Two 2.3-cm-diameter-disks spaced 1.8 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.6×105 V/m.

A. What is the voltage across the capacitor = 830 V (solved)

B. How much charge is on each disk? (Find Q1 and Q2)

C. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.3×107 m/s . What was the electron's speed as it left the negative plate?

I just need B and C. Thank you!!

V= (1/eo) Q d / A

V = 828 I get
eo = 8.85*10^-12
d = 1.8*10^-3
A = pi r^2 = pi (1.15*10^-2)^2

Ke leaving plate = (1/2) m v^2
Ke gained = electron charge * Voltage
total Ke= (1/2) m (10^7)^2

B. How much charge is on each disk? (Find Q1 and Q2)

Well, it seems like these disks are quite the charged individuals! To find the charge on each disk, we can use the formula Q = CV, where Q represents the charge, C represents the capacitance, and V represents the voltage.

First, let's find the capacitance. The formula for the capacitance of a parallel-plate capacitor is C = ε₀A/d, where C represents the capacitance, ε₀ represents the permittivity of free space, A represents the area of the plates, and d represents the distance between the plates.

Given that the diameter of each disk is 2.3 cm, we can find the radius by dividing the diameter by 2. So the radius, r, of each disk is 1.15 cm.

Now, we can find the area of each plate by using the formula A = πr^2. Substituting the value of r into the formula, we get A = π(1.15 cm)^2.

The permittivity of free space, ε₀, is a constant value of approximately 8.85 x 10^-12 F/m.

The distance between the plates, d, is given as 1.8 mm, which can be converted to meters by dividing by 1000. So, d = 1.8 mm / 1000 = 0.0018 m.

Now we can find the capacitance using the formula C = ε₀A/d. Substituting the values we have, we get C = (8.85 x 10^-12 F/m) * (π(1.15 cm)^2) / 0.0018 m.

Once you have found the capacitance, you can then use the formula Q = CV to calculate the charge on each disk.

B. To find the charge on each disk, we can use the formula for the capacitance of a parallel-plate capacitor:

C = ε₀A/d

Where:
C is the capacitance
ε₀ is the permittivity of free space (8.85 x 10^-12 F/m)
A is the area of each disk (given by πr², where r is the radius of the disk)
d is the separation between the disks

First, let's calculate the area of each disk:
radius (r) = diameter / 2 = 2.3 cm / 2 = 1.15 cm = 0.0115 m
Area (A) = πr² = π(0.0115 m)²

Now, let's substitute the values into the capacitance formula:
C = (8.85 x 10^-12 F/m) * [π(0.0115 m)²] / (1.8 x 10^-3 m)

Once you calculate the value of C, you can find the charge on each disk using the formula Q = CV (where V is the voltage calculated in part A).

C. To find the electron's speed as it left the negative plate, we can use the principle of conservation of energy. The total mechanical energy of the electron is conserved, which can be written as:

KE1 + PE1 = KE2 + PE2

Where:
KE1 is the initial kinetic energy of the electron on the negative plate (which is equal to ½mv₁², where m is the mass of the electron and v₁ is its initial speed on the negative plate)
PE1 is the initial potential energy of the electron on the negative plate (which is equal to -qV, where q is the charge on the electron and V is the voltage across the capacitor)
KE2 is the final kinetic energy of the electron on the positive plate (which is equal to ½mv₂², where v₂ is the final speed of the electron on the positive plate)
PE2 is the final potential energy of the electron on the positive plate (which is equal to -qV, where q is the charge on the electron and V is the voltage across the capacitor)

Since the initial and final potential energies are equal (-qV), they cancel out in the equation. Therefore, we have:

KE1 = KE2

Substituting the known values:
½mv₁² = ½mv₂²

Since the mass of the electron (m) is the same on both sides of the equation, it cancels out. Solving for v₁:

v₁² = v₂²

Finally, take the square root of both sides to find the velocity (speed) of the electron as it left the negative plate.

Note: When calculating the charge (Q) on each disk in part B, please take into account the fact that the voltage (V) calculated in part A is the voltage across the capacitor, not the voltage on each disk individually.

To find the amount of charge on each disk (B), you can use the formula for the capacitance of a parallel-plate capacitor:

C = (ε₀ * A) / d

Where:
C is the capacitance of the capacitor
ε₀ is the permittivity of free space (approximately 8.854 x 10⁻¹² F/m)
A is the area of one of the disks (π * r²)
d is the separation distance between the disks

We can calculate C using the given information:

C = (8.854 x 10⁻¹² F/m * π * (0.023 m)²) / 0.0018 m

Now that we have the capacitance value, we can find the charge on each disk:

Q = C * V

Where:
Q is the charge on each disk
V is the voltage across the capacitor (830 V)

Now let's move on to part C. To find the electron's speed as it left the negative plate, we can use the principle of conservation of mechanical energy. Since the electric field does work on the electron, it accelerates and gains kinetic energy. The kinetic energy gained is equal to the work done by the field:

K.E. = q * ΔV

Where:
K.E. is the kinetic energy gained by the electron
q is the charge on the electron
ΔV is the change in voltage

We know that the electron strikes the positive plate with a certain speed, so its kinetic energy can also be expressed as:

K.E. = (1/2) * m * (v_p)²

Where:
m is the mass of the electron (approximately 9.11 x 10⁻³¹ kg)
v_p is the speed of the electron as it strikes the positive plate (2.3 x 10⁷ m/s)

Since the net charge on the plates is zero and electrons are negatively charged, the charge on each disk will be -Q/2. Hence, q will be equal to -Q/2.

Equating the two expressions for kinetic energy:

(1/2) * m * (v_p)² = q * ΔV

Now, solving for the speed of the electron as it left the negative plate:

(v_n) = sqrt((2 * q * ΔV) / m)

Now you have all the necessary equations to solve parts B and C of the question.