given by:
E = σ/ε0
where σ is the charge per unit area on each plate and ε0 is the permittivity of free space (8.854 x 10^-12 F/m).
Now, substituting the values, we get:
E = (1.67 x 10^-7 C/m^2) / (8.854 x 10^-12 F/m)
E = 1.884 x 10^4 N/C
Next, we can calculate the voltage V:
V = E x d = (1.884 x 10^4 N/C) x (1.35 x 10^-2 m)
V = 2.544 x 10^2 V
Now, substitute the values for q (1.6 x 10^-19 C) and V (2.544 x 10^2 V) into the equation for PE:
PE = (1.6 x 10^-19 C) x (2.544 x 10^2 V)
PE = 4.070 x 10^-17 J
Since PE = KE, the kinetic energy of the electron just before reaching the positive plate is also 4.070 x 10^-17 J.
Now, to find the speed of the electron, we can use the equation:
KE = (1/2)mv^2
where m is the mass of the electron (9.11 x 10^-31 kg) and v is the velocity of the electron. Rearranging the equation, we have:
v^2 = (2KE) / m
v = √[(2KE) / m]
Substituting the values, we get:
v = √[(2 x 4.070 x 10^-17 J) / (9.11 x 10^-31 kg)]
v = 1.88 x 10^7 m/s
Therefore, the electron is moving at a speed of approximately 1.88 x 10^7 m/s just before it reaches the positive plate.