The plates of a parallel plate capacitor are separated by a distance of 1.2cm, and the electric field within the capacitor has a magnitude of 2.1x 10^6 V/m. An electron starts from rest at the negative plate and accelerates to the positive plate. What is the kinetic energy of the electron just as the electron reaches the positive plate?
The field strength E multiplied by the plate separation equals the potential energy change, per Coulomb. That would be 25,200 J/C.
Set that equal to (1/2)(m/e) V^2, the kinetic energy per Coulomb, and solve for V.
m and e are the electron mass and charge.
To find the kinetic energy of the electron just as it reaches the positive plate, we need to determine the potential difference (also known as the voltage) across the capacitor. Once we have the voltage, we can use the equation for kinetic energy to calculate the answer.
First, let's find the potential difference across the capacitor using the given electric field and plate separation distance.
The electric field between the plates of a parallel plate capacitor is given by the formula:
E = V/d
where E is the electric field, V is the potential difference, and d is the plate separation distance.
Rearranging the formula, we have:
V = E * d
Substituting the given values:
V = (2.1x10^6 V/m) * (1.2 cm) = (2.1x10^6 V/m) * (0.012 m) = 25,200 V
Now that we know the potential difference across the capacitor, we can determine the kinetic energy of the electron.
The kinetic energy of an object is given by the equation:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.
In this case, the electron starts from rest (v = 0) at the negative plate and accelerates to the positive plate. As it accelerates, it gains kinetic energy.
The mass of an electron is approximately 9.1x10^-31 kg.
Therefore, substituting the values into the equation:
KE = (1/2) * (9.1x10^-31 kg) * (v^2)
Since the electron starts from rest, its initial velocity is zero. As it travels through the electric field, it accelerates and gains velocity. At the positive plate, its final velocity can be determined using the formula for uniformly accelerated motion:
v^2 = u^2 + 2a * s
Where u is the initial velocity (zero in this case), a is the acceleration, and s is the distance traveled.
The acceleration experienced by the electron can be calculated using the formula:
a = E * e
where a is the acceleration, E is the electric field, and e is the elementary charge (1.6x10^-19 C).
Substituting the given electric field value:
a = (2.1x10^6 V/m) * (1.6x10^-19 C) = 3.36x10^-13 m/s^2
Since the electron starts from rest, u = 0.
Now, we need to find the distance traveled by the electron, s.
The distance traveled by the electron can be found using the plate separation distance:
s = d = 0.012 m
Now we can substitute the values into the equation for uniformly accelerated motion:
v^2 = u^2 + 2a * s
v^2 = 0 + 2 * (3.36x10^-13 m/s^2) * (0.012 m)
v^2 = 8.064x10^-15 m^2/s^2
Taking the square root of both sides:
v ≈ 2.841x10^-7 m/s
Finally, we can substitute the value of v into the equation for kinetic energy:
KE = (1/2) * (9.1x10^-31 kg) * (v^2)
KE = (1/2) * (9.1x10^-31 kg) * ((2.841x10^-7 m/s)^2)
KE ≈ 6.85x10^-17 J
Therefore, the kinetic energy of the electron just as it reaches the positive plate is approximately 6.85x10^-17 Joules.