A parallel plate capacitor is made of two 3.0 cm x 3.0 cm square plates that are separated by 1.0 mm are connected to a12.0 V battery. If and electron is released from the negative electrode, what speed will it reach by the time it arrives at the positive electrode?
A parallel plate capacitor has a plate area of 0.64cm^2. When the plates are in a vacuum, the capacitance of the capacitor is 4.9 pF. a) Calculate the value of the capacitance if the space between plates is filled with nylon? k of nylon is 3.4 b) What is the max potential difference that can be applied to the plates without causing dielectric breakdown?
To find the speed at which the electron will reach the positive electrode of the parallel plate capacitor, we can use the principles of electric potential energy and kinetic energy.
The potential energy gained by the electron when it moves from the negative to the positive electrode can be calculated using the equation:
U = qV
where U is the potential energy, q is the charge of the electron, and V is the potential difference across the capacitor plates.
We know the charge of an electron is 1.6 x 10^-19 coulombs and V is 12.0 volts. Therefore, the potential energy gained by the electron is:
U = (1.6 x 10^-19 C) x (12.0 V)
U ≈ 1.92 x 10^-18 joules
The potential energy gained by the electron is converted into kinetic energy. The kinetic energy can be calculated using the equation:
K.E. = (1/2)mv^2
where K.E. is the kinetic energy, m is the mass of the electron, and v is the speed of the electron.
The mass of an electron is approximately 9.11 x 10^-31 kg.
Equating the potential energy gained by the electron to the kinetic energy it possesses when it reaches the positive electrode:
(1/2)mv^2 = 1.92 x 10^-18 J
Solving for v:
v^2 = (2 * 1.92 x 10^-18 J) / (9.11 x 10^-31 kg)
v ≈ √(4.20 x 10^12 m^2/s^2)
v ≈ 2.05 x 10^6 m/s
Therefore, the speed at which the electron will reach the positive electrode of the parallel plate capacitor is approximately 2.05 x 10^6 m/s.