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.