An electron traveling at 7.00 x 10^6 m/s enters and passes through a parallel plate capacitor. determine the electric field generated by the capacitor. note that the electron just clears the corner of the positive plate. The distance of the parallel plate capacitor is 2.0 cm and the height it 1.5mm

PLEASE SHOW ALL WORK AND PLUG IN NUMBERS!!!

you show all the work, and plug in numbers, and I will check. I outlined this in detail already. I am not going to make a cut and paste answer for you. That is cheating.

To determine the electric field generated by the capacitor, we can use the formula:

E = V / d

where:
E is the electric field,
V is the voltage across the capacitor, and
d is the distance between the plates.

First, we need to find the voltage across the capacitor. Since the electron just clears the corner of the positive plate, it means that the potential energy gained by the electron is equal to the charge of the electron times the voltage difference between the plates.

The potential energy gained by the electron can be calculated using the formula:

PE = q * V

where:
PE is the potential energy,
q is the charge of the electron (1.6 x 10^-19 Coulombs), and
V is the voltage across the capacitor.

Since the electron gains kinetic energy from its motion, we can equate the potential energy gained to the kinetic energy of the electron, given by the equation:

KE = (1/2) * m * v^2

where:
KE is the kinetic energy,
m is the mass of the electron (9.11 x 10^-31 kg), and
v is the velocity of the electron (7.00 x 10^6 m/s).

Setting the potential energy equal to the kinetic energy, we have:

q * V = (1/2) * m * v^2

Now, rearranging the equation to solve for V:

V = [(1/2) * m * v^2] / q

Plugging in the known values:

V = [(1/2) * (9.11 x 10^-31 kg) * (7.00 x 10^6 m/s)^2] / (1.6 x 10^-19 Coulombs)

Calculating V gives:

V ≈ 180 Volts

Now that we have the voltage across the capacitor, we can calculate the electric field using the formula mentioned earlier:

E = V / d

Plugging in the remaining known values:

E = (180 V) / (0.02 m)

Calculating E gives:

E = 9000 N/C

Therefore, the electric field generated by the capacitor is 9000 N/C.