Two parallel plates are charged to a potential difference of 10.0 volts. Determine the maximum speed of an electron that starts from rest at the negative plate and travels towards the positive plate.
kinetic energy gained = Qelectron * 10 volts
(1/2)(mass of electron) v^2 = that kinetic energy you just found
Thank you Damon! So I equal the KE formula to (-1.6 x 10^-19)(10) which is (-1.6 x 10^-18)?
yes
To determine the maximum speed of an electron traveling between two parallel plates that are charged with a potential difference, we can make use of the conservation of energy.
The electric potential difference between the two plates can be represented by ΔV = 10.0 volts.
Let's assume that the distance between the two plates is d. The formula for the electric potential difference is ΔV = Ed, where E is the electric field strength between the plates.
Due to symmetry, the electric field strength between two parallel plates is constant, so we can write:
E = ΔV / d = 10.0 volts / d
Now, let's use the electric field to determine the acceleration of the electron. The force experienced by an electron in an electric field is given by the equation F = qE, where q is the charge of the electron.
The acceleration experienced by the electron is given by Newton's second law, F = ma, so we can write:
qE = ma
The charge of an electron is q = -e, where e is the elementary charge.
Therefore, -eE = ma
Since the mass of an electron is m = 9.11 x 10^-31 kg, substituting these values into the equation, we have:
- (1.6 x 10^-19 C) * (10.0 volts / d) = (9.11 x 10^-31 kg) * a
Simplifying the equation, we find:
a = 1.76 x 10^11 m/s^2 / d
Now, we can use the equations of motion to determine the maximum speed of the electron.
The equation for velocity in terms of acceleration and displacement is:
v^2 = u^2 + 2ad
The initial velocity of the electron, u, is 0 m/s since it starts from rest.
The displacement, d, is the distance between the two plates.
Now, rearranging the equation, we get:
v^2 = 0^2 + 2 * (1.76 x 10^11 m/s^2 / d) * d
Simplifying further, we find:
v^2 = 3.52 x 10^11 m^2/s^2
Taking the square root of both sides:
v = 5.93 x 10^5 m/s
Hence, the maximum speed of the electron is approximately 5.93 x 10^5 m/s.